From 6d8cef8cd1d2ed8ea9327d1c0cddfe05b8e8eb2f Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 29 Jul 2018 11:24:25 +0800
Subject: [PATCH 001/172] auto commit

---
 notes/Linux.md           |  2 +-
 notes/剑指 offer 题解.md | 12 ++++++------
 2 files changed, 7 insertions(+), 7 deletions(-)

diff --git a/notes/Linux.md b/notes/Linux.md
index 705b9b0d..6325e86e 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -903,7 +903,7 @@ declare -x HOME="/home/dmtsai"
 declare -x HOSTNAME="study.centos.vbird"
 .....(其他省略).....
 
-$ export | cut -c 12
+$ export | cut -c 12-
 ```
 
 ## 排序指令
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index e1655922..05d43578 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -898,7 +898,7 @@ public double Power(double base, int exponent) {
 
 ## 题目描述
 
-输入数字 n，按顺序打印出从 1 最大的 n 位十进制数。比如输入 3，则打印出 1、2、3 一直到最大的 3 位数即 999。
+输入数字 n，按顺序打印出从 1 到最大的 n 位十进制数。比如输入 3，则打印出 1、2、3 一直到最大的 3 位数即 999。
 
 ## 解题思路
 
@@ -911,16 +911,16 @@ public void print1ToMaxOfNDigits(int n) {
     if (n <= 0)
         return;
     char[] number = new char[n];
-    print1ToMaxOfNDigits(number, -1);
+    print1ToMaxOfNDigits(number, 0);
 }
 
 private void print1ToMaxOfNDigits(char[] number, int digit) {
-    if (digit == number.length - 1) {
+    if (digit == number.length) {
         printNumber(number);
         return;
     }
     for (int i = 0; i < 10; i++) {
-        number[digit + 1] = (char) (i + '0');
+        number[digit] = (char) (i + '0');
         print1ToMaxOfNDigits(number, digit + 1);
     }
 }
@@ -1050,13 +1050,13 @@ public boolean match(char[] str, char[] pattern) {
 
 ```html
 []  ： 字符集合
-()  ： 分组，在这里是为了让表达式更清晰
+()  ： 分组
 ?   ： 重复 0 ~ 1
 +   ： 重复 1 ~ n
 *   ： 重复 0 ~ n
 .   ： 任意字符
 \\. ： 转义后的 .
-\\d ： 任意数字
+\\d ： 数字
 ```
 
 ```java

From 65a50dbc7edfae216060432242280f66e39e05a7 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 29 Jul 2018 12:38:35 +0800
Subject: [PATCH 002/172] auto commit

---
 notes/剑指 offer 题解.md                      |   2 +-
 pics/f94389e9-55b1-4f49-9d37-00ed05900ae0.png | Bin 0 -> 12288 bytes
 2 files changed, 1 insertion(+), 1 deletion(-)
 create mode 100644 pics/f94389e9-55b1-4f49-9d37-00ed05900ae0.png

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 05d43578..9cf7dc2a 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -170,7 +170,7 @@ Given target = 20, return false.
 
 当前元素的查找区间为左下角的所有元素，例如元素 12 的查找区间如下：
 
-<div align="center"> <img src="../pics//026d3cb4-67f7-4a83-884d-8032f57ec446.png" width="250"/> </div><br>
+<div align="center"> <img src="../pics//f94389e9-55b1-4f49-9d37-00ed05900ae0.png" width="250"/> </div><br>
 
 复杂度：O(M + N) + O(1)
 
diff --git a/pics/f94389e9-55b1-4f49-9d37-00ed05900ae0.png b/pics/f94389e9-55b1-4f49-9d37-00ed05900ae0.png
new file mode 100644
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From dc45cfc14785d4f4a7efb40690fae05c325986a7 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 29 Jul 2018 14:12:07 +0800
Subject: [PATCH 003/172] auto commit

---
 notes/剑指 offer 题解.md                      |  20 ++++++++++++------
 pics/27ff9548-edb6-4465-92c8-7e6386e0b185.png | Bin 0 -> 8192 bytes
 pics/280f7728-594f-4811-a03a-fa8d32c013da.png | Bin 0 -> 7168 bytes
 pics/2858f8ad-aedb-45a5-a706-e98c96d690fa.jpg | Bin 0 -> 23552 bytes
 pics/6539b9a4-2b24-4d10-8c94-2eb5aba1e296.png | Bin 0 -> 17408 bytes
 pics/ea2304ce-268b-4238-9486-4d8f8aea8ca4.png | Bin 0 -> 23552 bytes
 pics/faecea49-9974-40db-9821-c8636137df61.jpg | Bin 0 -> 18432 bytes
 7 files changed, 14 insertions(+), 6 deletions(-)
 create mode 100644 pics/27ff9548-edb6-4465-92c8-7e6386e0b185.png
 create mode 100644 pics/280f7728-594f-4811-a03a-fa8d32c013da.png
 create mode 100644 pics/2858f8ad-aedb-45a5-a706-e98c96d690fa.jpg
 create mode 100644 pics/6539b9a4-2b24-4d10-8c94-2eb5aba1e296.png
 create mode 100644 pics/ea2304ce-268b-4238-9486-4d8f8aea8ca4.png
 create mode 100644 pics/faecea49-9974-40db-9821-c8636137df61.jpg

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 9cf7dc2a..e36e7a85 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1,4 +1,5 @@
 <!-- GFM-TOC -->
+* [1. 前言](#1-前言)
 * [2. 实现 Singleton](#2-实现-singleton)
 * [3. 数组中重复的数字](#3-数组中重复的数字)
 * [4. 二维数组中的查找](#4-二维数组中的查找)
@@ -80,6 +81,13 @@
 <!-- GFM-TOC -->
 
 
+# 1. 前言
+
+本文的绘图可通过以下途径免费获得并使用：
+
+- [ProcessOn](https://www.processon.com/view/5a3e4c7be4b0909c1aa18b49)
+- [DrawIO](https://drive.google.com/file/d/1nSSCpPUC05MFoeFuf_aeTtkm7dG5-bJ1/view?usp=sharing)
+
 # 2. 实现 Singleton
 
 [单例模式](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/%E8%AE%BE%E8%AE%A1%E6%A8%A1%E5%BC%8F.md)
@@ -456,7 +464,7 @@ public int pop() throws Exception {
 
 如果使用递归求解，会重复计算一些子问题。例如，计算 f(10) 需要计算 f(9) 和 f(8)，计算 f(9) 需要计算 f(8) 和 f(7)，可以看到 f(8) 被重复计算了。
 
-<div align="center"> <img src="../pics//a0df8edc-581b-4977-95c2-d7025795b899.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//faecea49-9974-40db-9821-c8636137df61.jpg" width="300"/> </div><br>
 
 递归是将一个问题划分成多个子问题求解，动态规划也是如此，但是动态规划会把子问题的解缓存起来，从而避免重复求解子问题。
 
@@ -941,11 +949,11 @@ private void printNumber(char[] number) {
 
 ① 如果该节点不是尾节点，那么可以直接将下一个节点的值赋给该节点，然后令该节点指向下下个节点，再删除下一个节点，时间复杂度为 O(1)。
 
-<div align="center"> <img src="../pics//41392d76-dd1d-4712-85d9-e8bb46b04a2d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//27ff9548-edb6-4465-92c8-7e6386e0b185.png" width="600"/> </div><br>
 
 ② 否则，就需要先遍历链表，找到节点的前一个节点，然后让前一个节点指向 null，时间复杂度为 O(N)。
 
-<div align="center"> <img src="../pics//db4921d4-184b-48ba-a3cf-1d1141e3ba2d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//280f7728-594f-4811-a03a-fa8d32c013da.png" width="600"/> </div><br>
 
 综上，如果进行 N 次操作，那么大约需要操作节点的次数为 N-1+N=2N-1，其中 N-1 表示 N-1 个不是尾节点的每个节点以 O(1) 的时间复杂度操作节点的总次数，N 表示 1 个尾节点以 O(N) 的时间复杂度操作节点的总次数。(2N-1)/N \~ 2，因此该算法的平均时间复杂度为 O(1)。
 
@@ -1103,7 +1111,7 @@ public void reOrderArray(int[] nums) {
 
 设链表的长度为 N。设两个指针 P1 和 P2，先让 P1 移动 K 个节点，则还有 N - K 个节点可以移动。此时让 P1 和 P2 同时移动，可以知道当 P1 移动到链表结尾时，P2 移动到 N - K 个节点处，该位置就是倒数第 K 个节点。
 
-<div align="center"> <img src="../pics//207c1801-2335-4b1b-b65c-126a0ba966cb.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//ea2304ce-268b-4238-9486-4d8f8aea8ca4.png" width="500"/> </div><br>
 
 ```java
 public ListNode FindKthToTail(ListNode head, int k)
@@ -1140,7 +1148,7 @@ public ListNode FindKthToTail(ListNode head, int k)
 
 在相遇点，slow 要到环的入口点还需要移动 z 个节点，如果让 fast 重新从头开始移动，并且速度变为每次移动一个节点，那么它到环入口点还需要移动 x 个节点。在上面已经推导出 x=z，因此 fast 和 slow 将在环入口点相遇。
 
-<div align="center"> <img src="../pics//71363383-2d06-4c63-8b72-c01c2186707d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//2858f8ad-aedb-45a5-a706-e98c96d690fa.jpg" width="600"/> </div><br>
 
 ```java
 public ListNode EntryNodeOfLoop(ListNode pHead)
@@ -1346,7 +1354,7 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2)
 
 下图的矩阵顺时针打印结果为：1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
 
-<div align="center"> <img src="../pics//0f373947-c68f-45b4-a59e-086154745ac5.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//6539b9a4-2b24-4d10-8c94-2eb5aba1e296.png" width="300"/> </div><br>
 
 ## 解题思路
 
diff --git a/pics/27ff9548-edb6-4465-92c8-7e6386e0b185.png b/pics/27ff9548-edb6-4465-92c8-7e6386e0b185.png
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HcmV?d00001


From e525c61651f1792b9cfb8898e156df8930b5dfea Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 29 Jul 2018 15:24:29 +0800
Subject: [PATCH 004/172] auto commit

---
 notes/剑指 offer 题解.md | 19 ++++++++++++++++---
 1 file changed, 16 insertions(+), 3 deletions(-)

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index e36e7a85..47d5d94e 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1640,6 +1640,19 @@ private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
 
 输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的 head。
 
+```java
+public class RandomListNode
+{
+    int label;
+    RandomListNode next = null;
+    RandomListNode random = null;
+
+    RandomListNode(int label) {
+        this.label = label;
+    }
+}
+```
+
 <div align="center"> <img src="../pics//a01d1516-8168-461a-a24b-620b9cfc40f4.png" width="300"/> </div><br>
 
 ## 解题思路
@@ -1820,7 +1833,7 @@ private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
 
 多数投票问题，可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。
 
-使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0 ，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
+使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
 
 ```java
 public int MoreThanHalfNum_Solution(int[] nums)
@@ -1924,7 +1937,7 @@ public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
         if (maxHeap.size() > k)
             maxHeap.poll();
     }
-    return new ArrayList<>(maxHeap) ;
+    return new ArrayList<>(maxHeap);
 }
 ```
 
@@ -2005,7 +2018,7 @@ public char FirstAppearingOnce()
 
 ## 题目描述
 
-{6,-3,-2,7,-15,1,2,2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
+{6, -3, -2, 7, -15, 1, 2, 2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
 
 ## 解题思路
 

From 8a42a419be5a062cb59bc19e536e4b3de826e7cb Mon Sep 17 00:00:00 2001
From: fighterhit <fighterhit@163.com>
Date: Mon, 30 Jul 2018 15:26:33 +0800
Subject: [PATCH 005/172] Update Linux.md

---
 notes/Linux.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Linux.md b/notes/Linux.md
index 6325e86e..8f27d75c 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -1078,7 +1078,7 @@ dmtsai pts/0 192.168.1.100 Thu Jul 9 17:23 - 23:36 (06:12)
 dmtsai pts/0 192.168.1.100 Thu Jul 9 08:02 - 08:17 (00:14)
 dmtsai tty1 Fri May 29 11:55 - 12:11 (00:15)
 
-$ last -n 5 | awk '{print $1 "\t" $3}
+$ last -n 5 | awk '{print $1 "\t" $3}'
 ```
 
 可以根据字段的某些条件进行匹配，例如匹配字段小于某个值的那一行数据。

From f8600a4611fe45f847e6a1e869aad7a1d43bcda4 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 30 Jul 2018 23:22:52 +0800
Subject: [PATCH 006/172] auto commit

---
 README.md                        |   8 ++-
 notes/剑指 offer 题解.md         |  41 +++++++++---
 notes/{安全性.md => 攻击技术.md} |   0
 notes/系统设计基础.md            | 104 +++++++++++++++++++++++++++++++
 4 files changed, 141 insertions(+), 12 deletions(-)
 rename notes/{安全性.md => 攻击技术.md} (100%)
 create mode 100644 notes/系统设计基础.md

diff --git a/README.md b/README.md
index 63067ac8..4b1411b5 100644
--- a/README.md
+++ b/README.md
@@ -105,6 +105,10 @@ Leetcode 上数据库题目的解题记录。
 
 ## 系统设计 :bulb:
 
+> [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
+
+性能、伸缩性、扩展性、可用性、安全性
+
 > [分布式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/分布式.md)
 
 分布式锁、分布式事务、CAP、BASE、Paxos、Raft
@@ -113,9 +117,9 @@ Leetcode 上数据库题目的解题记录。
 
 负载均衡、Session 管理
 
-> [安全性](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/安全性.md)
+> [攻击技术](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/攻击技术.md)
 
-XSS、CSRF、SQL 注入、DDos
+XSS、CSRF、SQL 注入、DDoS
 
 > [消息队列](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/消息队列.md)
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 05d43578..47d5d94e 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1,4 +1,5 @@
 <!-- GFM-TOC -->
+* [1. 前言](#1-前言)
 * [2. 实现 Singleton](#2-实现-singleton)
 * [3. 数组中重复的数字](#3-数组中重复的数字)
 * [4. 二维数组中的查找](#4-二维数组中的查找)
@@ -80,6 +81,13 @@
 <!-- GFM-TOC -->
 
 
+# 1. 前言
+
+本文的绘图可通过以下途径免费获得并使用：
+
+- [ProcessOn](https://www.processon.com/view/5a3e4c7be4b0909c1aa18b49)
+- [DrawIO](https://drive.google.com/file/d/1nSSCpPUC05MFoeFuf_aeTtkm7dG5-bJ1/view?usp=sharing)
+
 # 2. 实现 Singleton
 
 [单例模式](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/%E8%AE%BE%E8%AE%A1%E6%A8%A1%E5%BC%8F.md)
@@ -170,7 +178,7 @@ Given target = 20, return false.
 
 当前元素的查找区间为左下角的所有元素，例如元素 12 的查找区间如下：
 
-<div align="center"> <img src="../pics//026d3cb4-67f7-4a83-884d-8032f57ec446.png" width="250"/> </div><br>
+<div align="center"> <img src="../pics//f94389e9-55b1-4f49-9d37-00ed05900ae0.png" width="250"/> </div><br>
 
 复杂度：O(M + N) + O(1)
 
@@ -456,7 +464,7 @@ public int pop() throws Exception {
 
 如果使用递归求解，会重复计算一些子问题。例如，计算 f(10) 需要计算 f(9) 和 f(8)，计算 f(9) 需要计算 f(8) 和 f(7)，可以看到 f(8) 被重复计算了。
 
-<div align="center"> <img src="../pics//a0df8edc-581b-4977-95c2-d7025795b899.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//faecea49-9974-40db-9821-c8636137df61.jpg" width="300"/> </div><br>
 
 递归是将一个问题划分成多个子问题求解，动态规划也是如此，但是动态规划会把子问题的解缓存起来，从而避免重复求解子问题。
 
@@ -941,11 +949,11 @@ private void printNumber(char[] number) {
 
 ① 如果该节点不是尾节点，那么可以直接将下一个节点的值赋给该节点，然后令该节点指向下下个节点，再删除下一个节点，时间复杂度为 O(1)。
 
-<div align="center"> <img src="../pics//41392d76-dd1d-4712-85d9-e8bb46b04a2d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//27ff9548-edb6-4465-92c8-7e6386e0b185.png" width="600"/> </div><br>
 
 ② 否则，就需要先遍历链表，找到节点的前一个节点，然后让前一个节点指向 null，时间复杂度为 O(N)。
 
-<div align="center"> <img src="../pics//db4921d4-184b-48ba-a3cf-1d1141e3ba2d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//280f7728-594f-4811-a03a-fa8d32c013da.png" width="600"/> </div><br>
 
 综上，如果进行 N 次操作，那么大约需要操作节点的次数为 N-1+N=2N-1，其中 N-1 表示 N-1 个不是尾节点的每个节点以 O(1) 的时间复杂度操作节点的总次数，N 表示 1 个尾节点以 O(N) 的时间复杂度操作节点的总次数。(2N-1)/N \~ 2，因此该算法的平均时间复杂度为 O(1)。
 
@@ -1103,7 +1111,7 @@ public void reOrderArray(int[] nums) {
 
 设链表的长度为 N。设两个指针 P1 和 P2，先让 P1 移动 K 个节点，则还有 N - K 个节点可以移动。此时让 P1 和 P2 同时移动，可以知道当 P1 移动到链表结尾时，P2 移动到 N - K 个节点处，该位置就是倒数第 K 个节点。
 
-<div align="center"> <img src="../pics//207c1801-2335-4b1b-b65c-126a0ba966cb.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//ea2304ce-268b-4238-9486-4d8f8aea8ca4.png" width="500"/> </div><br>
 
 ```java
 public ListNode FindKthToTail(ListNode head, int k)
@@ -1140,7 +1148,7 @@ public ListNode FindKthToTail(ListNode head, int k)
 
 在相遇点，slow 要到环的入口点还需要移动 z 个节点，如果让 fast 重新从头开始移动，并且速度变为每次移动一个节点，那么它到环入口点还需要移动 x 个节点。在上面已经推导出 x=z，因此 fast 和 slow 将在环入口点相遇。
 
-<div align="center"> <img src="../pics//71363383-2d06-4c63-8b72-c01c2186707d.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//2858f8ad-aedb-45a5-a706-e98c96d690fa.jpg" width="600"/> </div><br>
 
 ```java
 public ListNode EntryNodeOfLoop(ListNode pHead)
@@ -1346,7 +1354,7 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2)
 
 下图的矩阵顺时针打印结果为：1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
 
-<div align="center"> <img src="../pics//0f373947-c68f-45b4-a59e-086154745ac5.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//6539b9a4-2b24-4d10-8c94-2eb5aba1e296.png" width="300"/> </div><br>
 
 ## 解题思路
 
@@ -1632,6 +1640,19 @@ private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
 
 输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的 head。
 
+```java
+public class RandomListNode
+{
+    int label;
+    RandomListNode next = null;
+    RandomListNode random = null;
+
+    RandomListNode(int label) {
+        this.label = label;
+    }
+}
+```
+
 <div align="center"> <img src="../pics//a01d1516-8168-461a-a24b-620b9cfc40f4.png" width="300"/> </div><br>
 
 ## 解题思路
@@ -1812,7 +1833,7 @@ private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
 
 多数投票问题，可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。
 
-使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0 ，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
+使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
 
 ```java
 public int MoreThanHalfNum_Solution(int[] nums)
@@ -1916,7 +1937,7 @@ public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
         if (maxHeap.size() > k)
             maxHeap.poll();
     }
-    return new ArrayList<>(maxHeap) ;
+    return new ArrayList<>(maxHeap);
 }
 ```
 
@@ -1997,7 +2018,7 @@ public char FirstAppearingOnce()
 
 ## 题目描述
 
-{6,-3,-2,7,-15,1,2,2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
+{6, -3, -2, 7, -15, 1, 2, 2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
 
 ## 解题思路
 
diff --git a/notes/安全性.md b/notes/攻击技术.md
similarity index 100%
rename from notes/安全性.md
rename to notes/攻击技术.md
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
new file mode 100644
index 00000000..e132aa2c
--- /dev/null
+++ b/notes/系统设计基础.md
@@ -0,0 +1,104 @@
+<!-- GFM-TOC -->
+* [一、性能](#一性能)
+* [二、伸缩性](#二伸缩性)
+* [三、扩展性](#三扩展性)
+* [四、可用性](#四可用性)
+* [五、安全性](#五安全性)
+<!-- GFM-TOC -->
+
+
+# 一、性能
+
+## 性能指标
+
+### 1. 响应时间
+
+指从某个请求发出到接收到响应消耗的时间。
+
+在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
+
+### 2. 吞吐量
+
+指系统在单位时间内可以处理的请求数量，通常使用每秒的请求数来衡量。
+
+### 3. 并发用户数
+
+指系统能同时处理的并发用户请求数量。
+
+在没有并发存在的系统中，请求被顺序执行，此时响应时间为吞吐量的倒数。例如系统支持的吞吐量为 100 req/s，那么平均响应时间应该为 0.01s。
+
+目前的大型系统都支持多线程来处理并发请求，多线程能够提高吞吐量以及缩短响应时间，主要有两个原因：
+
+- 多 CPU
+- IO 等待时间
+
+使用 IO 多路复用等方式，系统在等待一个 IO 操作完成的这段时间内不需要被阻塞，可以去处理其它请求。通过将这个等待时间利用起来，使得 CPU 利用率大大提高。
+
+并发用户数不是越高越好，因为如果并发用户数太高，系统来不及处理这么多的请求，会使得过多的请求需要等待，那么响应时间就会大大提高。
+
+## 性能优化
+
+### 1. 集群
+
+将多台服务器组成集群，使用负载均衡将请求转发到集群中，避免单一服务器的负载压力过大导致性能降低。
+
+### 2. 缓存
+
+缓存能够提高性能的原因如下：
+
+- 缓存数据通常位于内存等介质中，这种介质对于读操作特别快；
+- 缓存数据可以位于靠近用户的地理位置上；
+- 可以将计算结果进行缓存，从而避免重复计算。
+
+### 3. 异步
+
+某些流程可以将操作转换为消息，将消息发送到消息队列之后立即返回，之后这个操作会被异步处理。
+
+# 二、伸缩性
+
+指不断向集群中添加服务器来缓解不断上升的用户并发访问压力和不断增长的数据存储需求。
+
+## 伸缩性与性能
+
+如果系统存在性能问题，那么单个用户的请求总是很慢的；
+
+如果系统存在伸缩性问题，那么单个用户的请求可能会很快，但是在并发数很高的情况下系统会很慢。
+
+## 实现伸缩性
+
+应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
+
+关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
+
+对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
+
+# 三、扩展性
+
+指的是添加新功能时对现有系统的其它应用无影响，这就要求不同应用具备低耦合的特点。
+
+实现可扩展主要有两种方式：
+
+- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
+- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务器框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
+
+# 四、可用性
+
+## 冗余
+
+保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
+
+应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
+
+存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
+
+## 监控
+
+对 CPU、内存、磁盘、网络等系统负载信息进行监控，当某个数据达到一定阈值时通知运维人员，从而在系统发生故障之前及时发现问题。
+
+## 服务降级
+
+服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
+
+# 五、安全性
+
+要求系统的应对各种攻击手段时能够有可靠的应对措施。

From 75e5fd8be63458f7fd7f519d46bb2a811ad087b3 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 31 Jul 2018 00:30:14 +0800
Subject: [PATCH 007/172] auto commit

---
 README.md                                     |   4 +
 notes/缓存.md                                 | 265 ++++++++++++++++++
 pics/15313ed8-a520-4799-a300-2b6b36be314f.jpg | Bin 0 -> 33792 bytes
 pics/66402828-fb2b-418f-83f6-82153491bcfe.jpg | Bin 0 -> 38912 bytes
 pics/68b110b9-76c6-4ee2-b541-4145e65adb3e.jpg | Bin 0 -> 34816 bytes
 5 files changed, 269 insertions(+)
 create mode 100644 notes/缓存.md
 create mode 100644 pics/15313ed8-a520-4799-a300-2b6b36be314f.jpg
 create mode 100644 pics/66402828-fb2b-418f-83f6-82153491bcfe.jpg
 create mode 100644 pics/68b110b9-76c6-4ee2-b541-4145e65adb3e.jpg

diff --git a/README.md b/README.md
index 4b1411b5..5e49a332 100644
--- a/README.md
+++ b/README.md
@@ -121,6 +121,10 @@ Leetcode 上数据库题目的解题记录。
 
 XSS、CSRF、SQL 注入、DDoS
 
+> [缓存](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/缓存.md)
+
+缓存特征、缓存位置、缓存问题、数据分布、一致性哈希、LRU、CDN
+
 > [消息队列](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/消息队列.md)
 
 消息处理模型、使用场景、可靠性
diff --git a/notes/缓存.md b/notes/缓存.md
new file mode 100644
index 00000000..e60373dd
--- /dev/null
+++ b/notes/缓存.md
@@ -0,0 +1,265 @@
+<!-- GFM-TOC -->
+* [一、缓存特征](#一缓存特征)
+* [二、LRU](#二lru)
+* [三、缓存位置](#三缓存位置)
+* [四、CDN](#四cdn)
+* [五、缓存问题](#五缓存问题)
+* [六、数据分布](#六数据分布)
+* [七、一致性哈希](#七一致性哈希)
+<!-- GFM-TOC -->
+
+
+# 一、缓存特征
+
+## 命中率
+
+当某个请求能够通过访问缓存而得到响应时，称为缓存命中。
+
+缓存命中率越高，缓存的利用率也就越高。
+
+## 最大空间
+
+缓存通常位于内存中，内存的空间通常比磁盘空间小的多，因此缓存的最大空间不可能非常大。
+
+当缓存存放的数据量超过最大空间时，就需要淘汰部分数据来存放新到达的数据。
+
+## 淘汰策略
+
+- FIFO（First In First Out）：先进先出策略，在实时性的场景下，需要经常访问最新的数据，那么就可以使用 FIFO，使最先进入的数据（最晚的数据）被淘汰。
+
+- LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最远的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
+
+参考资料：
+
+- [缓存那些事](https://tech.meituan.com/cache_about.html)
+
+# 二、LRU
+
+以下是一个基于 双向队列 + HashMap 的 LRU 算法实现，对算法的解释如下：
+
+- 最基本的思路是当访问某个节点时，将其从原来的位置删除，并重新插入到链表头部，这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就删除链表尾部的节点。
+- 为了使删除操作时间复杂度为 O(1)，那么就不能采用遍历的方式找到某个节点。HashMap 存储这 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
+
+```java
+public class LRU<K, V> implements Iterable<K> {
+
+    private Node head;
+    private Node tail;
+    private HashMap<K, Node> map;
+    private int maxSize;
+
+    private class Node {
+
+        Node pre;
+        Node next;
+        K k;
+        V v;
+
+        public Node(K k, V v) {
+            this.k = k;
+            this.v = v;
+        }
+    }
+
+    public LRU(int maxSize) {
+
+        this.maxSize = maxSize;
+        this.map = new HashMap<>(maxSize * 4 / 3);
+
+        head = new Node(null, null);
+        tail = new Node(null, null);
+
+        head.next = tail;
+        tail.pre = head;
+    }
+
+    public V get(K key) {
+
+        if (!map.containsKey(key)) {
+            return null;
+        }
+
+        Node node = map.get(key);
+        unlink(node);
+        appendHead(node);
+
+        return node.v;
+    }
+
+    public void put(K key, V value) {
+
+        if (map.containsKey(key)) {
+            Node node = map.get(key);
+            unlink(node);
+        }
+
+        Node node = new Node(key, value);
+        map.put(key, node);
+        appendHead(node);
+
+        if (map.size() > maxSize) {
+            Node toRemove = removeTail();
+            map.remove(toRemove);
+        }
+    }
+
+    private void unlink(Node node) {
+        Node pre = node.pre;
+        node.pre = node.next;
+        node.next = pre;
+    }
+
+    private void appendHead(Node node) {
+        node.next = head.next;
+        head.next = node;
+    }
+
+    private Node removeTail() {
+        Node node = tail.pre;
+        node.pre = tail;
+        return node;
+    }
+
+    @Override
+    public Iterator<K> iterator() {
+
+        return new Iterator<K>() {
+
+            private Node cur = head.next;
+
+            @Override
+            public boolean hasNext() {
+                return cur != tail;
+            }
+
+            @Override
+            public K next() {
+                Node node = cur;
+                cur = cur.next;
+                return node.k;
+            }
+        };
+    }
+}
+```
+
+# 三、缓存位置
+
+## 浏览器
+
+当 HTTP 响应允许进行缓存时，浏览器会将 HTML、CSS、JavaScript、图片等静态资源进行缓存。
+
+## ISP
+
+网络服务提供商（ISP）是网络访问的第一跳，通过将数据缓存在 ISP 中能够大大提高用户的访问速度。
+
+## 反向代理
+
+反向代理位于服务器之前，请求与响应都需要经过反向代理。通过将数据缓存在反向代理，在用户请求时就可以直接使用缓存进行响应。
+
+## 本地缓存
+
+使用 Guava Cache 将数据缓存在服务器本地内存中，服务器代码可以直接读取本地内存中的缓存，速度非常快。
+
+## 分布式缓存
+
+使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存，本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+
+## 数据库缓存
+
+MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查询效率。
+
+# 四、CDN
+
+内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源其它数据分发给用户。
+
+CDN 主要有以下优点：
+
+- 更快地将数据分发给用户；
+- 通过部署多台服务器，从而提高系统整体的带宽性能；
+- 多台服务器可以看成是一种冗余机制，从而具有高可用性。
+
+<div align="center"> <img src="../pics//15313ed8-a520-4799-a300-2b6b36be314f.jpg"/> </div><br>
+
+参考资料：
+
+- [内容分发网络](https://zh.wikipedia.org/wiki/%E5%85%A7%E5%AE%B9%E5%82%B3%E9%81%9E%E7%B6%B2%E8%B7%AF)
+- [How Aspiration CDN helps to improve your website loading speed?](https://www.aspirationhosting.com/aspiration-cdn/)
+
+# 五、缓存问题
+
+## 缓存穿透
+
+指的是对某个一定不存在的数据进行请求，该请求将会穿透缓存来到数据库。
+
+解决方案：
+
+- 对这些不存在的数据缓存一个空数据；
+- 对这类请求进行过滤。
+
+## 缓存雪崩
+
+指的是由于数据没有被加载到缓存中，或者缓存数据在同一时间大面积失效（过期），又或者缓存服务器宕机，导致大量的请求都去到达数据库。
+
+在存在缓存的系统中，系统非常依赖于缓存，缓存分担了很大一部分的数据请求。当发生缓存雪崩时，数据库无法处理这么大的请求，导致数据库崩溃。
+
+解决方案：
+
+- 为了防止缓存在同一时间大面积过期导致的缓存雪崩，可以通过观察用户行为，合理设置缓存过期时间来实现；
+- 为了防止缓存服务器宕机出现的缓存雪崩，可以使用分布式缓存，分布式缓存中每一个节点只缓存部分的数据，当某个节点宕机时可以保证其它节点的缓存仍然可用。
+
+## 缓存一致性
+
+缓存一致性要求数据更新的同时缓存数据也能够实时更新。
+
+解决方案：
+
+- 在数据更新的同时立即去更新缓存；
+- 在读缓存之前先判断缓存是否是最新的，如果不是最新的先进行更新。
+
+要保证缓存一致性需要付出很大的代价，缓存数据最好是那些对一致性要求不高的数据，允许缓存数据存在一些脏数据。
+
+# 六、数据分布
+
+## 哈希分布
+
+哈希分布就是将数据计算哈希值之后，按照哈希值分配到不同的节点上。例如有 N 个节点，数据的主键为 key，则将该数据分配的节点序号为：hash(key)%N。
+
+传统的哈希分布算法存在一个问题：当节点数量变化时，也就是 N 值变化，那么几乎所有的数据都需要重新分布，将导致大量的数据迁移。
+
+## 顺序分布
+
+将数据划分为多个连续的部分，按数据的 ID 或者时间分布到不同节点上。例如 User 表的 ID 范围为 1 \~ 7000，使用顺序分布可以将其划分成多个子表，对应的主键范围为 1 \~ 1000，1001 \~ 2000，...，6001 \~ 7000。
+
+顺序分布相比于哈希分布的主要优点如下：
+
+- 能保持数据原有的顺序；
+- 并且能够准确控制每台服务器存储的数据量，从而使得存储空间的利用率最大。
+
+参考资料：
+
+- 大规模分布式存储系统
+
+# 七、一致性哈希
+
+Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据失效的问题。
+
+## 基本原理
+
+将哈希空间 [0, 2<sup>n</sup>-1] 看成一个哈希环，每个服务器节点都配置到哈希环上。每个数据对象通过哈希取模得到哈希值之后，存放到哈希环中顺时针方向第一个大于等于该哈希值的节点上。
+
+<div align="center"> <img src="../pics//68b110b9-76c6-4ee2-b541-4145e65adb3e.jpg"/> </div><br>
+
+一致性哈希在增加或者删除节点时只会影响到哈希环中相邻的节点，例如下图中新增节点 X，只需要将它前一个节点 C 上的数据重新进行分布即可，对于节点 A、B、D 都没有影响。
+
+<div align="center"> <img src="../pics//66402828-fb2b-418f-83f6-82153491bcfe.jpg"/> </div><br>
+
+## 虚拟节点
+
+上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
+
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点号，从而使得数据分布也更加均匀。
+
+参考资料：
+
+- [一致性哈希算法](https://my.oschina.net/jayhu/blog/732849)
diff --git a/pics/15313ed8-a520-4799-a300-2b6b36be314f.jpg b/pics/15313ed8-a520-4799-a300-2b6b36be314f.jpg
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literal 0
HcmV?d00001


From d78b96587829957eb46f5046ed6a640b8d58e9ff Mon Sep 17 00:00:00 2001
From: HenryChan98 <chanhenry154@yahoo.com>
Date: Tue, 31 Jul 2018 14:35:08 +0800
Subject: [PATCH 008/172] Update SQL exercises

---
 other/sql 经典练习题.sql | 51 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)

diff --git a/other/sql 经典练习题.sql b/other/sql 经典练习题.sql
index 72965fa2..a4b4f964 100644
--- a/other/sql 经典练习题.sql	
+++ b/other/sql 经典练习题.sql	
@@ -430,3 +430,54 @@ where SCORE.SNO = STUDENT.SNO and SSEX = '男' and CNO = (
 
 
 
+-- 46、使用游标方式来同时查询每位同学的名字，他所选课程及成绩。
+
+declare
+ cursor student_cursor is
+  select S.SNO,S.SNAME,C.CNAME,SC.DEGREE as DEGREE
+  from STUDENT S, COURSE C, SCORE SC
+  where S.SNO=SC.SNO
+  and SC.CNO=C.CNO;
+
+  student_row student_cursor%ROWTYPE;
+
+begin
+  open student_cursor;
+   loop
+    fetch student_cursor INTO student_row;
+    exit when student_cursor%NOTFOUND;
+     dbms_output.put_line( student_row.SNO || '' || 
+
+student_row.SNAME|| '' || student_row.CNAME || '' ||
+
+student_row.DEGREE);
+   end loop;
+  close student_cursor;
+END;
+/ 
+
+
+-- 47、 声明触发器指令，每当有同学转换班级时执行触发器显示当前和之前所在班级。
+
+CREATE OR REPLACE TRIGGER display_class_changes 
+AFTER DELETE OR INSERT OR UPDATE ON student 
+FOR EACH ROW 
+WHEN (NEW.sno > 0) 
+
+BEGIN 
+   
+   dbms_output.put_line('Old class: ' || :OLD.class); 
+   dbms_output.put_line('New class: ' || :NEW.class); 
+END; 
+/ 
+
+
+Update student
+set class=95031
+where sno=109;
+
+
+-- 48、 删除已设置的触发器指令
+
+DROP TRIGGER display_class_changes;
+

From b1bbd3d41895c488ead8768b6e882e2c102e7e12 Mon Sep 17 00:00:00 2001
From: fighter <fighterhit@163.com>
Date: Tue, 31 Jul 2018 17:30:04 +0800
Subject: [PATCH 009/172] change XSRF to CSRF

---
 notes/攻击技术.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/攻击技术.md b/notes/攻击技术.md
index 22af60ea..c99071a0 100644
--- a/notes/攻击技术.md
+++ b/notes/攻击技术.md
@@ -141,7 +141,7 @@ http://www.examplebank.com/withdraw?account=AccoutName&amount=1000&for=PayeeName
 
 ### 1. 检查 Referer 首部字段
 
-Referer 首部字段位于 HTTP 报文中，用于标识请求来源的地址。检查这个首部字段并要求请求来源的地址在同一个域名下，可以极大的防止 XSRF 攻击。
+Referer 首部字段位于 HTTP 报文中，用于标识请求来源的地址。检查这个首部字段并要求请求来源的地址在同一个域名下，可以极大的防止 CSRF 攻击。
 
 这种办法简单易行，工作量低，仅需要在关键访问处增加一步校验。但这种办法也有其局限性，因其完全依赖浏览器发送正确的 Referer 字段。虽然 HTTP 协议对此字段的内容有明确的规定，但并无法保证来访的浏览器的具体实现，亦无法保证浏览器没有安全漏洞影响到此字段。并且也存在攻击者攻击某些浏览器，篡改其 Referer 字段的可能。
 

From 37b833c347b8d0e03802ad5c3e44845de43a54a1 Mon Sep 17 00:00:00 2001
From: cyc2018 <zhengyc101@163.com>
Date: Tue, 31 Jul 2018 10:34:59 +0000
Subject: [PATCH 010/172] Updates SUMMARY.md

Auto commit by GitBook Editor
---
 SUMMARY.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/SUMMARY.md b/SUMMARY.md
index e60f018e..3ae54443 100644
--- a/SUMMARY.md
+++ b/SUMMARY.md
@@ -30,7 +30,7 @@
   * [一致性](/notes/一致性.md)
   * [分布式问题分析](/notes/分布式问题分析.md)
 
-
+This file is used to generate gitbook catalogue.
 
 
 

From 6cf167f64bac2e18ed31365bf2304113791941e3 Mon Sep 17 00:00:00 2001
From: cyc2018 <zhengyc101@163.com>
Date: Tue, 31 Jul 2018 10:36:23 +0000
Subject: [PATCH 011/172] Updates SUMMARY.md

Auto commit by GitBook Editor
---
 SUMMARY.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/SUMMARY.md b/SUMMARY.md
index 3ae54443..dde02f47 100644
--- a/SUMMARY.md
+++ b/SUMMARY.md
@@ -1,3 +1,5 @@
+This file is used to generate gitbook catalogue.
+
 # Summary
 
 * 算法
@@ -30,7 +32,5 @@
   * [一致性](/notes/一致性.md)
   * [分布式问题分析](/notes/分布式问题分析.md)
 
-This file is used to generate gitbook catalogue.
-
 
 

From 529152680cfa299f2d2c3e651e4e6b96f4ec77e1 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Tue, 31 Jul 2018 18:37:56 +0800
Subject: [PATCH 012/172] Update SUMMARY.md

---
 SUMMARY.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/SUMMARY.md b/SUMMARY.md
index dde02f47..69b6103c 100644
--- a/SUMMARY.md
+++ b/SUMMARY.md
@@ -1,4 +1,4 @@
-This file is used to generate gitbook catalogue.
+This file used to generate gitbook catalogue.
 
 # Summary
 

From 7515d4a8df2e8f4c5860071098250f522581e680 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 31 Jul 2018 23:30:59 +0800
Subject: [PATCH 013/172] auto commit

---
 notes/Docker.md                               | 102 ++++++++++++++++++
 ...011f3ef6-d824-4d43-8b2c-36dab8eaaa72-1.png | Bin 0 -> 32768 bytes
 pics/71f61bc3-582d-4c27-8bdd-dc7fb135bf8f.png | Bin 0 -> 24576 bytes
 pics/7e873b60-44dc-4911-b080-defd5b8f0b49.png | Bin 0 -> 17408 bytes
 pics/docker-filesystems-busyboxrw.png         | Bin 0 -> 32768 bytes
 5 files changed, 102 insertions(+)
 create mode 100644 notes/Docker.md
 create mode 100644 pics/011f3ef6-d824-4d43-8b2c-36dab8eaaa72-1.png
 create mode 100644 pics/71f61bc3-582d-4c27-8bdd-dc7fb135bf8f.png
 create mode 100644 pics/7e873b60-44dc-4911-b080-defd5b8f0b49.png
 create mode 100644 pics/docker-filesystems-busyboxrw.png

diff --git a/notes/Docker.md b/notes/Docker.md
new file mode 100644
index 00000000..f230f762
--- /dev/null
+++ b/notes/Docker.md
@@ -0,0 +1,102 @@
+<!-- GFM-TOC -->
+* [一、解决的问题](#一解决的问题)
+* [二、与虚拟机的比较](#二与虚拟机的比较)
+* [三、优势](#三优势)
+* [四、使用场景](#四使用场景)
+* [五、镜像与容器](#五镜像与容器)
+<!-- GFM-TOC -->
+
+
+<div align="center"> <img src="../pics//011f3ef6-d824-4d43-8b2c-36dab8eaaa72-1.png"/> </div><br>
+
+# 一、解决的问题
+
+由于不同的机器有不同的操作系统，以及不同的库和组件，在将一个应用部署到多台机器上需要进行大量的环境配置操作。
+
+Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程进行隔离，被隔离的进程独立于宿主操作系统和其它隔离的进程。使用 Docker 可以不修改应用程序代码，不需要开发人员学习特定环境下的技术，就能够将现有的应用程序部署在其他机器中。
+
+参考资料：
+
+- [DOCKER 101: INTRODUCTION TO DOCKER WEBINAR RECAP](https://blog.docker.com/2017/08/docker-101-introduction-docker-webinar-recap/)
+- [Docker 入门教程](http://www.ruanyifeng.com/blog/2018/02/docker-tutorial.html)
+
+# 二、与虚拟机的比较
+
+虚拟机也是一种虚拟化技术，它与 Docker 最大的区别在于它是通过模拟硬件，并在硬件上安装操作系统来实现。
+
+<div align="center"> <img src="../pics//71f61bc3-582d-4c27-8bdd-dc7fb135bf8f.png"/> </div><br>
+
+<div align="center"> <img src="../pics//7e873b60-44dc-4911-b080-defd5b8f0b49.png"/> </div><br>
+
+## 启动速度
+
+启动虚拟机需要启动虚拟机的操作系统，再启动相应的应用，这个过程会非常慢；
+
+而启动 Docker 相当于启动宿主操作系统上的一个进程。
+
+## 占用资源
+
+虚拟机是一个完整的操作系统，需要占用大量的磁盘空间、内存和 CPU，一台机器只能开启几十个的虚拟机。
+
+而 Docker 只是一个进程，只需要将应用以及相应的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
+
+参考资料：
+
+- [Docker container vs Virtual machine](http://www.bogotobogo.com/DevOps/Docker/Docker_Container_vs_Virtual_Machine.php)
+
+# 三、优势
+
+除了启动速度快以及占用资源少之外，Docker 具有以下优势：
+
+## 更容易迁移
+
+Docker 可以提供一致性的运行环境，可以在不同的机器上进行迁移，而不用担心环境变化导致无法运行。
+
+## 更容易维护
+
+Docker 使用分层技术和镜像，使得应用可以更容易复用重复部分。复用程度越高，维护工作也越容易。
+
+## 更容易扩展
+
+可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像得到我们想要的镜像非常容易。
+
+参考资料：
+
+- [为什么要使用 Docker？](https://yeasy.gitbooks.io/docker_practice/introduction/why.html)
+
+# 四、使用场景
+
+## 持续集成
+
+持续集成指的是频繁地将代码集成到主干上，这样能够更快地发现错误。
+
+Docker 具有轻量级以及隔离性的特点，在将代码集成到一个 Docker 中不会对其它 Docker 产生影响。
+
+## 提供可伸缩的云服务
+
+根据应用的负载情况，可以很容易地增加或者减少 Docker。
+
+## 搭建微服务架构
+
+Docker 轻量级的特点使得它很适合用于部署、维护、组合微服务。
+
+参考资料：
+
+- [What is Docker](https://www.docker.com/what-docker)
+- [持续集成是什么？](http://www.ruanyifeng.com/blog/2015/09/continuous-integration.html)
+
+# 五、镜像与容器
+
+镜像是一种静态的结构，可以看成面向对象里面的类，而容器是镜像的一个实例。
+
+镜像包含着容器运行时所需要的代码以及其它组件，它是一种分层结构，每一层都是只读的（read-only layers）。构建镜像时，会一层一层构建，前一层是后一层的基础。镜像的这种分层存储结构很适合镜像的复用以及定制。
+
+在构建容器时，通过在镜像的基础上添加一个可写层（writable layer），用来保存着容器运行过程中的修改。
+
+<div align="center"> <img src="../pics//docker-filesystems-busyboxrw.png"/> </div><br>
+
+参考资料：
+
+- [How to Create Docker Container using Dockerfile](https://linoxide.com/linux-how-to/dockerfile-create-docker-container/)
+- [理解 Docker（2）：Docker 镜像](http://www.cnblogs.com/sammyliu/p/5877964.html)
+
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From 5499e2b898a8dcad2003b9a95b9568fc03d6f187 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 31 Jul 2018 23:32:22 +0800
Subject: [PATCH 014/172] auto commit

---
 README.md | 4 ++++
 1 file changed, 4 insertions(+)

diff --git a/README.md b/README.md
index 5e49a332..e6c2eea2 100644
--- a/README.md
+++ b/README.md
@@ -135,6 +135,10 @@ XSS、CSRF、SQL 注入、DDoS
 
 一些 Git 的使用和概念。
 
+> [Docker](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Docker.md)
+
+Docker 基本原理。
+
 > [正则表达式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/正则表达式.md)
 
 参考 正则表达式必知必会。

From ec0f8feef7c9fb2e283921f6305c048915c8ba83 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 1 Aug 2018 00:14:36 +0800
Subject: [PATCH 015/172] auto commit

---
 README.md | 4 ++++
 1 file changed, 4 insertions(+)

diff --git a/README.md b/README.md
index e6c2eea2..fe167481 100644
--- a/README.md
+++ b/README.md
@@ -143,6 +143,10 @@ Docker 基本原理。
 
 参考 正则表达式必知必会。
 
+> [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
+
+参考 正则表达式必知必会。
+
 ## 编码实践 :speak_no_evil:
 
 > [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)

From 872c20998733315311b2cbbefbbee131f4c0ca07 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 1 Aug 2018 00:15:26 +0800
Subject: [PATCH 016/172] auto commit

---
 notes/构建工具.md                             | 147 ++++++++++++++++++
 pics/897503d0-59e3-4752-903d-529fbdb72fee.jpg | Bin 0 -> 20480 bytes
 2 files changed, 147 insertions(+)
 create mode 100644 notes/构建工具.md
 create mode 100644 pics/897503d0-59e3-4752-903d-529fbdb72fee.jpg

diff --git a/notes/构建工具.md b/notes/构建工具.md
new file mode 100644
index 00000000..ae1cac20
--- /dev/null
+++ b/notes/构建工具.md
@@ -0,0 +1,147 @@
+<!-- GFM-TOC -->
+* [一、什么是构建工具](#一什么是构建工具)
+* [二、Java 主流构建工具](#二java-主流构建工具)
+* [三、Maven](#三maven)
+<!-- GFM-TOC -->
+
+
+# 一、什么是构建工具
+
+构建工具用于构建项目的自动化工具，主要包含以下工作：
+
+## 依赖管理
+
+不再需要手动导入 Jar 依赖包，并且可以自动处理依赖关系，也就是说某个依赖如果依赖于其它依赖，构建工具可以帮助我们自动处理这种依赖管理；
+
+## 运行单元测试
+
+不再需要在项目代码中添加测试代码。
+
+## 将源代码转化为可执行文件
+
+包含预处理、编译、汇编、链接等步骤。
+
+## 将可执行文件进行打包
+
+不再需要使用 IDE 将应用程序打包成 Jar 包。
+
+## 发布到生成服务器上
+
+不再需要通过 FTP 将 Jar 包上传到服务器上。
+
+参考资料：
+
+- [What is a build tool?](https://stackoverflow.com/questions/7249871/what-is-a-build-tool)
+
+# 二、Java 主流构建工具
+
+主要包括 Ant、Maven 和 Gradle。
+
+<div align="center"> <img src="../pics//897503d0-59e3-4752-903d-529fbdb72fee.jpg"/> </div><br>
+
+Gradle 和 Maven 的区别是，它使用 Groovy 这种特定领域语言（DSL）来管理构建脚本，而不再使用 XML 这种标记性语言。因为项目如果庞大的话，XML 很容易就变得臃肿。
+
+例如要在项目中引入 Junit，Maven 的代码如下：
+
+```xml
+<?xml version="1.0" encoding="UTF-8"?>
+<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
+  xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
+   <modelVersion>4.0.0</modelVersion>
+ 
+   <groupId>jizg.study.maven.hello</groupId>
+   <artifactId>hello-first</artifactId>
+   <version>0.0.1-SNAPSHOT</version>
+
+   <dependencies>
+          <dependency>
+               <groupId>junit</groupId>
+               <artifactId>junit</artifactId>
+               <version>4.10</version>
+               <scope>test</scope>
+          </dependency>
+   </dependencies>
+</project>
+```
+
+而 Gradle 只需要几行代码：
+
+```java
+dependencies {
+    testCompile "junit:junit:4.10"
+}
+```
+
+参考资料：
+
+- [Java Build Tools Comparisons: Ant vs Maven vs Gradle](https://programmingmitra.blogspot.com/2016/05/java-build-tools-comparisons-ant-vs.html)
+- [maven 2 gradle](http://sagioto.github.io/maven2gradle/)
+- [新一代构建工具 gradle](https://www.imooc.com/learn/833)
+
+# 三、Maven
+
+## 概述
+
+提供了项目对象模型（POM）文件来管理项目的构建。
+
+## 仓库
+
+仓库的搜索顺序为：本地仓库、中央仓库、远程仓库。
+
+- 本地仓库用来存储项目的依赖库；
+- 中央仓库是下载依赖库的默认位置；
+- 远程仓库，因为并非所有的库存储在中央仓库，或者中央仓库访问速度很慢，远程仓库是中央仓库的补充。
+
+## POM
+
+POM 代表项目对象模型，它是一个 XML 文件，保存在项目根目录的 pom.xml 文件中。
+
+```xml
+<dependency>
+    <groupId>junit</groupId>
+    <artifactId>junit</artifactId>
+    <version>4.12</version>
+    <scope>test</scope>
+</dependency>
+```
+
+[groupId, artifactId, version, packaging, classfier] 称为一个项目的坐标，其中 groupId、artifactId、version 必须定义，packaging 可选（默认为 Jar），classfier 不能直接定义的，需要结合插件使用。
+
+- groupId：项目组 Id，必须全球唯一；
+- artifactId：项目 Id，即项目名；
+- version：项目版本；
+- packaging：项目打包方式。
+
+## 依赖原则
+
+### 依赖路径最短优先原则
+
+```html
+A -> B -> C -> X(1.0)
+A -> D -> X(2.0)
+```
+由于 X(2.0) 路径最短，所以使用 X(2.0)。
+
+### 声明顺序优先原则
+
+```html
+A -> B -> X(1.0)
+A -> C -> X(2.0)
+```
+
+在 POM 中最先声明的优先，上面的两个依赖如果先声明 B，那么最后使用 X(1.0)。
+
+### 覆写优先原则
+
+子 POM 内声明的依赖优先于父 POM 中声明的依赖。
+
+## 解决依赖冲突
+
+找到 Maven 加载的 Jar 包版本，使用 `mvn dependency:tree` 查看依赖树，根据依赖原则来调整依赖在 POM 文件的声明顺序。
+
+参考资料：
+
+- [POM Reference](http://maven.apache.org/pom.html#Dependency_Version_Requirement_Specification)
+
+
+
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HcmV?d00001


From 35c0302ca043ad642515a59abeddff24c83b9b00 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 1 Aug 2018 00:16:52 +0800
Subject: [PATCH 017/172] auto commit

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index fe167481..c06de1d0 100644
--- a/README.md
+++ b/README.md
@@ -145,7 +145,7 @@ Docker 基本原理。
 
 > [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
 
-参考 正则表达式必知必会。
+构建工具的基本概念、主流构建工具介绍
 
 ## 编码实践 :speak_no_evil:
 

From 5c6d152dba5f4cdcd78827db3605e206424b6713 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 1 Aug 2018 00:20:53 +0800
Subject: [PATCH 018/172] auto commit

---
 notes/构建工具.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/notes/构建工具.md b/notes/构建工具.md
index ae1cac20..c0e4acd5 100644
--- a/notes/构建工具.md
+++ b/notes/构建工具.md
@@ -7,15 +7,15 @@
 
 # 一、什么是构建工具
 
-构建工具用于构建项目的自动化工具，主要包含以下工作：
+构建工具是用于构建项目的自动化工具，主要包含以下工作：
 
 ## 依赖管理
 
-不再需要手动导入 Jar 依赖包，并且可以自动处理依赖关系，也就是说某个依赖如果依赖于其它依赖，构建工具可以帮助我们自动处理这种依赖管理；
+不再需要手动导入 Jar 依赖包，并且可以自动处理依赖关系，也就是说某个依赖如果依赖于其它依赖，构建工具可以帮助我们自动处理这种依赖管理。
 
 ## 运行单元测试
 
-不再需要在项目代码中添加测试代码。
+不再需要在项目代码中添加测试代码，从而污染项目代码。
 
 ## 将源代码转化为可执行文件
 
@@ -25,7 +25,7 @@
 
 不再需要使用 IDE 将应用程序打包成 Jar 包。
 
-## 发布到生成服务器上
+## 发布到生产服务器上
 
 不再需要通过 FTP 将 Jar 包上传到服务器上。
 

From 428975aef07fa091034df033790aade5e189e78b Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Wed, 1 Aug 2018 00:56:41 +0800
Subject: [PATCH 019/172] =?UTF-8?q?Java=20=E5=87=BD=E6=95=B0=E9=87=8D?=
 =?UTF-8?q?=E5=86=99=E6=97=B6=EF=BC=8C=E5=AD=90=E7=B1=BB=E7=9A=84=E8=BF=94?=
 =?UTF-8?q?=E5=9B=9E=E5=80=BC=E7=B1=BB=E5=9E=8B=E8=A6=81=E7=AD=89=E4=BA=8E?=
 =?UTF-8?q?=E6=88=96=E8=80=85=E5=B0=8F=E4=BA=8E=E7=88=B6=E7=B1=BB=E7=9A=84?=
 =?UTF-8?q?=E8=BF=94=E5=9B=9E=E5=80=BC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 基础.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 907eea00..564d643e 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -549,9 +549,9 @@ SuperExtendExample.func()
 
 ## 重写与重载
 
-- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法；
+- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法，子类的返回值类型要等于或者小于父类的返回值；
 
-- 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载。
+- 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载，但如果返回值类型等于或者小于父类的返回值则算是重写。
 
 # 五、Object 通用方法
 

From 34f1d045617c9db1c36e71f1d44f268ae455252a Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Wed, 1 Aug 2018 09:53:19 +0800
Subject: [PATCH 020/172] Update README.md

---
 README.md | 22 ++++++++++++----------
 1 file changed, 12 insertions(+), 10 deletions(-)

diff --git a/README.md b/README.md
index c06de1d0..fe49083c 100644
--- a/README.md
+++ b/README.md
@@ -175,7 +175,9 @@ Google 开源项目的代码风格规范。
 
 笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
 
-欢迎提交对本仓库的改进建议~
+如果想要提交一个仓库现在还没有的全新内容，可以先将相应的文档放到 other 目录下。
+
+欢迎在 Issue 中提交对本仓库的改进建议~
 
 **Authorization**
 
@@ -183,21 +185,21 @@ Google 开源项目的代码风格规范。
 
 转载使用请注明出处，谢谢！
 
-**Uploading**
-
-笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
-
-进行 Markdown 文档转换是因为 Github 使用的 GFM 不支持 MathJax 公式和 TOC 标记，所以需要替换 MathJax 公式为 CodeCogs 的云服务和重新生成 TOC 目录。
-
-这里提供了笔者实现的 GFM 文档转换工具的链接：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
-
 **Typesetting**
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
 
 笔记不使用 `![]()` 这种方式来引用图片，而是用 `<img>` 标签。一方面是为了能够控制图片以合适的大小显示，另一方面是因为 GFM 不支持 `<center> ![]() </center>` 让图片居中显示，只能使用 `<div align="center"> <img src=""/> </div>` 达到居中的效果。
 
-这里提供了笔者实现的中英混排文档在线排版工具的链接：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
+笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费试用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
+
+**Uploading**
+
+笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
+
+进行 Markdown 文档转换是因为 Github 使用的 GFM 不支持 MathJax 公式和 TOC 标记，所以需要替换 MathJax 公式为 CodeCogs 的云服务和重新生成 TOC 目录。
+
+笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
 **Statement**
 

From a255b9913befb2158d8310c116742158bb34cf32 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Wed, 1 Aug 2018 09:56:51 +0800
Subject: [PATCH 021/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index fe49083c..294ac656 100644
--- a/README.md
+++ b/README.md
@@ -191,7 +191,7 @@ Google 开源项目的代码风格规范。
 
 笔记不使用 `![]()` 这种方式来引用图片，而是用 `<img>` 标签。一方面是为了能够控制图片以合适的大小显示，另一方面是因为 GFM 不支持 `<center> ![]() </center>` 让图片居中显示，只能使用 `<div align="center"> <img src=""/> </div>` 达到居中的效果。
 
-笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费试用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
+笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
 
 **Uploading**
 

From fa13575bfb2e56f4479f9a5501d446a722e00d90 Mon Sep 17 00:00:00 2001
From: fighter <fighterhit@163.com>
Date: Wed, 1 Aug 2018 18:20:48 +0800
Subject: [PATCH 022/172] fix typos

---
 notes/MySQL.md | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index d49fc952..0ab28ad8 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -95,15 +95,15 @@ VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
 
 ## 时间和日期
 
-MySQL 提供了两种相似的日期时间类型：DATATIME 和 TIMESTAMP。
+MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
 
-### 1. DATATIME
+### 1. DATETIME
 
 能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
 
 它与时区无关。
 
-默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATATIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
+默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
 
 ### 2. TIMESTAMP
 
@@ -329,7 +329,7 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 一个大查询如果一次性执行的话，可能一次锁住很多数据、占满整个事务日志、耗尽系统资源、阻塞很多小的但重要的查询。
 
 ```sql
-DELEFT FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
+DELETE FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
 ```
 
 ```sql
@@ -351,7 +351,7 @@ do {
 - 查询本身效率也可能会有所提升。例如下面的例子中，使用 IN() 代替连接查询，可以让 MySQL 按照 ID 顺序进行查询，这可能比随机的连接要更高效。
 
 ```sql
-SELECT * FROM tab
+SELECT * FROM tag
 JOIN tag_post ON tag_post.tag_id=tag.id
 JOIN post ON tag_post.post_id=post.id
 WHERE tag.tag='mysql';

From ef16ad9ed6152cbcf28a3636c846632ca104bd34 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 2 Aug 2018 00:24:52 +0800
Subject: [PATCH 023/172] auto commit

---
 notes/HTTP.md           | 81 ++++++++++++++++++++++++++---------------
 notes/Java 基础.md      |  2 +-
 notes/MySQL.md          |  6 +--
 notes/计算机操作系统.md | 31 +++++++++++-----
 4 files changed, 77 insertions(+), 43 deletions(-)

diff --git a/notes/HTTP.md b/notes/HTTP.md
index 7f54bd1a..243dc283 100644
--- a/notes/HTTP.md
+++ b/notes/HTTP.md
@@ -360,9 +360,9 @@ console.log(document.cookie);
 
 ### 5. Secure 和 HttpOnly
 
-标记为 Secure 的 Cookie 只应通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
+标记为 Secure 的 Cookie 只能通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
 
-标记为 HttpOnly 的 Cookie 不能被 JavaScript 脚本调用。因为跨站脚本攻击 (XSS) 常常使用 JavaScript 的 `Document.cookie` API 窃取用户的 Cookie 信息，因此使用 HttpOnly 标记可以在一定程度上避免 XSS 攻击。
+标记为 HttpOnly 的 Cookie 不能被 JavaScript 脚本调用。跨站脚本攻击 (XSS) 常常使用 JavaScript 的 `Document.cookie` API 窃取用户的 Cookie 信息，因此使用 HttpOnly 标记可以在一定程度上避免 XSS 攻击。
 
 ```html
 Set-Cookie: id=a3fWa; Expires=Wed, 21 Oct 2015 07:28:00 GMT; Secure; HttpOnly
@@ -382,15 +382,15 @@ Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径
 
 除了可以将用户信息通过 Cookie 存储在用户浏览器中，也可以利用 Session 存储在服务器端，存储在服务器端的信息更加安全。
 
-Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在内存型数据库中，比如 Redis。
+Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在内存型数据库中，比如 Redis，效率会更高。
 
 使用 Session 维护用户登录的过程如下：
 
 - 用户进行登录时，用户提交包含用户名和密码的表单，放入 HTTP 请求报文中；
 - 服务器验证该用户名和密码；
-- 如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 ID 称为 Session ID；
+- 如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 Key 称为 Session ID；
 - 服务器返回的响应报文的 Set-Cookie 首部字段包含了这个 Session ID，客户端收到响应报文之后将该 Cookie 值存入浏览器中；
-- 客户端之后对同一个服务器进行请求时会包含该 Cookie 值，服务器收到之后提取出 Session ID，从 Redis 中取出用户信息，继续之后的业务操作。
+- 客户端之后对同一个服务器进行请求时会包含该 Cookie 值，服务器收到之后提取出 Session ID，从 Redis 中取出用户信息，继续之前的业务操作。
 
 应该注意 Session ID 的安全性问题，不能让它被恶意攻击者轻易获取，那么就不能产生一个容易被猜到的 Session ID 值。此外，还需要经常重新生成 Session ID。在对安全性要求极高的场景下，例如转账等操作，除了使用 Session 管理用户状态之外，还需要对用户进行重新验证，比如重新输入密码，或者使用短信验证码等方式。
 
@@ -409,7 +409,7 @@ Session 可以存储在服务器上的文件、数据库或者内存中。也可
 ### 1. 优点
 
 - 缓解服务器压力；
-- 降低客户端获取资源的延迟（缓存资源比服务器上的资源离客户端更近）。
+- 降低客户端获取资源的延迟：缓存通常位于内存中，读取缓存的速度更快。并且缓存在地理位置上也有可能比源服务器来得近，例如浏览器缓存。
 
 ### 2. 实现方法
 
@@ -460,7 +460,10 @@ max-age 指令出现在响应报文中，表示缓存资源在缓存服务器中
 Cache-Control: max-age=31536000
 ```
 
-Expires 首部字段也可以用于告知缓存服务器该资源什么时候会过期。在 HTTP/1.1 中，会优先处理 Cache-Control : max-age 指令；而在 HTTP/1.0 中，Cache-Control : max-age 指令会被忽略掉。
+Expires 首部字段也可以用于告知缓存服务器该资源什么时候会过期。
+
+- 在 HTTP/1.1 中，会优先处理 max-age 指令；
+- 在 HTTP/1.0 中，max-age 指令会被忽略掉。
 
 ```html
 Expires: Wed, 04 Jul 2012 08:26:05 GMT
@@ -496,13 +499,16 @@ If-Modified-Since: Wed, 21 Oct 2015 07:28:00 GMT
 
 ### 1. 短连接与长连接
 
-当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源，如果每进行一次 HTTP 通信就要断开一次 TCP 连接，连接建立和断开的开销会很大。长连接只需要建立一次 TCP 连接就能进行多次 HTTP 通信。
+当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源。如果每进行一次 HTTP 通信就要断开一次 TCP 连接，连接建立和断开的开销会很大。
 
-从 HTTP/1.1 开始默认是长连接的，如果要断开连接，需要由客户端或者服务器端提出断开，使用 Connection : close；而在 HTTP/1.1 之前默认是短连接的，如果需要长连接，则使用 Connection : Keep-Alive。
+长连接只需要建立一次 TCP 连接就能进行多次 HTTP 通信。
+
+- 从 HTTP/1.1 开始默认是长连接的，如果要断开连接，需要由客户端或者服务器端提出断开，使用 `Connection : close`；
+- 在 HTTP/1.1 之前默认是短连接的，如果需要使用长连接，则使用 `Connection : Keep-Alive`。
 
 ### 2. 流水线
 
-默认情况下，HTTP 请求是按顺序发出的，下一个请求只有在当前请求收到相应之后才会被发出。由于会受到网络延迟和带宽的限制，在下一个请求被发送到服务器之前，可能需要等待很长时间。
+默认情况下，HTTP 请求是按顺序发出的，下一个请求只有在当前请求收到响应之后才会被发出。由于会受到网络延迟和带宽的限制，在下一个请求被发送到服务器之前，可能需要等待很长时间。
 
 流水线是在同一条长连接上发出连续的请求，而不用等待响应返回，这样可以避免连接延迟。
 
@@ -512,17 +518,17 @@ If-Modified-Since: Wed, 21 Oct 2015 07:28:00 GMT
 
 ### 1. 类型
 
-**（一）服务端驱动型内容协商** 
+**（一）服务端驱动型** 
 
 客户端设置特定的 HTTP 首部字段，例如 Accept、Accept-Charset、Accept-Encoding、Accept-Language、Content-Languag，服务器根据这些字段返回特定的资源。
 
 它存在以下问题：
 
 - 服务器很难知道客户端浏览器的全部信息；
-- 客户端提供的信息相当冗长（HTTP/2 协议的首部压缩机制缓解了这个问题），并且存在隐私风险（HTTP 指纹识别技术）。
+- 客户端提供的信息相当冗长（HTTP/2 协议的首部压缩机制缓解了这个问题），并且存在隐私风险（HTTP 指纹识别技术）；
 - 给定的资源需要返回不同的展现形式，共享缓存的效率会降低，而服务器端的实现会越来越复杂。
 
-**（二）代理驱动型协商** 
+**（二）代理驱动型** 
 
 服务器返回 300 Multiple Choices 或者 406 Not Acceptable，客户端从中选出最合适的那个资源。
 
@@ -538,9 +544,11 @@ Vary: Accept-Language
 
 ## 内容编码
 
-内容编码将实体主体进行压缩，从而减少传输的数据量。常用的内容编码有：gzip、compress、deflate、identity。
+内容编码将实体主体进行压缩，从而减少传输的数据量。
 
-浏览器发送 Accept-Encoding 首部，其中包含有它所支持的压缩算法，以及各自的优先级，服务器则从中选择一种，使用该算法对响应的消息主体进行压缩，并且发送 Content-Encoding 首部来告知浏览器它选择了哪一种算法。由于该内容协商过程是基于编码类型来选择资源的展现形式的，在响应中，Vary 首部中至少要包含 Content-Encoding，这样的话，缓存服务器就可以对资源的不同展现形式进行缓存。
+常用的内容编码有：gzip、compress、deflate、identity。
+
+浏览器发送 Accept-Encoding 首部，其中包含有它所支持的压缩算法，以及各自的优先级。服务器则从中选择一种，使用该算法对响应的消息主体进行压缩，并且发送 Content-Encoding 首部来告知浏览器它选择了哪一种算法。由于该内容协商过程是基于编码类型来选择资源的展现形式的，在响应的 Vary 首部至少要包含 Content-Encoding。
 
 ## 范围请求
 
@@ -622,10 +630,14 @@ HTTP/1.1 使用虚拟主机技术，使得一台服务器拥有多个域名，
 - 网络访问控制
 - 访问日志记录
 
-代理服务器分为正向代理和反向代理两种，用户察觉得到正向代理的存在；而反向代理一般位于内部网络中，用户察觉不到。
+代理服务器分为正向代理和反向代理两种：
+
+- 用户察觉得到正向代理的存在。
 
 <div align="center"> <img src="../pics//a314bb79-5b18-4e63-a976-3448bffa6f1b.png" width=""/> </div><br>
 
+- 而反向代理一般位于内部网络中，用户察觉不到。
+
 <div align="center"> <img src="../pics//2d09a847-b854-439c-9198-b29c65810944.png" width=""/> </div><br>
 
 ### 2. 网关
@@ -634,7 +646,7 @@ HTTP/1.1 使用虚拟主机技术，使得一台服务器拥有多个域名，
 
 ### 3. 隧道
 
-使用 SSL 等加密手段，为客户端和服务器之间建立一条安全的通信线路。
+使用 SSL 等加密手段，在客户端和服务器之间建立一条安全的通信线路。
 
 # 六、HTTPs
 
@@ -644,7 +656,7 @@ HTTP 有以下安全性问题：
 - 不验证通信方的身份，通信方的身份有可能遭遇伪装；
 - 无法证明报文的完整性，报文有可能遭篡改。
 
-HTTPs 并不是新协议，而是让 HTTP 先和 SSL（Secure Sockets Layer）通信，再由 SSL 和 TCP 通信。也就是说 HTTPs 使用了隧道进行通信。
+HTTPs 并不是新协议，而是让 HTTP 先和 SSL（Secure Sockets Layer）通信，再由 SSL 和 TCP 通信，也就是说 HTTPs 使用了隧道进行通信。
 
 通过使用 SSL，HTTPs 具有了加密（防窃听）、认证（防伪装）和完整性保护（防篡改）。
 
@@ -727,7 +739,11 @@ HTTP/2.0 将报文分成 HEADERS 帧和 DATA 帧，它们都是二进制格式
 
 <div align="center"> <img src="../pics//86e6a91d-a285-447a-9345-c5484b8d0c47.png" width="400"/> </div><br>
 
-在通信过程中，只会有一个 TCP 连接存在，它承载了任意数量的双向数据流（Stream）。一个数据流都有一个唯一标识符和可选的优先级信息，用于承载双向信息。消息（Message）是与逻辑请求或响应消息对应的完整的一系列帧。帧（Fram）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
+在通信过程中，只会有一个 TCP 连接存在，它承载了任意数量的双向数据流（Stream）。
+
+- 一个数据流都有一个唯一标识符和可选的优先级信息，用于承载双向信息。
+- 消息（Message）是与逻辑请求或响应消息对应的完整的一系列帧。
+- 帧（Fram）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
 
 <div align="center"> <img src="../pics//af198da1-2480-4043-b07f-a3b91a88b815.png" width="600"/> </div><br>
 
@@ -739,7 +755,11 @@ HTTP/2.0 在客户端请求一个资源时，会把相关的资源一起发送
 
 ## 首部压缩
 
-HTTP/1.1 的首部带有大量信息，而且每次都要重复发送。HTTP/2.0 要求客户端和服务器同时维护和更新一个包含之前见过的首部字段表，从而避免了重复传输。不仅如此，HTTP/2.0 也使用 Huffman 编码对首部字段进行压缩。
+HTTP/1.1 的首部带有大量信息，而且每次都要重复发送。
+
+HTTP/2.0 要求客户端和服务器同时维护和更新一个包含之前见过的首部字段表，从而避免了重复传输。
+
+不仅如此，HTTP/2.0 也使用 Huffman 编码对首部字段进行压缩。
 
 <div align="center"> <img src="../pics//_u4E0B_u8F7D.png" width="600"/> </div><br>
 
@@ -751,7 +771,9 @@ GET 用于获取资源，而 POST 用于传输实体主体。
 
 ## 参数
 
-GET 和 POST 的请求都能使用额外的参数，但是 GET 的参数是以查询字符串出现在 URL 中，而 POST 的参数存储在实体主体中。
+GET 和 POST 的请求都能使用额外的参数，但是 GET 的参数是以查询字符串出现在 URL 中，而 POST 的参数存储在实体主体中。不能因为 POST 参数存储在实体主体中就认为它的安全性更高，因为照样可以通过一些抓包工具（Fiddler）查看。
+
+因为 URL 只支持 ASCII 码，因此 GET 的参数中如果存在中文等字符就需要先进行编码。例如 `中文` 会转换为 `%E4%B8%AD%E6%96%87`，而空格会转换为 `%20`。POST 参考支持标准字符集。
 
 ```
 GET /test/demo_form.asp?name1=value1&name2=value2 HTTP/1.1
@@ -763,10 +785,6 @@ Host: w3schools.com
 name1=value1&name2=value2
 ```
 
-不能因为 POST 参数存储在实体主体中就认为它的安全性更高，因为照样可以通过一些抓包工具（Fiddler）查看。
-
-因为 URL 只支持 ASCII 码，因此 GET 的参数中如果存在中文等字符就需要先进行编码，例如`中文`会转换为`%E4%B8%AD%E6%96%87`，而空格会转换为`%20`。POST 支持标准字符集。
-
 ## 安全
 
 安全的 HTTP 方法不会改变服务器状态，也就是说它只是可读的。
@@ -779,9 +797,13 @@ GET 方法是安全的，而 POST 却不是，因为 POST 的目的是传送实
 
 ## 幂等性
 
-幂等的 HTTP 方法，同样的请求被执行一次与连续执行多次的效果是一样的，服务器的状态也是一样的。换句话说就是，幂等方法不应该具有副作用（统计用途除外）。在正确实现的条件下，GET，HEAD，PUT 和 DELETE 等方法都是幂等的，而 POST 方法不是。所有的安全方法也都是幂等的。
+幂等的 HTTP 方法，同样的请求被执行一次与连续执行多次的效果是一样的，服务器的状态也是一样的。换句话说就是，幂等方法不应该具有副作用（统计用途除外）。
 
-GET /pageX HTTP/1.1 是幂等的。连续调用多次，客户端接收到的结果都是一样的：
+所有的安全方法也都是幂等的。
+
+在正确实现的条件下，GET，HEAD，PUT 和 DELETE 等方法都是幂等的，而 POST 方法不是。
+
+GET /pageX HTTP/1.1 是幂等的，连续调用多次，客户端接收到的结果都是一样的：
 
 ```
 GET /pageX HTTP/1.1
@@ -790,7 +812,7 @@ GET /pageX HTTP/1.1
 GET /pageX HTTP/1.1
 ```
 
-POST /add_row HTTP/1.1 不是幂等的。如果调用多次，就会增加多行记录：
+POST /add_row HTTP/1.1 不是幂等的，如果调用多次，就会增加多行记录：
 
 ```
 POST /add_row HTTP/1.1   -> Adds a 1nd row
@@ -820,7 +842,8 @@ DELETE /idX/delete HTTP/1.1   -> Returns 404
 
 > XMLHttpRequest 是一个 API，它为客户端提供了在客户端和服务器之间传输数据的功能。它提供了一个通过 URL 来获取数据的简单方式，并且不会使整个页面刷新。这使得网页只更新一部分页面而不会打扰到用户。XMLHttpRequest 在 AJAX 中被大量使用。
 
-在使用 XMLHttpRequest 的 POST 方法时，浏览器会先发送 Header 再发送 Data。但并不是所有浏览器会这么做，例如火狐就不会。而 GET 方法 Header 和 Data 会一起发送。
+- 在使用 XMLHttpRequest 的 POST 方法时，浏览器会先发送 Header 再发送 Data。但并不是所有浏览器会这么做，例如火狐就不会。
+- 而 GET 方法 Header 和 Data 会一起发送。
 
 # 九、HTTP/1.0 与 HTTP/1.1 的区别
 
diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 907eea00..cda00ea0 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -549,7 +549,7 @@ SuperExtendExample.func()
 
 ## 重写与重载
 
-- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法；
+- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法，子类的返回值类型要等于或者小于父类的返回值；
 
 - 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载。
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index d49fc952..f6aa853b 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -95,15 +95,15 @@ VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
 
 ## 时间和日期
 
-MySQL 提供了两种相似的日期时间类型：DATATIME 和 TIMESTAMP。
+MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
 
-### 1. DATATIME
+### 1. DATETIME
 
 能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
 
 它与时区无关。
 
-默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATATIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
+默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
 
 ### 2. TIMESTAMP
 
diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 52c74dd7..5c9f63a7 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -1,7 +1,7 @@
 <!-- GFM-TOC -->
 * [一、概述](#一概述)
-    * [操作系统基本特征](#操作系统基本特征)
-    * [操作系统基本功能](#操作系统基本功能)
+    * [基本特征](#基本特征)
+    * [基本功能](#基本功能)
     * [系统调用](#系统调用)
     * [大内核和微内核](#大内核和微内核)
     * [中断分类](#中断分类)
@@ -35,7 +35,7 @@
 
 # 一、概述
 
-## 操作系统基本特征
+## 基本特征
 
 ### 1. 并发
 
@@ -63,7 +63,7 @@
 
 异步指进程不是一次性执行完毕，而是走走停停，以不可知的速度向前推进。
 
-## 操作系统基本功能
+## 基本功能
 
 ### 1. 进程管理
 
@@ -152,19 +152,27 @@ Linux 的系统调用主要有以下这些：
 
 一个进程中可以有多个线程，它们共享进程资源。
 
+QQ 和浏览器是两个进程，浏览器进程里面有很多线程，例如 HTTP 请求线程、事件响应线程、渲染线程等等，线程的并发执行使得在浏览器中点击一个新链接从而发起 HTTP 请求时，浏览器还可以响应用户的其它事件。
+
 <div align="center"> <img src="../pics//3cd630ea-017c-488d-ad1d-732b4efeddf5.png"/> </div><br>
 
 ### 3. 区别
 
-- 拥有资源：进程是资源分配的基本单位，但是线程不拥有资源，线程可以访问隶属进程的资源。
+（一）拥有资源
 
-- 调度：线程是独立调度的基本单位，在同一进程中，线程的切换不会引起进程切换，从一个进程内的线程切换到另一个进程中的线程时，会引起进程切换。
+进程是资源分配的基本单位，但是线程不拥有资源，线程可以访问隶属进程的资源。
 
-- 系统开销：由于创建或撤销进程时，系统都要为之分配或回收资源，如内存空间、I/O 设备等，所付出的开销远大于创建或撤销线程时的开销。类似地，在进行进程切换时，涉及当前执行进程 CPU 环境的保存及新调度进程 CPU 环境的设置，而线程切换时只需保存和设置少量寄存器内容，开销很小。
+（二）调度
 
-- 通信方面：进程间通信 (IPC) 需要进程同步和互斥手段的辅助，以保证数据的一致性。而线程间可以通过直接读/写同一进程中的数据段（如全局变量）来进行通信。
+线程是独立调度的基本单位，在同一进程中，线程的切换不会引起进程切换，从一个进程内的线程切换到另一个进程中的线程时，会引起进程切换。
 
-举例：QQ 和浏览器是两个进程，浏览器进程里面有很多线程，例如 HTTP 请求线程、事件响应线程、渲染线程等等，线程的并发执行使得在浏览器中点击一个新链接从而发起 HTTP 请求时，浏览器还可以响应用户的其它事件。
+（三）系统开销
+
+由于创建或撤销进程时，系统都要为之分配或回收资源，如内存空间、I/O 设备等，所付出的开销远大于创建或撤销线程时的开销。类似地，在进行进程切换时，涉及当前执行进程 CPU 环境的保存及新调度进程 CPU 环境的设置，而线程切换时只需保存和设置少量寄存器内容，开销很小。
+
+（四）通信方面
+
+进程间通信 (IPC) 需要进程同步和互斥手段的辅助，以保证数据的一致性。而线程间可以通过直接读/写同一进程中的数据段（如全局变量）来进行通信。
 
 ## 进程状态的切换
 
@@ -211,7 +219,10 @@ Linux 的系统调用主要有以下这些：
 
 将所有就绪进程按 FCFS 的原则排成一个队列，每次调度时，把 CPU 时间分配给队首进程，该进程可以执行一个时间片。当时间片用完时，由计时器发出时钟中断，调度程序便停止该进程的执行，并将它送往就绪队列的末尾，同时继续把 CPU 时间分配给队首的进程。
 
-时间片轮转算法的效率和时间片的大小有很大关系。因为进程切换都要保存进程的信息并且载入新进程的信息，如果时间片太小，会导致进程切换得太频繁，在进程切换上就会花过多时间。
+时间片轮转算法的效率和时间片的大小有很大关系：
+
+- 因为进程切换都要保存进程的信息并且载入新进程的信息，如果时间片太小，会导致进程切换得太频繁，在进程切换上就会花过多时间。
+- 而如果时间片过长，那么实时性就不能得到保证。
 
 <div align="center"> <img src="../pics//8c662999-c16c-481c-9f40-1fdba5bc9167.png"/> </div><br>
 

From 4ab241fd8175ec4105727c790f01d4a37f6f440c Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 2 Aug 2018 00:25:48 +0800
Subject: [PATCH 024/172] auto commit

---
 notes/Java 基础.md | 2 +-
 notes/MySQL.md     | 4 ++--
 2 files changed, 3 insertions(+), 3 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 564d643e..cda00ea0 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -551,7 +551,7 @@ SuperExtendExample.func()
 
 - 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法，子类的返回值类型要等于或者小于父类的返回值；
 
-- 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载，但如果返回值类型等于或者小于父类的返回值则算是重写。
+- 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载。
 
 # 五、Object 通用方法
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index 0ab28ad8..f6aa853b 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -329,7 +329,7 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 一个大查询如果一次性执行的话，可能一次锁住很多数据、占满整个事务日志、耗尽系统资源、阻塞很多小的但重要的查询。
 
 ```sql
-DELETE FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
+DELEFT FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
 ```
 
 ```sql
@@ -351,7 +351,7 @@ do {
 - 查询本身效率也可能会有所提升。例如下面的例子中，使用 IN() 代替连接查询，可以让 MySQL 按照 ID 顺序进行查询，这可能比随机的连接要更高效。
 
 ```sql
-SELECT * FROM tag
+SELECT * FROM tab
 JOIN tag_post ON tag_post.tag_id=tag.id
 JOIN post ON tag_post.post_id=post.id
 WHERE tag.tag='mysql';

From 2abde21661dcefbddebdfe96719d7cb12d658bf2 Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Thu, 2 Aug 2018 08:03:42 +0800
Subject: [PATCH 025/172] =?UTF-8?q?Update=20Java=20=E8=99=9A=E6=8B=9F?=
 =?UTF-8?q?=E6=9C=BA.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 虚拟机.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 3bb9f289..d67ef5bd 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -102,7 +102,7 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ## 直接内存
 
-在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
+在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因此避免了在 Java 堆和 Native 堆中来回复制数据。
 
 # 二、垃圾收集
 
@@ -146,7 +146,7 @@ Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC
 
 ### 3. 引用类型
 
-无论是通过引用计算算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
+无论是通过引用计数算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
 
 Java 具有四种强度不同的引用类型。
 
@@ -256,7 +256,7 @@ finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。
 一般将 Java 堆分为新生代和老年代。
 
 - 新生代使用：复制算法
-- 老年代使用：标记 - 清理 或者 标记 - 整理 算法
+- 老年代使用：标记 - 清除 或者 标记 - 整理 算法
 
 ## 垃圾收集器
 

From ef9dc26fb6af0862f532d3efc26915b2c6c4f4be Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Thu, 2 Aug 2018 08:10:35 +0800
Subject: [PATCH 026/172] =?UTF-8?q?Update=20Java=20=E8=99=9A=E6=8B=9F?=
 =?UTF-8?q?=E6=9C=BA.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 虚拟机.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index d67ef5bd..906f4ef6 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -102,7 +102,7 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ## 直接内存
 
-在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因此避免了在 Java 堆和 Native 堆中来回复制数据。
+在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因为其避免了在 Java 堆和 Native 堆中来回复制数据。
 
 # 二、垃圾收集
 

From 726366858b1c19bf495e95d2b8d092fdd654e0d6 Mon Sep 17 00:00:00 2001
From: luocaodan <luocaodan@buaa.edu.cn>
Date: Thu, 2 Aug 2018 18:28:38 +0800
Subject: [PATCH 027/172] =?UTF-8?q?Update=20Java=20=E5=B9=B6=E5=8F=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index ab3528f2..b704f70c 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1419,7 +1419,7 @@ public final int incrementAndGet() {
 }
 ```
 
-以下代码是 getAndAddInt() 源码，var1 指示内存地址，var2 指示旧值，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果 var2==var5，那么就更新内存地址为 var1 的变量为 var5+var4。可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
+以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果 该字段内存地址中的值==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
 
 ```java
 public final int getAndAddInt(Object var1, long var2, int var4) {

From 1d99676008f1f907397315a3fa9ef1a6eb01257d Mon Sep 17 00:00:00 2001
From: Zoctan <752481828@qq.com>
Date: Thu, 2 Aug 2018 22:40:04 +0800
Subject: [PATCH 028/172] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=20=E2=80=9D=E5=8F=82?=
 =?UTF-8?q?=E6=95=B0=E4=BC=A0=E9=80=92=E2=80=9C=20=E4=B8=AD=E4=B8=80?=
 =?UTF-8?q?=E7=82=B9=E5=86=85=E5=AE=B9?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 基础.md | 16 +++++++++++++++-
 1 file changed, 15 insertions(+), 1 deletion(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index cda00ea0..675b0918 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -223,7 +223,7 @@ System.out.println(s4 == s5);  // true
 
 Java 的参数是以值传递的形式传入方法中，而不是引用传递。
 
-Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。
+Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。而在方法中改变对象的成员变量值也会使原对象的成员变量发生变化，因为是改变同一个地址指向的内容。
 
 ```java
 public class Dog {
@@ -236,6 +236,10 @@ public class Dog {
     String getName() {
         return name;
     }
+    
+    void setName(String name) {
+        this.name = name;
+    }
 
     String getObjectAddress() {
         return super.toString();
@@ -251,6 +255,9 @@ public class PassByValueExample {
         func(dog);
         System.out.println(dog.getObjectAddress()); // Dog@4554617c
         System.out.println(dog.getName());          // A
+        func2(dog);
+        System.out.println(dog.getObjectAddress()); // Dog@4554617c
+        System.out.println(dog.getName());          // C
     }
 
     private static void func(Dog dog) {
@@ -259,6 +266,13 @@ public class PassByValueExample {
         System.out.println(dog.getObjectAddress()); // Dog@74a14482
         System.out.println(dog.getName());          // B
     }
+    
+    private static void func2(Dog dog) {
+        System.out.println(dog.getObjectAddress()); // Dog@4554617c
+        dog.setName("C");
+        System.out.println(dog.getObjectAddress()); // Dog@4554617c
+        System.out.println(dog.getName());          // C
+    }
 }
 ```
 

From fd0e8c11a843e112518a85969cce083a8a233a84 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:03:37 +0800
Subject: [PATCH 029/172] add group

---
 README.md       |   2 +-
 other/Group.md  |   9 +++++++++
 other/group.jpg | Bin 0 -> 114283 bytes
 3 files changed, 10 insertions(+), 1 deletion(-)
 create mode 100644 other/Group.md
 create mode 100644 other/group.jpg

diff --git a/README.md b/README.md
index 294ac656..a2816aeb 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link"> <img src="https://img.shields.io/badge/>-gitter-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="other/Group.md"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->
diff --git a/other/Group.md b/other/Group.md
new file mode 100644
index 00000000..59b35913
--- /dev/null
+++ b/other/Group.md
@@ -0,0 +1,9 @@
+创建交流群的主要目的是为了为大家提供一个交流平台，方便大家在学习的过程中互相讨论。
+
+这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以提升自己对知识的掌握程度。
+
+至于交流群和 Issue 有什么区别，主要是两方面：一是交流群实时性高一些，二是交流群会更活跃一些。另外，Issue 主要是用来发布一些项目中的错误和一些改进建议，当然也可以发布一些可以讨论的问题，交流群可以讨论的内容比较广，例如在阅读本项目过程中不理解的地方可以在交流群中寻求别人的帮助、新技术的讨论、招聘信息、学习和工作的感受等等。
+
+交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
+
+<center><img src="group.jpg" width="300px"></center> </br>
diff --git a/other/group.jpg b/other/group.jpg
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HcmV?d00001


From 2a96c66636b64fb220c1b0070bf2984c671f5bb1 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:04:38 +0800
Subject: [PATCH 030/172] fix

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index a2816aeb..c015a100 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="other/Group.md"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->

From 5fefd1a67f3b1d19d10b74d40a8ba379884a799f Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:05:50 +0800
Subject: [PATCH 031/172] fix

---
 other/Group.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index 59b35913..388b0ac6 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -6,4 +6,4 @@
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 
-<center><img src="group.jpg" width="300px"></center> </br>
+<div align="cent"><img src="group.jpg" width="300px"></div> </br>

From 77f2be3635e017caf238bbbdc53093fde762b813 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:06:41 +0800
Subject: [PATCH 032/172] fix

---
 other/Group.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index 388b0ac6..15d2b7d0 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -6,4 +6,4 @@
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 
-<div align="cent"><img src="group.jpg" width="300px"></div> </br>
+<div align="center"><img src="group.jpg" width="300px"></div> </br>

From 274026e13816b13e067d859d0a9c55071d30b7e6 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:07:16 +0800
Subject: [PATCH 033/172] fix

---
 other/Group.md | 3 ++-
 1 file changed, 2 insertions(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index 15d2b7d0..ae3a89f6 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -6,4 +6,5 @@
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 
-<div align="center"><img src="group.jpg" width="300px"></div> </br>
+
+</br> <div align="center"><img src="group.jpg" width="300px"></div> </br>

From d0fa9aff51e474dab7db2fb64c3c841c40d40a02 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Thu, 2 Aug 2018 23:10:09 +0800
Subject: [PATCH 034/172] fix

---
 other/Group.md | 3 ++-
 1 file changed, 2 insertions(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index ae3a89f6..5e7bbc90 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -2,9 +2,10 @@
 
 这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以提升自己对知识的掌握程度。
 
+因为笔者白天要上班，因此不能及时进行回复，大部分时间会处于潜水状态。
+
 至于交流群和 Issue 有什么区别，主要是两方面：一是交流群实时性高一些，二是交流群会更活跃一些。另外，Issue 主要是用来发布一些项目中的错误和一些改进建议，当然也可以发布一些可以讨论的问题，交流群可以讨论的内容比较广，例如在阅读本项目过程中不理解的地方可以在交流群中寻求别人的帮助、新技术的讨论、招聘信息、学习和工作的感受等等。
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 
-
 </br> <div align="center"><img src="group.jpg" width="300px"></div> </br>

From 480a8070cd9daf6954749c7de45fbb9846874337 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Fri, 3 Aug 2018 00:12:55 +0800
Subject: [PATCH 035/172] auto commit

---
 notes/Java 基础.md      | 16 +-----------
 notes/Java 并发.md      |  2 +-
 notes/Java 虚拟机.md    |  6 ++---
 notes/Linux.md          | 58 ++++++++++++++---------------------------
 notes/MySQL.md          | 47 +++++++++++----------------------
 notes/Redis.md          |  3 +++
 notes/计算机操作系统.md | 34 ++++++++++++++++--------
 7 files changed, 66 insertions(+), 100 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 675b0918..0bce5622 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -223,7 +223,7 @@ System.out.println(s4 == s5);  // true
 
 Java 的参数是以值传递的形式传入方法中，而不是引用传递。
 
-Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。而在方法中改变对象的成员变量值也会使原对象的成员变量发生变化，因为是改变同一个地址指向的内容。
+以下代码中 Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。但是如果在方法中改变对象的字段值会改变原对象该字段值，因为改变的是同一个地址指向的内容。
 
 ```java
 public class Dog {
@@ -236,10 +236,6 @@ public class Dog {
     String getName() {
         return name;
     }
-    
-    void setName(String name) {
-        this.name = name;
-    }
 
     String getObjectAddress() {
         return super.toString();
@@ -255,9 +251,6 @@ public class PassByValueExample {
         func(dog);
         System.out.println(dog.getObjectAddress()); // Dog@4554617c
         System.out.println(dog.getName());          // A
-        func2(dog);
-        System.out.println(dog.getObjectAddress()); // Dog@4554617c
-        System.out.println(dog.getName());          // C
     }
 
     private static void func(Dog dog) {
@@ -266,13 +259,6 @@ public class PassByValueExample {
         System.out.println(dog.getObjectAddress()); // Dog@74a14482
         System.out.println(dog.getName());          // B
     }
-    
-    private static void func2(Dog dog) {
-        System.out.println(dog.getObjectAddress()); // Dog@4554617c
-        dog.setName("C");
-        System.out.println(dog.getObjectAddress()); // Dog@4554617c
-        System.out.println(dog.getName());          // C
-    }
 }
 ```
 
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index b704f70c..627b824f 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1419,7 +1419,7 @@ public final int incrementAndGet() {
 }
 ```
 
-以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果 该字段内存地址中的值==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
+以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果 该字段内存地址中的值 ==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
 
 ```java
 public final int getAndAddInt(Object var1, long var2, int var4) {
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 906f4ef6..3bb9f289 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -102,7 +102,7 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ## 直接内存
 
-在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因为其避免了在 Java 堆和 Native 堆中来回复制数据。
+在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
 
 # 二、垃圾收集
 
@@ -146,7 +146,7 @@ Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC
 
 ### 3. 引用类型
 
-无论是通过引用计数算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
+无论是通过引用计算算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
 
 Java 具有四种强度不同的引用类型。
 
@@ -256,7 +256,7 @@ finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。
 一般将 Java 堆分为新生代和老年代。
 
 - 新生代使用：复制算法
-- 老年代使用：标记 - 清除 或者 标记 - 整理 算法
+- 老年代使用：标记 - 清理 或者 标记 - 整理 算法
 
 ## 垃圾收集器
 
diff --git a/notes/Linux.md b/notes/Linux.md
index 8f27d75c..ebcc347b 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -11,7 +11,6 @@
     * [GNU](#gnu)
     * [开源协议](#开源协议)
 * [二、磁盘](#二磁盘)
-    * [HDD](#hdd)
     * [磁盘接口](#磁盘接口)
     * [磁盘的文件名](#磁盘的文件名)
 * [三、分区](#三分区)
@@ -185,21 +184,6 @@ GNU 计划，译为革奴计划，它的目标是创建一套完全自由的操
 
 # 二、磁盘
 
-## HDD
-
-Hard Disk Drives(HDD) 俗称硬盘，具有以下结构：
-
-- 盘面（Platter）：一个硬盘有多个盘面；
-- 磁道（Track）：盘面上的圆形带状区域，一个盘面可以有多个磁道；
-- 扇区（Track Sector）：磁道上的一个弧段，一个磁道可以有多个扇区，它是最小的物理储存单位，目前主要有 512 bytes 与 4 K 两种大小；
-- 磁头（Head）：与盘面非常接近，能够将盘面上的磁场转换为电信号（读），或者将电信号转换为盘面的磁场（写）；
-- 制动手臂（Actuator arm）：用于在磁道之间移动磁头；
-- 主轴（Spindle）：使整个盘面转动。
-
-<div align="center"> <img src="../pics//014fbc4d-d873-4a12-b160-867ddaed9807.jpg" width=""/> </div><br>
-
-[Decoding UCS Invicta – Part 1](https://blogs.cisco.com/datacenter/decoding-ucs-invicta-part-1)
-
 ## 磁盘接口
 
 ### 1. IDE
@@ -1025,10 +1009,10 @@ g/re/p（globally search a regular expression and print)，使用正则表示式
 
 ```html
 $ grep [-acinv] [--color=auto] 搜寻字符串 filename
--c ： 计算找到个数
+-c ： 统计个数
 -i ： 忽略大小写
 -n ： 输出行号
--v ： 反向选择，亦即显示出没有 搜寻字符串 内容的那一行
+-v ： 反向选择，也就是显示出没有 搜寻字符串 内容的那一行
 --color=auto ：找到的关键字加颜色显示
 ```
 
@@ -1068,7 +1052,7 @@ $ printf '%10s %5i %5i %5i %8.2f \n' $(cat printf.txt)
 
 awk 每次处理一行，处理的最小单位是字段，每个字段的命名方式为：\$n，n 为字段号，从 1 开始，\$0 表示一整行。
 
-示例 1：取出登录用户的用户名和 ip
+示例：取出登录用户的用户名和 IP
 
 ```html
 $ last -n 5
@@ -1077,7 +1061,9 @@ dmtsai pts/0 192.168.1.100 Thu Jul 9 23:36 - 02:58 (03:22)
 dmtsai pts/0 192.168.1.100 Thu Jul 9 17:23 - 23:36 (06:12)
 dmtsai pts/0 192.168.1.100 Thu Jul 9 08:02 - 08:17 (00:14)
 dmtsai tty1 Fri May 29 11:55 - 12:11 (00:15)
+```
 
+```html
 $ last -n 5 | awk '{print $1 "\t" $3}'
 ```
 
@@ -1087,7 +1073,7 @@ $ last -n 5 | awk '{print $1 "\t" $3}'
 $ awk '条件类型 1 {动作 1} 条件类型 2 {动作 2} ...' filename
 ```
 
-示例 2：/etc/passwd 文件第三个字段为 UID，对 UID 小于 10 的数据进行处理。
+示例：/etc/passwd 文件第三个字段为 UID，对 UID 小于 10 的数据进行处理。
 
 ```text
 $ cat /etc/passwd | awk 'BEGIN {FS=":"} $3 < 10 {print $1 "\t " $3}'
@@ -1104,7 +1090,7 @@ awk 变量：
 | NR | 目前所处理的是第几行数据 |
 | FS | 目前的分隔字符，默认是空格键 |
 
-示例 3：输出正在处理的行号，并显示每一行有多少字段
+示例：显示正在处理的行号以及每一行有多少字段
 
 ```html
 $ last -n 5 | awk '{print $1 "\t lines: " NR "\t columns: " NF}'
@@ -1125,19 +1111,19 @@ dmtsai lines: 5 columns: 9
 
 示例一：查看自己的进程
 
-```
+```sh
 # ps -l
 ```
 
 示例二：查看系统所有进程
 
-```
+```sh
 # ps aux
 ```
 
 示例三：查看特定的进程
 
-```
+```sh
 # ps aux | grep threadx
 ```
 
@@ -1147,7 +1133,7 @@ dmtsai lines: 5 columns: 9
 
 示例：两秒钟刷新一次
 
-```
+```sh
 # top -d 2
 ```
 
@@ -1157,7 +1143,7 @@ dmtsai lines: 5 columns: 9
 
 示例：查看所有进程树
 
-```
+```sh
 # pstree -A
 ```
 
@@ -1167,14 +1153,12 @@ dmtsai lines: 5 columns: 9
 
 示例：查看特定端口的进程
 
-```
+```sh
 # netstat -anp | grep port
 ```
 
 ## 进程状态
 
-<div align="center"> <img src="../pics//76a49594323247f21c9b3a69945445ee.png" width=""/> </div><br>
-
 | 状态 | 说明 |
 | :---: | --- |
 | R | running or runnable (on run queue) |
@@ -1183,9 +1167,11 @@ dmtsai lines: 5 columns: 9
 | Z | zombie (terminated but not reaped by its parent) |
 | T | stopped (either by a job control signal or because it is being traced) |
 
+<div align="center"> <img src="../pics//76a49594323247f21c9b3a69945445ee.png" width=""/> </div><br>
+
 ## SIGCHLD
 
-当一个子进程改变了它的状态时：停止运行，继续运行或者退出，有两件事会发生在父进程中：
+当一个子进程改变了它的状态时（停止运行，继续运行或者退出），有两件事会发生在父进程中：
 
 - 得到 SIGCHLD 信号；
 - waitpid() 或者 wait() 调用会返回。
@@ -1194,7 +1180,7 @@ dmtsai lines: 5 columns: 9
 
 其中子进程发送的 SIGCHLD 信号包含了子进程的信息，包含了进程 ID、进程状态、进程使用 CPU 的时间等。
 
-在子进程退出时，它的进程描述符不会立即释放，这是为了让父进程得到子进程信息。父进程通过 wait() 和 waitpid() 来获得一个已经退出的子进程的信息。
+在子进程退出时，它的进程描述符不会立即释放，这是为了让父进程得到子进程信息，父进程通过 wait() 和 waitpid() 来获得一个已经退出的子进程的信息。
 
 ## wait()
 
@@ -1206,11 +1192,7 @@ pid_t wait(int *status)
 
 如果成功，返回被收集的子进程的进程 ID；如果调用进程没有子进程，调用就会失败，此时返回 -1，同时 errno 被置为 ECHILD。
 
-参数 status 用来保存被收集的子进程退出时的一些状态，如果对这个子进程是如何死掉的毫不在意，只想把这个子进程消灭掉，可以设置这个参数为 NULL：
-
-```c
-pid = wait(NULL);
-```
+参数 status 用来保存被收集的子进程退出时的一些状态，如果对这个子进程是如何死掉的毫不在意，只想把这个子进程消灭掉，可以设置这个参数为 NULL。
 
 ## waitpid()
 
@@ -1238,9 +1220,9 @@ options 参数主要有 WNOHANG 和 WUNTRACED 两个选项，WNOHANG 可以使 w
 
 僵尸进程通过 ps 命令显示出来的状态为 Z（zombie）。
 
-系统所能使用的进程号是有限的，如果大量的产生僵尸进程，将因为没有可用的进程号而导致系统不能产生新的进程。
+系统所能使用的进程号是有限的，如果产生大量僵尸进程，将因为没有可用的进程号而导致系统不能产生新的进程。
 
-要消灭系统中大量的僵尸进程，只需要将其父进程杀死，此时僵尸进程就会变成孤儿进程，从而被 init 所收养，这样 init 就会释放所有的僵死进程所占有的资源，从而结束僵尸进程。
+要消灭系统中大量的僵尸进程，只需要将其父进程杀死，此时僵尸进程就会变成孤儿进程，从而被 init 所收养，这样 init 就会释放所有的僵尸进程所占有的资源，从而结束僵尸进程。
 
 # 参考资料
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index f6aa853b..1c575fae 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -9,7 +9,7 @@
     * [字符串](#字符串)
     * [时间和日期](#时间和日期)
 * [三、索引](#三索引)
-    * [B Tree 原理](#b-tree-原理)
+    * [B+ Tree 原理](#b-tree-原理)
     * [索引分类](#索引分类)
     * [索引的优点](#索引的优点)
     * [索引优化](#索引优化)
@@ -125,49 +125,33 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 索引是在存储引擎层实现的，而不是在服务器层实现的，所以不同存储引擎具有不同的索引类型和实现。
 
-## B Tree 原理
+## B+ Tree 原理
 
-### 1. B-Tree
+### 1. 数据结构
 
-<div align="center"> <img src="../pics//06976908-98ab-46e9-a632-f0c2760ec46c.png"/> </div><br>
+B Tree 指的是 Balance Tree，也就是平衡树。平衡树时一颗查找树，并且所有叶子节点位于同一层。
 
-定义一条数据记录为一个二元组 [key, data]，B-Tree 是满足下列条件的数据结构：
+B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具有 B Tree 的平衡性，并且通过顺序访问指针来提高区间查询的性能。
 
-- 所有叶节点具有相同的深度，也就是说 B-Tree 是平衡的；
-- 一个节点中的 key 从左到右非递减排列；
-- 如果某个指针的左右相邻 key 分别是 key<sub>i</sub> 和 key<sub>i+1</sub>，且不为 null，则该指针指向节点的所有 key 大于等于 key<sub>i</sub> 且小于等于 key<sub>i+1</sub>。
-
-查找算法：首先在根节点进行二分查找，如果找到则返回对应节点的 data，否则在相应区间的指针指向的节点递归进行查找。
-
-由于插入删除新的数据记录会破坏 B-Tree 的性质，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作以保持 B-Tree 性质。
-
-### 2. B+Tree
-
-<div align="center"> <img src="../pics//7299afd2-9114-44e6-9d5e-4025d0b2a541.png"/> </div><br>
-
-与 B-Tree 相比，B+Tree 有以下不同点：
-
-- 每个节点的指针上限为 2d 而不是 2d+1（d 为节点的出度）；
-- 内节点不存储 data，只存储 key；
-- 叶子节点不存储指针。
-
-### 3. 顺序访问指针
+在 B+ Tree 中，一个节点中的 key 从左到右非递减排列，如果某个指针的左右相邻 key 分别是 key<sub>i</sub> 和 key<sub>i+1</sub>，且不为 null，则该指针指向节点的所有 key 大于等于 key<sub>i</sub> 且小于等于 key<sub>i+1</sub>。
 
 <div align="center"> <img src="../pics//061c88c1-572f-424f-b580-9cbce903a3fe.png"/> </div><br>
 
-一般在数据库系统或文件系统中使用的 B+Tree 结构都在经典 B+Tree 基础上进行了优化，在叶子节点增加了顺序访问指针，做这个优化的目的是为了提高区间访问的性能。
+### 2. 操作
 
-### 4. 优势
+操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
-红黑树等平衡树也可以用来实现索引，但是文件系统及数据库系统普遍采用 B Tree 作为索引结构，主要有以下两个原因：
+插入删除操作记录会破坏平衡树的平衡性，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作。
+
+### 3. 与红黑树的比较
+
+红黑树等平衡树也可以用来实现索引，但是文件系统及数据库系统普遍采用 B+ Tree 作为索引结构，主要有以下两个原因：
 
 （一）更少的检索次数
 
 平衡树检索数据的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
 
-红黑树的出度为 2，而 B Tree 的出度一般都非常大。红黑树的树高 h 很明显比 B Tree 大非常多，因此检索的次数也就更多。
-
-B+Tree 相比于 B-Tree 更适合外存索引，因为 B+Tree 内节点去掉了 data 域，因此可以拥有更大的出度，检索效率会更高。
+红黑树的出度为 2，而 B+ Tree 的出度一般都非常大。红黑树的树高 h 很明显比 B+ Tree 大非常多，因此检索的次数也就更多。
 
 （二）利用计算机预读特性
 
@@ -175,8 +159,6 @@ B+Tree 相比于 B-Tree 更适合外存索引，因为 B+Tree 内节点去掉了
 
 操作系统一般将内存和磁盘分割成固态大小的块，每一块称为一页，内存与磁盘以页为单位交换数据。数据库系统将索引的一个节点的大小设置为页的大小，使得一次 I/O 就能完全载入一个节点，并且可以利用预读特性，相邻的节点也能够被预先载入。
 
-更多内容请参考：[MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
-
 ## 索引分类
 
 ### 1. B+Tree 索引
@@ -442,3 +424,4 @@ MySQL 读写分离能提高性能的原因在于：
 - [服务端指南 数据存储篇 | MySQL（09） 分库与分表带来的分布式困境与应对之策](http://blog.720ui.com/2017/mysql_core_09_multi_db_table2/ "服务端指南 数据存储篇 | MySQL（09） 分库与分表带来的分布式困境与应对之策")
 - [How to create unique row ID in sharded databases?](https://stackoverflow.com/questions/788829/how-to-create-unique-row-id-in-sharded-databases)
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
+- [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
diff --git a/notes/Redis.md b/notes/Redis.md
index 247d0987..c54287c8 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -420,6 +420,8 @@ Reids 具体有 6 种淘汰策略：
 
 使用 Redis 缓存数据时，为了提高缓存命中率，需要保证缓存数据都是热点数据。可以将内存最大使用量设置为热点数据占用的内存量，然后启用 allkeys-lru 淘汰策略，将最近最少使用的数据淘汰。
 
+Redis 4.0 引入了 volatile-lfu 和 allkeys-lfu 淘汰策略，LFU 策略通过统计访问频率，将访问频率最少的键值对淘汰。
+
 # 八、持久化
 
 Redis 是内存型数据库，为了保证数据在断电后不会丢失，需要将内存中的数据持久化到硬盘上。
@@ -601,3 +603,4 @@ Redis 没有关系型数据库中的表这一概念来将同种类型的数据
 - [Redis 3.0 中文版- 分片](http://wiki.jikexueyuan.com/project/redis-guide)
 - [Redis 应用场景](http://www.scienjus.com/redis-use-case/)
 - [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
+- [Using Redis as an LRU cache](https://redis.io/topics/lru-cache)
diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 5c9f63a7..6bfb559c 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -23,6 +23,7 @@
     * [段页式](#段页式)
     * [分页与分段的比较](#分页与分段的比较)
 * [五、设备管理](#五设备管理)
+    * [磁盘结构](#磁盘结构)
     * [磁盘调度算法](#磁盘调度算法)
 * [六、链接](#六链接)
     * [编译系统](#编译系统)
@@ -304,7 +305,7 @@ void P2() {
 
 为了同步生产者和消费者的行为，需要记录缓冲区中物品的数量。数量可以使用信号量来进行统计，这里需要使用两个信号量：empty 记录空缓冲区的数量，full 记录满缓冲区的数量。其中，empty 信号量是在生产者进程中使用，当 empty 不为 0 时，生产者才可以放入物品；full 信号量是在消费者进程中使用，当 full 信号量不为 0 时，消费者才可以取走物品。
 
-注意，不能先对缓冲区进行加锁，再测试信号量。也就是说，不能先执行 down(mutex) 再执行 down(empty)。如果这么做了，那么可能会出现这种情况：生产者对缓冲区加锁后，执行 down(empty) 操作，发现 empty = 0，此时生产者睡眠。消费者不能进入临界区，因为生产者对缓冲区加锁了，也就无法执行 up(empty) 操作，empty 永远都为 0，那么生产者和消费者就会一直等待下去，造成死锁。
+注意，不能先对缓冲区进行加锁，再测试信号量。也就是说，不能先执行 down(mutex) 再执行 down(empty)。如果这么做了，那么可能会出现这种情况：生产者对缓冲区加锁后，执行 down(empty) 操作，发现 empty = 0，此时生产者睡眠。消费者不能进入临界区，因为生产者对缓冲区加锁了，消费者就无法执行 up(empty) 操作，empty 永远都为 0，导致生产者永远等待下，不会释放锁，消费者因此也会永远等待下去。
 
 ```c
 #define N 100
@@ -314,7 +315,7 @@ semaphore empty = N;
 semaphore full = 0;
 
 void producer() {
-    while(TRUE){
+    while(TRUE) {
         int item = produce_item();
         down(&empty);
         down(&mutex);
@@ -325,7 +326,7 @@ void producer() {
 }
 
 void consumer() {
-    while(TRUE){
+    while(TRUE) {
         down(&full);
         down(&mutex);
         int item = remove_item();
@@ -542,7 +543,7 @@ int pipe(int fd[2]);
 
 它具有以下限制：
 
-- 只支持半双工通信（单向传输）；
+- 只支持半双工通信（单向交替传输）；
 - 只能在父子进程中使用。
 
 <div align="center"> <img src="../pics//53cd9ade-b0a6-4399-b4de-7f1fbd06cdfb.png"/> </div><br>
@@ -707,7 +708,7 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 虚拟内存的目的是为了让物理内存扩充成更大的逻辑内存，从而让程序获得更多的可用内存。
 
-为了更好的管理内存，操作系统将内存抽象成地址空间。每个程序拥有自己的地址空间，这个地址空间被分割成多个块，每一块称为一页。这些页被映射到物理内存，但不需要映射到连续的物理内存，也不需要所有页都必须在物理内存中。当程序引用到一部分不在物理内存中的地址空间时，由硬件执行必要的映射，将缺失的部分装入物理内存并重新执行失败的指令。
+为了更好的管理内存，操作系统将内存抽象成地址空间。每个程序拥有自己的地址空间，这个地址空间被分割成多个块，每一块称为一页。这些页被映射到物理内存，但不需要映射到连续的物理内存，也不需要所有页都必须在物理内存中。当程序引用到不在物理内存中的页时，由硬件执行必要的映射，将缺失的部分装入物理内存并重新执行失败的指令。
 
 从上面的描述中可以看出，虚拟内存允许程序不用将地址空间中的每一页都映射到物理内存，也就是说一个程序不需要全部调入内存就可以运行，这使得有限的内存运行大程序称为可能。例如有一台计算机可以产生 16 位地址，那么一个程序的地址空间范围是 0\~64K。该计算机只有 32KB 的物理内存，虚拟内存技术允许该计算机运行一个 64K 大小的程序。
 
@@ -715,12 +716,11 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 ## 分页系统地址映射
 
-- 内存管理单元（MMU）：管理着地址空间和物理内存的转换。
-- 页表（Page table）：页（地址空间）和页框（物理内存空间）的映射表。例如下图中，页表的第 0 个表项为 010，表示第 0 个页映射到第 2 个页框。页表项的最后一位用来标记页是否在内存中。
+内存管理单元（MMU）管理着地址空间和物理内存的转换，其中的页表（Page table）存储着页（程序地址空间）和页框（物理内存空间）的映射表。
 
-下图的页表存放着 16 个页，这 16 个页需要用 4 个比特位来进行索引定位。因此对于虚拟地址（0010 000000000100），前 4 位是用来存储页面号，而后 12 位存储在页中的偏移量。
+下图的页表存放着 16 个页，这 16 个页需要用 4 个比特位来进行索引定位，也就是存储页面号，剩下 12 个比特位存储偏移量。
 
-（0010 000000000100）根据前 4 位得到页号为 2，读取表项内容为（110 1），它的前 3 为为页框号，最后 1 位表示该页在内存中。最后映射得到物理内存地址为（110 000000000100）。
+例如对于虚拟地址（0010 000000000100），前 4 位是存储页面号 2，读取表项内容为（110 1）。该页在内存中，并且页框的地址为 （110 000000000100）。
 
 <div align="center"> <img src="../pics//cf4386a1-58c9-4eca-a17f-e12b1e9770eb.png" width="500"/> </div><br>
 
@@ -828,11 +828,22 @@ FIFO 算法可能会把经常使用的页面置换出去，为了避免这一问
 
 # 五、设备管理
 
+## 磁盘结构
+
+- 盘面（Platter）：一个磁盘有多个盘面；
+- 磁道（Track）：盘面上的圆形带状区域，一个盘面可以有多个磁道；
+- 扇区（Track Sector）：磁道上的一个弧段，一个磁道可以有多个扇区，它是最小的物理储存单位，目前主要有 512 bytes 与 4 K 两种大小；
+- 磁头（Head）：与盘面非常接近，能够将盘面上的磁场转换为电信号（读），或者将电信号转换为盘面的磁场（写）；
+- 制动手臂（Actuator arm）：用于在磁道之间移动磁头；
+- 主轴（Spindle）：使整个盘面转动。
+
+<div align="center"> <img src="../pics//014fbc4d-d873-4a12-b160-867ddaed9807.jpg"/> </div><br>
+
 ## 磁盘调度算法
 
 读写一个磁盘块的时间的影响因素有：
 
-- 旋转时间（主轴旋转磁盘，使得磁头移动到适当的扇区上）
+- 旋转时间（主轴转动盘面，使得磁头移动到适当的扇区上）
 - 寻道时间（制动手臂移动，使得磁头移动到适当的磁道上）
 - 实际的数据传输时间
 
@@ -933,8 +944,9 @@ gcc -o hello hello.c
 - Tanenbaum A S, Bos H. Modern operating systems[M]. Prentice Hall Press, 2014.
 - 汤子瀛, 哲凤屏, 汤小丹. 计算机操作系统[M]. 西安电子科技大学出版社, 2001.
 - Bryant, R. E., & O’Hallaron, D. R. (2004). 深入理解计算机系统.
+- 史蒂文斯. UNIX 环境高级编程 [M]. 人民邮电出版社, 2014.
 - [Operating System Notes](https://applied-programming.github.io/Operating-Systems-Notes/)
-- [进程间的几种通信方式](http://blog.csdn.net/yufaw/article/details/7409596)
 - [Operating-System Structures](https://www.cs.uic.edu/\~jbell/CourseNotes/OperatingSystems/2_Structures.html)
 - [Processes](http://cse.csusb.edu/tongyu/courses/cs460/notes/process.php)
 - [Inter Process Communication Presentation[1]](https://www.slideshare.net/rkolahalam/inter-process-communication-presentation1)
+- [Decoding UCS Invicta – Part 1](https://blogs.cisco.com/datacenter/decoding-ucs-invicta-part-1)

From df8cd08303067a37e7b6a5326bad88cfdc7abbf5 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Fri, 3 Aug 2018 00:17:10 +0800
Subject: [PATCH 036/172] auto commit

---
 other/README.md | 3 +++
 1 file changed, 3 insertions(+)
 create mode 100644 other/README.md

diff --git a/other/README.md b/other/README.md
new file mode 100644
index 00000000..39876f02
--- /dev/null
+++ b/other/README.md
@@ -0,0 +1,3 @@
+- 其他人添加的全新内容
+- 主页 README 引用的图片
+- 微信群描述文件
\ No newline at end of file

From c4cd8a30ffde6049ee703943e5fc9a53a93c54ff Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 3 Aug 2018 00:20:20 +0800
Subject: [PATCH 037/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index c015a100..39cedf7b 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-wechat_group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->

From ae14046536061e69d46858fb168ac6bb53cbb716 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 3 Aug 2018 00:20:56 +0800
Subject: [PATCH 038/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 39cedf7b..56627328 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-wechat_group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-wechat-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->

From 54073390ecefa3457a027026a7aefed7c19ef2a6 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 3 Aug 2018 00:22:26 +0800
Subject: [PATCH 039/172] Update Group.md

---
 other/Group.md | 10 +++++++---
 1 file changed, 7 insertions(+), 3 deletions(-)

diff --git a/other/Group.md b/other/Group.md
index 5e7bbc90..fee7b8bf 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -1,10 +1,14 @@
-创建交流群的主要目的是为了为大家提供一个交流平台，方便大家在学习的过程中互相讨论。
+创建交流群的主要目的是为了给大家提供一个交流平台，方便大家在学习的过程中互相讨论。
 
-这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以提升自己对知识的掌握程度。
+这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以帮助大家对知识的掌握程度。
 
 因为笔者白天要上班，因此不能及时进行回复，大部分时间会处于潜水状态。
 
-至于交流群和 Issue 有什么区别，主要是两方面：一是交流群实时性高一些，二是交流群会更活跃一些。另外，Issue 主要是用来发布一些项目中的错误和一些改进建议，当然也可以发布一些可以讨论的问题，交流群可以讨论的内容比较广，例如在阅读本项目过程中不理解的地方可以在交流群中寻求别人的帮助、新技术的讨论、招聘信息、学习和工作的感受等等。
+至于交流群和 Issue 有什么区别，主要是两方面：一是交流群实时性高一些，二是交流群会更活跃一些。
+
+另外，Issue 主要是用来发布一些项目中的错误和一些改进建议，当然也可以发布一些可以讨论的问题。
+
+交流群可以讨论的内容比较广，例如在阅读本项目过程中不理解的地方可以在交流群中寻求别人的帮助、新技术的讨论、招聘信息、学习和工作的感受等等。
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 

From 781259816ddebe5a41bf08ba169156b81e99c1e7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 3 Aug 2018 13:01:18 +0800
Subject: [PATCH 040/172] Update README.md

---
 README.md | 42 +++++++++++++++++++++---------------------
 1 file changed, 21 insertions(+), 21 deletions(-)

diff --git a/README.md b/README.md
index 56627328..87d4d460 100644
--- a/README.md
+++ b/README.md
@@ -15,59 +15,59 @@
 
 > [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
-目录根据原书第二版进行编排。
+目录根据原书第二版进行编排，代码和原书有所不同，尽量实现地比原书更加简洁。
 
 > [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
 
-做了一个大致分类，并对每种分类题型的解题思路做了总结。
+对题目做了一个大致分类，并对每种题型的解题思路做了总结。
 
 > [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
 
-主要参考 Robert Sedgewick 的算法书进行实现，源代码以及测试代码可在另一个仓库获取。
+排序、并查集、栈和队列、红黑树、散列表。
 
 ## 操作系统 :computer:
 
 > [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
-参考 现代操作系统、Unix 环境高级编程、深入理解计算机系统。
+进程管理、内存管理、设备管理、链接。
 
 > [Linux](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Linux.md)
 
-参考 鸟哥的 Linux 私房菜。
+基本实现原理以及基本操作。
 
 ## 网络 :cloud:
 
 > [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
-参考 谢希仁的计算机网络、计算机网络 自顶向下方法、TCP/IP 详解。
+物理层、链路层、网络层、运输层、应用层。
 
 > [HTTP](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/HTTP.md)
 
-参考 图解 HTTP，更多的是参考网上的文档，比如 MDN、维基百科等。
+方法、状态码、Cookie、缓存、连接管理、HTTPs、HTTP 2.0。
 
 > [Socket](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Socket.md)
 
-参考 Unix 网络编程。
+I/O 模型、I/O 多路复用。
 
 ## 面向对象 :couple:
 
 > [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
-参考 Head First 设计模式、设计模式 可复用面向对象软件的基础，实现了 Gof 的 23 种设计模式。
+实现了 Gof 的 23 种设计模式。
 
 > [面向对象思想](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/面向对象思想.md)
 
-内容包括三大原则（继承、封装、多态）、类图、设计原则。
+三大原则（继承、封装、多态）、类图、设计原则。
 
 ## 数据库 :floppy_disk:
 
 > [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
-参考 数据库系统原理。
+事务、锁、隔离级别、MVCC、间隙锁、范式。
 
 > [SQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/SQL.md)
 
-参考 SQL 必知必会。
+SQL 基本语法。
 
 > [Leetcode-Database 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode-Database%20题解.md)
 
@@ -75,33 +75,33 @@ Leetcode 上数据库题目的解题记录。
 
 > [MySQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/MySQL.md)
 
-参考 高性能 MySQL。
+存储引擎、索引、查询优化、切分、复制。
 
 > [Redis](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Redis.md)
 
-参考 Redis 设计与实现、Redis 实战。
+五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
 ## Java :coffee:
 
 > [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
-参考 Effective Java、Java 编程思想，也有部分内容参考官方文档以及 StackOverflow。
+不会涉及很多基本语法介绍，主要是一些实现原理以及关键特性。
 
 > [Java 容器](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20容器.md)
 
-包含容器源码分析。
+源码分析：ArrayListVector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
 
 > [Java 并发](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20并发.md)
 
-参考 Java 编程思想、深入理解 Java 虚拟机。
+线程使用方式、两种互斥同步方法、线程协作、JUC、线程安全、内存模型、锁优化。
 
 > [Java 虚拟机](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20虚拟机.md)
 
-参考 深入理解 Java 虚拟机。
+运行时数据区域、垃圾收集、类加载。
 
 > [Java I/O](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20IO.md)
 
-包含 NIO 的原理以及实例。
+NIO 的原理以及实例。
 
 ## 系统设计 :bulb:
 
@@ -141,11 +141,11 @@ Docker 基本原理。
 
 > [正则表达式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/正则表达式.md)
 
-参考 正则表达式必知必会。
+正则表达式基本语法。
 
 > [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
 
-构建工具的基本概念、主流构建工具介绍
+构建工具的基本概念、主流构建工具介绍。
 
 ## 编码实践 :speak_no_evil:
 

From 9cd8b3ae9631d73b5d94d4022b468e8a60682c04 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Fri, 3 Aug 2018 14:40:33 +0800
Subject: [PATCH 041/172] auto commit

---
 BOOKLIST.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/BOOKLIST.md b/BOOKLIST.md
index bdd8eb92..4805a7dc 100644
--- a/BOOKLIST.md
+++ b/BOOKLIST.md
@@ -79,4 +79,4 @@
 
 - [JavaScript 语言精粹](https://book.douban.com/subject/3590768/)
 - [利用 Python 进行数据分析](https://book.douban.com/subject/25779298/)
-- [概率论与数理统计](https://book.douban.com/subject/2201479/)
\ No newline at end of file
+- [概率论与数理统计](https://book.douban.com/subject/2201479/)

From 2f4f1c814411214475372d73aac7fdf836fe55cd Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Fri, 3 Aug 2018 14:44:06 +0800
Subject: [PATCH 044/172] auto commit

---
 .gitignore | 1 +
 1 file changed, 1 insertion(+)

diff --git a/.gitignore b/.gitignore
index e43b0f98..a56b7514 100644
--- a/.gitignore
+++ b/.gitignore
@@ -1 +1,2 @@
 .DS_Store
+ 

From 4c69f7a7cc83c552655e27f72a7d91cd42ef1658 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Fri, 3 Aug 2018 15:31:36 +0800
Subject: [PATCH 045/172] auto commit

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 87d4d460..39b43f30 100644
--- a/README.md
+++ b/README.md
@@ -15,7 +15,7 @@
 
 > [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
-目录根据原书第二版进行编排，代码和原书有所不同，尽量实现地比原书更加简洁。
+目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
 
 > [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
 

From 0e7655e7aeea75fe027849315d45691280067e82 Mon Sep 17 00:00:00 2001
From: Elong <peielcn@gmail.com>
Date: Fri, 3 Aug 2018 18:42:53 +0800
Subject: [PATCH 046/172] =?UTF-8?q?Update=20Java=20=E5=9F=BA=E7=A1=80.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

equals()方法的文字描述修改
---
 notes/Java 基础.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 0bce5622..1e20ea55 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -586,7 +586,7 @@ protected void finalize() throws Throwable {}
 **1. equals() 与 == 的区别** 
 
 - 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
-- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价。
+- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 则是根据引用的对象的 equals() 方法的具体实现来进行比较。
 
 ```java
 Integer x = new Integer(1);

From bb6d5e49790664dd8b756cc11660fd4878c2b966 Mon Sep 17 00:00:00 2001
From: luocaodan <luocaodan@buaa.edu.cn>
Date: Sat, 4 Aug 2018 11:23:03 +0800
Subject: [PATCH 047/172] =?UTF-8?q?Update=20Java=20=E8=99=9A=E6=8B=9F?=
 =?UTF-8?q?=E6=9C=BA.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

应该是jre目录下，而不是jdk目录
---
 notes/Java 虚拟机.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 3bb9f289..dbe4e070 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -627,9 +627,9 @@ public static void main(String[] args) {
 
 从 Java 开发人员的角度看，类加载器可以划分得更细致一些：
 
-- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JAVA_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
+- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JRE_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
 
-- 扩展类加载器（Extension ClassLoader）这个类加载器是由 ExtClassLoader（sun.misc.Launcher$ExtClassLoader）实现的。它负责将 &lt;JAVA_HOME>/lib/ext 或者被 java.ext.dir 系统变量所指定路径中的所有类库加载到内存中，开发者可以直接使用扩展类加载器。
+- 扩展类加载器（Extension ClassLoader）这个类加载器是由 ExtClassLoader（sun.misc.Launcher$ExtClassLoader）实现的。它负责将 &lt;JRE_HOME>/lib/ext 或者被 java.ext.dir 系统变量所指定路径中的所有类库加载到内存中，开发者可以直接使用扩展类加载器。
 
 - 应用程序类加载器（Application ClassLoader）这个类加载器是由 AppClassLoader（sun.misc.Launcher$AppClassLoader）实现的。由于这个类加载器是 ClassLoader 中的 getSystemClassLoader() 方法的返回值，因此一般称为系统类加载器。它负责加载用户类路径（ClassPath）上所指定的类库，开发者可以直接使用这个类加载器，如果应用程序中没有自定义过自己的类加载器，一般情况下这个就是程序中默认的类加载器。
 

From b9f9e9807e771074fe13dd04b048782ac6a4444f Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 4 Aug 2018 16:59:09 +0800
Subject: [PATCH 048/172] auto commit

---
 .gitignore           | 2 +-
 notes/Java 基础.md   | 2 +-
 notes/Java 并发.md   | 4 +++-
 notes/Java 虚拟机.md | 2 +-
 4 files changed, 6 insertions(+), 4 deletions(-)

diff --git a/.gitignore b/.gitignore
index a56b7514..c91a56b6 100644
--- a/.gitignore
+++ b/.gitignore
@@ -1,2 +1,2 @@
 .DS_Store
- 
+*.txt
diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 1e20ea55..f5133a6f 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -586,7 +586,7 @@ protected void finalize() throws Throwable {}
 **1. equals() 与 == 的区别** 
 
 - 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
-- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 则是根据引用的对象的 equals() 方法的具体实现来进行比较。
+- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用的对象的 equals() 方法的具体实现来进行比较。
 
 ```java
 Integer x = new Integer(1);
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 627b824f..89ab7cac 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1419,7 +1419,9 @@ public final int incrementAndGet() {
 }
 ```
 
-以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果 该字段内存地址中的值 ==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
+以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果该字段内存地址中的值 ==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。
+
+可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
 
 ```java
 public final int getAndAddInt(Object var1, long var2, int var4) {
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index dbe4e070..ec33515e 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -627,7 +627,7 @@ public static void main(String[] args) {
 
 从 Java 开发人员的角度看，类加载器可以划分得更细致一些：
 
-- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JRE_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
+- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JAVA_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
 
 - 扩展类加载器（Extension ClassLoader）这个类加载器是由 ExtClassLoader（sun.misc.Launcher$ExtClassLoader）实现的。它负责将 &lt;JRE_HOME>/lib/ext 或者被 java.ext.dir 系统变量所指定路径中的所有类库加载到内存中，开发者可以直接使用扩展类加载器。
 

From 336fa4f50f8b14518edc95f8a79bb1d59c40521e Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 4 Aug 2018 17:00:36 +0800
Subject: [PATCH 049/172] auto commit

---
 notes/Java 虚拟机.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index ec33515e..69952f2e 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -627,9 +627,9 @@ public static void main(String[] args) {
 
 从 Java 开发人员的角度看，类加载器可以划分得更细致一些：
 
-- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JAVA_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
+- 启动类加载器（Bootstrap ClassLoader）此类加载器负责将存放在 &lt;JRE_HOME>\lib 目录中的，或者被 -Xbootclasspath 参数所指定的路径中的，并且是虚拟机识别的（仅按照文件名识别，如 rt.jar，名字不符合的类库即使放在 lib 目录中也不会被加载）类库加载到虚拟机内存中。启动类加载器无法被 Java 程序直接引用，用户在编写自定义类加载器时，如果需要把加载请求委派给启动类加载器，直接使用 null 代替即可。
 
-- 扩展类加载器（Extension ClassLoader）这个类加载器是由 ExtClassLoader（sun.misc.Launcher$ExtClassLoader）实现的。它负责将 &lt;JRE_HOME>/lib/ext 或者被 java.ext.dir 系统变量所指定路径中的所有类库加载到内存中，开发者可以直接使用扩展类加载器。
+- 扩展类加载器（Extension ClassLoader）这个类加载器是由 ExtClassLoader（sun.misc.Launcher$ExtClassLoader）实现的。它负责将 &lt;JAVA_HOME>/lib/ext 或者被 java.ext.dir 系统变量所指定路径中的所有类库加载到内存中，开发者可以直接使用扩展类加载器。
 
 - 应用程序类加载器（Application ClassLoader）这个类加载器是由 AppClassLoader（sun.misc.Launcher$AppClassLoader）实现的。由于这个类加载器是 ClassLoader 中的 getSystemClassLoader() 方法的返回值，因此一般称为系统类加载器。它负责加载用户类路径（ClassPath）上所指定的类库，开发者可以直接使用这个类加载器，如果应用程序中没有自定义过自己的类加载器，一般情况下这个就是程序中默认的类加载器。
 

From fc97755cac62e8b12c3f0b0c45b06418947fa4b7 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 4 Aug 2018 18:32:17 +0800
Subject: [PATCH 050/172] auto commit

---
 notes/Java 虚拟机.md                          |  38 +++++++++---------
 pics/926c7438-c5e1-4b94-840a-dcb24ff1dafe.png | Bin 0 -> 19456 bytes
 pics/c9ad2bf4-5580-4018-bce4-1b9a71804d9c.png | Bin 0 -> 30720 bytes
 3 files changed, 19 insertions(+), 19 deletions(-)
 create mode 100644 pics/926c7438-c5e1-4b94-840a-dcb24ff1dafe.png
 create mode 100644 pics/c9ad2bf4-5580-4018-bce4-1b9a71804d9c.png

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 69952f2e..4382317b 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -1,7 +1,7 @@
 <!-- GFM-TOC -->
 * [一、运行时数据区域](#一运行时数据区域)
     * [程序计数器](#程序计数器)
-    * [虚拟机栈](#虚拟机栈)
+    * [Java 虚拟机栈](#java-虚拟机栈)
     * [本地方法栈](#本地方法栈)
     * [堆](#堆)
     * [方法区](#方法区)
@@ -23,19 +23,19 @@
 
 # 一、运行时数据区域
 
-<div align="center"> <img src="../pics//540631a4-6018-40a5-aed7-081e2eeeaeea.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//c9ad2bf4-5580-4018-bce4-1b9a71804d9c.png" width="500"/> </div><br>
 
 ## 程序计数器
 
 记录正在执行的虚拟机字节码指令的地址（如果正在执行的是本地方法则为空）。
 
-## 虚拟机栈
+## Java 虚拟机栈
 
-每个 Java 方法在执行的同时会创建一个栈帧用于存储局部变量表、操作数栈、常量池引用等信息。每一个方法从调用直至执行完成的过程，就对应着一个栈帧在 Java 虚拟机栈中入栈和出栈的过程。
+每个 Java 方法在执行的同时会创建一个栈帧用于存储局部变量表、操作数栈、常量池引用等信息，从调用直至执行完成的过程，就对应着一个栈帧在 Java 虚拟机栈中入栈和出栈的过程。
 
-<div align="center"> <img src="../pics//f5757d09-88e7-4bbd-8cfb-cecf55604854.png" width=""/> </div><br>
+<div align="center"> <img src="../pics//926c7438-c5e1-4b94-840a-dcb24ff1dafe.png" width=""/> </div><br>
 
-可以通过 -Xss 这个虚拟机参数来指定一个程序的 Java 虚拟机栈内存大小：
+可以通过 -Xss 这个虚拟机参数来指定每个线程的 Java 虚拟机栈内存大小：
 
 ```java
 java -Xss=512M HackTheJava
@@ -56,27 +56,22 @@ java -Xss=512M HackTheJava
 
 ## 堆
 
-所有对象实例都在这里分配内存。
+所有对象实例都在这里分配内存，是垃圾收集的主要区域（"GC 堆"）。
 
-是垃圾收集的主要区域（"GC 堆"）。现代的垃圾收集器基本都是采用分代收集算法，主要思想是针对不同的对象采取不同的垃圾回收算法。虚拟机把 Java 堆分成以下三块：
+现代的垃圾收集器基本都是采用分代收集算法，针对不同的对象采取不同的垃圾回收算法，可以将堆分成两块：
 
 - 新生代（Young Generation）
 - 老年代（Old Generation）
-- 永久代（Permanent Generation）
 
-当一个对象被创建时，它首先进入新生代，之后有可能被转移到老年代中。
-
-新生代存放着大量的生命很短的对象，因此新生代在三个区域中垃圾回收的频率最高。为了更高效地进行垃圾回收，把新生代继续划分成以下三个空间：
+新生代可以继续划分成以下三个空间：
 
 - Eden（伊甸园）
 - From Survivor（幸存者）
 - To Survivor
 
-<div align="center"> <img src="../pics//ppt_img.gif" width=""/> </div><br>
+堆不需要连续内存，并且可以动态增加其内存，增加失败会抛出 OutOfMemoryError 异常。
 
-Java 堆不需要连续内存，并且可以动态增加其内存，增加失败会抛出 OutOfMemoryError 异常。
-
-可以通过 -Xms 和 -Xmx 两个虚拟机参数来指定一个程序的 Java 堆内存大小，第一个参数设置初始值，第二个参数设置最大值。
+可以通过 -Xms 和 -Xmx 两个虚拟机参数来指定一个程序的堆内存大小，第一个参数设置初始值，第二个参数设置最大值。
 
 ```java
 java -Xms=1M -Xmx=2M HackTheJava
@@ -86,11 +81,13 @@ java -Xms=1M -Xmx=2M HackTheJava
 
 用于存放已被加载的类信息、常量、静态变量、即时编译器编译后的代码等数据。
 
-和 Java 堆一样不需要连续的内存，并且可以动态扩展，动态扩展失败一样会抛出 OutOfMemoryError 异常。
+和堆一样不需要连续的内存，并且可以动态扩展。
+
+动态扩展失败一样会抛出 OutOfMemoryError 异常。
 
 对这块区域进行垃圾回收的主要目标是对常量池的回收和对类的卸载，但是一般比较难实现。
 
-JDK 1.7 之前，HotSpot 虚拟机把它当成永久代来进行垃圾回收，JDK 1.8 之后，取消了永久代，用 metaspace（元数据）区替代。
+JDK 1.7 之前，HotSpot 虚拟机把它当成永久代来进行垃圾回收。但是从 JDK 1.7 开始，已经把原本放在永久代的字符串常量池移到 Native Method 中。
 
 ## 运行时常量池
 
@@ -102,7 +99,9 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ## 直接内存
 
-在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
+在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。
+
+这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
 
 # 二、垃圾收集
 
@@ -751,6 +750,7 @@ public class FileSystemClassLoader extends ClassLoader {
 # 参考资料
 
 - 周志明. 深入理解 Java 虚拟机 [M]. 机械工业出版社, 2011.
+- [Chapter 2. The Structure of the Java Virtual Machine](https://docs.oracle.com/javase/specs/jvms/se8/html/jvms-2.html#jvms-2.5.4)
 - [Jvm memory](https://www.slideshare.net/benewu/jvm-memory)
 - [Memory Architecture Of JVM(Runtime Data Areas)](https://hackthejava.wordpress.com/2015/01/09/memory-architecture-by-jvmruntime-data-areas/)
 - [JVM Run-Time Data Areas](https://www.programcreek.com/2013/04/jvm-run-time-data-areas/)
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From e610e0fb102c9b59c86a9d153f157866dd2db73a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 4 Aug 2018 19:21:15 +0800
Subject: [PATCH 051/172] auto commit

---
 notes/Java 虚拟机.md               |  69 +++++++++++++----------------
 pics/JNI-Java-Native-Interface.jpg | Bin 0 -> 39936 bytes
 2 files changed, 31 insertions(+), 38 deletions(-)
 create mode 100644 pics/JNI-Java-Native-Interface.jpg

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 4382317b..136ca5ad 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -23,7 +23,7 @@
 
 # 一、运行时数据区域
 
-<div align="center"> <img src="../pics//c9ad2bf4-5580-4018-bce4-1b9a71804d9c.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//c9ad2bf4-5580-4018-bce4-1b9a71804d9c.png" width="400"/> </div><br>
 
 ## 程序计数器
 
@@ -33,7 +33,7 @@
 
 每个 Java 方法在执行的同时会创建一个栈帧用于存储局部变量表、操作数栈、常量池引用等信息，从调用直至执行完成的过程，就对应着一个栈帧在 Java 虚拟机栈中入栈和出栈的过程。
 
-<div align="center"> <img src="../pics//926c7438-c5e1-4b94-840a-dcb24ff1dafe.png" width=""/> </div><br>
+<div align="center"> <img src="../pics//926c7438-c5e1-4b94-840a-dcb24ff1dafe.png" width="450"/> </div><br>
 
 可以通过 -Xss 这个虚拟机参数来指定每个线程的 Java 虚拟机栈内存大小：
 
@@ -52,7 +52,9 @@ java -Xss=512M HackTheJava
 
 与 Java 虚拟机栈类似，它们之间的区别只不过是本地方法栈为本地方法服务。
 
-<div align="center"> <img src="../pics//JNIFigure1.gif" width="350"/> </div><br>
+本地方法一般是用其它语言（C、C++ 或汇编语言等）编写的, 并且被编译为基于本机硬件和操作系统的程序。
+
+<div align="center"> <img src="../pics//JNI-Java-Native-Interface.jpg" width="350"/> </div><br>
 
 ## 堆
 
@@ -105,7 +107,9 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 # 二、垃圾收集
 
-程序计数器、虚拟机栈和本地方法栈这三个区域属于线程私有的，只存在于线程的生命周期内，线程结束之后也会消失，因此不需要对这三个区域进行垃圾回收。垃圾回收主要是针对 Java 堆和方法区进行。
+垃圾回收主要是针对堆和方法区进行。
+
+程序计数器、虚拟机栈和本地方法栈这三个区域属于线程私有的，只存在于线程的生命周期内，线程结束之后也会消失，因此不需要对这三个区域进行垃圾回收。
 
 ## 判断一个对象是否存活
 
@@ -115,6 +119,8 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 两个对象出现循环引用的情况下，此时引用计数器永远不为 0，导致无法对它们进行回收。
 
+正因为循环引用的存在，因此 Java 虚拟机不使用引用计数算法。
+
 ```java
 public class ReferenceCountingGC {
     public Object instance = null;
@@ -128,8 +134,6 @@ public class ReferenceCountingGC {
 }
 ```
 
-正因为循环引用的存在，因此 Java 虚拟机不使用引用计数算法。
-
 ### 2. 可达性分析算法
 
 通过 GC Roots 作为起始点进行搜索，能够到达到的对象都是存活的，不可达的对象可被回收。
@@ -151,7 +155,7 @@ Java 具有四种强度不同的引用类型。
 
 **（一）强引用** 
 
-被强引用关联的对象不会被垃圾收集器回收。
+被强引用关联的对象不会被回收。
 
 使用 new 一个新对象的方式来创建强引用。
 
@@ -161,7 +165,7 @@ Object obj = new Object();
 
 **（二）软引用** 
 
-被软引用关联的对象，只有在内存不够的情况下才会被回收。
+被软引用关联的对象只有在内存不够的情况下才会被回收。
 
 使用 SoftReference 类来创建软引用。
 
@@ -173,7 +177,7 @@ obj = null;  // 使对象只被软引用关联
 
 **（三）弱引用** 
 
-被弱引用关联的对象一定会被垃圾收集器回收，也就是说它只能存活到下一次垃圾收集。
+被弱引用关联的对象一定会被回收，也就是说它只能存活到下一次垃圾收集。
 
 使用 WeakReference 类来实现弱引用。
 
@@ -187,7 +191,7 @@ obj = null;
 
 又称为幽灵引用或者幻影引用。一个对象是否有虚引用的存在，完全不会对其生存时间构成影响，也无法通过虚引用取得一个对象实例。
 
-为一个对象设置虚引用关联的唯一目的就是能在这个对象被收集器回收时收到一个系统通知。
+为一个对象设置虚引用关联的唯一目的就是能在这个对象被回收时收到一个系统通知。
 
 使用 PhantomReference 来实现虚引用。
 
@@ -199,20 +203,20 @@ obj = null;
 
 ### 4. 方法区的回收
 
-因为方法区主要存放永久代对象，而永久代对象的回收率比新生代差很多，因此在方法区上进行回收性价比不高。
+因为方法区主要存放永久代对象，而永久代对象的回收率比新生代低很多，因此在方法区上进行回收性价比不高。
 
 主要是对常量池的回收和对类的卸载。
 
+在大量使用反射、动态代理、CGLib 等 ByteCode 框架、动态生成 JSP 以及 OSGi 这类频繁自定义 ClassLoader 的场景都需要虚拟机具备类卸载功能，以保证不会出现内存溢出。
+
 类的卸载条件很多，需要满足以下三个条件，并且满足了也不一定会被卸载：
 
-- 该类所有的实例都已经被回收，也就是 Java 堆中不存在该类的任何实例。
+- 该类所有的实例都已经被回收，也就是堆中不存在该类的任何实例。
 - 加载该类的 ClassLoader 已经被回收。
-- 该类对应的 java.lang.Class 对象没有在任何地方被引用，也就无法在任何地方通过反射访问该类方法。
+- 该类对应的 Class 对象没有在任何地方被引用，也就无法在任何地方通过反射访问该类方法。
 
 可以通过 -Xnoclassgc 参数来控制是否对类进行卸载。
 
-在大量使用反射、动态代理、CGLib 等 ByteCode 框架、动态生成 JSP 以及 OSGi 这类频繁自定义 ClassLoader 的场景都需要虚拟机具备类卸载功能，以保证不会出现内存溢出。
-
 ### 5. finalize()
 
 finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。但是 try-finally 等方式可以做的更好，并且该方法运行代价高昂，不确定性大，无法保证各个对象的调用顺序，因此最好不要使用。
@@ -246,16 +250,18 @@ finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。
 
 主要不足是只使用了内存的一半。
 
-现在的商业虚拟机都采用这种收集算法来回收新生代，但是并不是将内存划分为大小相等的两块，而是分为一块较大的 Eden 空间和两块较小的 Survivor 空间，每次使用 Eden 空间和其中一块 Survivor。在回收时，将 Eden 和 Survivor 中还存活着的对象一次性复制到另一块 Survivor 空间上，最后清理 Eden 和使用过的那一块 Survivor。HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了内存的利用率达到 90%。如果每次回收有多于 10% 的对象存活，那么一块 Survivor 空间就不够用了，此时需要依赖于老年代进行分配担保，也就是借用老年代的空间存储放不下的对象。
+现在的商业虚拟机都采用这种收集算法来回收新生代，但是并不是将新生代划分为大小相等的两块，而是分为一块较大的 Eden 空间和两块较小的 Survivor 空间，每次使用 Eden 空间和其中一块 Survivor。在回收时，将 Eden 和 Survivor 中还存活着的对象一次性复制到另一块 Survivor 空间上，最后清理 Eden 和使用过的那一块 Survivor。
+
+HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了内存的利用率达到 90%。如果每次回收有多于 10% 的对象存活，那么一块 Survivor 空间就不够用了，此时需要依赖于老年代进行分配担保，也就是借用老年代的空间存储放不下的对象。
 
 ### 4. 分代收集
 
 现在的商业虚拟机采用分代收集算法，它根据对象存活周期将内存划分为几块，不同块采用适当的收集算法。
 
-一般将 Java 堆分为新生代和老年代。
+一般将堆分为新生代和老年代。
 
 - 新生代使用：复制算法
-- 老年代使用：标记 - 清理 或者 标记 - 整理 算法
+- 老年代使用：标记 - 清除 或者 标记 - 整理 算法
 
 ## 垃圾收集器
 
@@ -263,8 +269,8 @@ finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。
 
 以上是 HotSpot 虚拟机中的 7 个垃圾收集器，连线表示垃圾收集器可以配合使用。
 
-- 单线程与并行（多线程）：单线程指的是垃圾收集器只使用一个线程进行收集，而并行使用多个线程。
-- 串行与并发：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并发指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
+- 单线程与多线程：单线程指的是垃圾收集器只使用一个线程进行收集，而多线程使用多个线程；
+- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并形指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
 
 ### 1. Serial 收集器
 
@@ -290,15 +296,15 @@ Serial 翻译为串行，也就是说它以串行的方式执行。
 
 ### 3. Parallel Scavenge 收集器
 
-与 ParNew 一样是并行的多线程收集器。
+与 ParNew 一样是多线程收集器。
 
 其它收集器关注点是尽可能缩短垃圾收集时用户线程的停顿时间，而它的目标是达到一个可控制的吞吐量，它被称为“吞吐量优先”收集器。这里的吞吐量指 CPU 用于运行用户代码的时间占总时间的比值。
 
 停顿时间越短就越适合需要与用户交互的程序，良好的响应速度能提升用户体验。而高吞吐量则可以高效率地利用 CPU 时间，尽快完成程序的运算任务，主要适合在后台运算而不需要太多交互的任务。
 
-提供了两个参数用于精确控制吞吐量，分别是控制最大垃圾收集停顿时间 -XX:MaxGCPauseMillis 参数以及直接设置吞吐量大小的 -XX:GCTimeRatio 参数（值为大于 0 且小于 100 的整数）。缩短停顿时间是以牺牲吞吐量和新生代空间来换取的：新生代空间变小，垃圾回收变得频繁，导致吞吐量下降。
+缩短停顿时间是以牺牲吞吐量和新生代空间来换取的：新生代空间变小，垃圾回收变得频繁，导致吞吐量下降。
 
-还提供了一个参数 -XX:+UseAdaptiveSizePolicy，这是一个开关参数，打开参数后，就不需要手工指定新生代的大小（-Xmn）、Eden 和 Survivor 区的比例（-XX:SurvivorRatio）、晋升老年代对象年龄（-XX:PretenureSizeThreshold）等细节参数了，虚拟机会根据当前系统的运行情况收集性能监控信息，动态调整这些参数以提供最合适的停顿时间或者最大的吞吐量，这种方式称为 GC 自适应的调节策略（GC Ergonomics）。
+可以通过一个开关参数打卡 GC 自适应的调节策略（GC Ergonomics），就不需要手工指定新生代的大小（-Xmn）、Eden 和 Survivor 区的比例、晋升老年代对象年龄等细节参数了。虚拟机会根据当前系统的运行情况收集性能监控信息，动态调整这些参数以提供最合适的停顿时间或者最大的吞吐量，这种方式称为 。
 
 ### 4. Serial Old 收集器
 
@@ -323,8 +329,6 @@ Serial 翻译为串行，也就是说它以串行的方式执行。
 
 CMS（Concurrent Mark Sweep），Mark Sweep 指的是标记 - 清除算法。
 
-特点：并发收集、低停顿。
-
 分为以下四个流程：
 
 - 初始标记：仅仅只是标记一下 GC Roots 能直接关联到的对象，速度很快，需要停顿。
@@ -344,7 +348,7 @@ CMS（Concurrent Mark Sweep），Mark Sweep 指的是标记 - 清除算法。
 
 G1（Garbage-First），它是一款面向服务端应用的垃圾收集器，在多 CPU 和大内存的场景下有很好的性能。HotSpot 开发团队赋予它的使命是未来可以替换掉 CMS 收集器。
 
-Java 堆被分为新生代、老年代和永久代，其它收集器进行收集的范围都是整个新生代或者老年代，而 G1 可以直接对新生代和老年代一起回收。
+堆被分为新生代和老年代，其它收集器进行收集的范围都是整个新生代或者老年代，而 G1 可以直接对新生代和老年代一起回收。
 
 <div align="center"> <img src="../pics//4cf711a8-7ab2-4152-b85c-d5c226733807.png" width="600"/> </div><br>
 
@@ -372,18 +376,6 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 
 更详细内容请参考：[Getting Started with the G1 Garbage Collector](http://www.oracle.com/webfolder/technetwork/tutorials/obe/java/G1GettingStarted/index.html)
 
-### 8. 比较
-
-| 收集器 | 单线程/并行 | 串行/并发 | 新生代/老年代 | 收集算法 | 目标 | 适用场景 |
-| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
-|  **Serial**  | 单线程 | 串行 | 新生代 | 复制 | 响应速度优先 | 单 CPU 环境下的 Client 模式 |
-|  **Serial Old**  | 单线程 | 串行 | 老年代 | 标记-整理 | 响应速度优先 | 单 CPU 环境下的 Client 模式、CMS 的后备预案 |
-|  **ParNew**  |  并行 |串行 | 新生代 | 复制算法 | 响应速度优先 | 多 CPU 环境时在 Server 模式下与 CMS 配合 |
-|  **Parallel Scavenge**  | 并行 | 串行 | 新生代 | 复制算法 | 吞吐量优先 | 在后台运算而不需要太多交互的任务 |
-|  **Parallel Old**  | 并行 | 串行 | 老年代 | 标记-整理 | 吞吐量优先 | 在后台运算而不需要太多交互的任务 |
-|  **CMS**  | 并行 | 并发 | 老年代 | 标记-清除 | 响应速度优先 | 集中在互联网站或 B/S 系统服务端上的 Java 应用 |
-|  **G1**  | 并行 | 并发 | 新生代 + 老年代 | 标记-整理 + 复制算法 | 响应速度优先 | 面向服务端应用，将来替换 CMS |
-
 ## 内存分配与回收策略
 
 ### 1. Minor GC 和 Full GC
@@ -752,6 +744,7 @@ public class FileSystemClassLoader extends ClassLoader {
 - 周志明. 深入理解 Java 虚拟机 [M]. 机械工业出版社, 2011.
 - [Chapter 2. The Structure of the Java Virtual Machine](https://docs.oracle.com/javase/specs/jvms/se8/html/jvms-2.html#jvms-2.5.4)
 - [Jvm memory](https://www.slideshare.net/benewu/jvm-memory)
+- [JNI Part1: Java Native Interface Introduction and “Hello World” application](http://electrofriends.com/articles/jni/jni-part1-java-native-interface/)
 - [Memory Architecture Of JVM(Runtime Data Areas)](https://hackthejava.wordpress.com/2015/01/09/memory-architecture-by-jvmruntime-data-areas/)
 - [JVM Run-Time Data Areas](https://www.programcreek.com/2013/04/jvm-run-time-data-areas/)
 - [Android on x86: Java Native Interface and the Android Native Development Kit](http://www.drdobbs.com/architecture-and-design/android-on-x86-java-native-interface-and/240166271)
diff --git a/pics/JNI-Java-Native-Interface.jpg b/pics/JNI-Java-Native-Interface.jpg
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HcmV?d00001


From cef2ed9ef8e1111f8ce9d71b4a782a8240a15efe Mon Sep 17 00:00:00 2001
From: luocaodan <luocaodan@buaa.edu.cn>
Date: Sat, 4 Aug 2018 23:06:47 +0800
Subject: [PATCH 052/172] =?UTF-8?q?Update=20=E7=AE=97=E6=B3=95.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/算法.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/算法.md b/notes/算法.md
index df5a8049..f440111c 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -2050,7 +2050,7 @@ public class SparseVector {
 
 如果只有一个圆盘，那么只需要进行一次移动操作。
 
-从上面的讨论可以知道，n 圆盘需要移动 (n-1)+1+(n-1) = 2n-1 次。
+从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然a<sub>n</sub> = 2<sup>n</sup> - 1，n 圆盘需要移动 2<sup>n</sup> - 1 次。
 
 <div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
 

From 288fb85d924314e5dac93ac033ad0a5573c83487 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 5 Aug 2018 15:29:23 +0800
Subject: [PATCH 053/172] auto commit

---
 notes/Java 基础.md                            |  34 ++---
 notes/Java 容器.md                            | 139 +++++++++++++++---
 notes/算法.md                                 |   2 +-
 pics/49495c95-52e5-4c9a-b27b-92cf235ff5ec.png | Bin 0 -> 10240 bytes
 pics/NP4z3i8m38Ntd28NQ4_0KCJ2q044Oez.png      | Bin 0 -> 22528 bytes
 pics/SoWkIImgAStDuUBAp2j9BKfBJ4vLy0G.png      | Bin 0 -> 14336 bytes
 pics/SoWkIImgAStDuUBAp2j9BKfBJ4vLy4q.png      | Bin 0 -> 12288 bytes
 7 files changed, 134 insertions(+), 41 deletions(-)
 create mode 100644 pics/49495c95-52e5-4c9a-b27b-92cf235ff5ec.png
 create mode 100644 pics/NP4z3i8m38Ntd28NQ4_0KCJ2q044Oez.png
 create mode 100644 pics/SoWkIImgAStDuUBAp2j9BKfBJ4vLy0G.png
 create mode 100644 pics/SoWkIImgAStDuUBAp2j9BKfBJ4vLy4q.png

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index f5133a6f..bcb94acc 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -4,7 +4,7 @@
     * [缓存池](#缓存池)
 * [二、String](#二string)
     * [概览](#概览)
-    * [String 不可变的好处](#string-不可变的好处)
+    * [不可变的好处](#不可变的好处)
     * [String, StringBuffer and StringBuilder](#string,-stringbuffer-and-stringbuilder)
     * [String.intern()](#stringintern)
 * [三、运算](#三运算)
@@ -153,7 +153,7 @@ public final class String
     private final char value[];
 ```
 
-## String 不可变的好处
+## 不可变的好处
 
 **1. 可以缓存 hash 值** 
 
@@ -212,7 +212,7 @@ String s5 = "bbb";
 System.out.println(s4 == s5);  // true
 ```
 
-在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被放在堆中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
+在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到 Native Method 中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
 
 - [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning) 
 - [深入解析 String#intern](https://tech.meituan.com/in_depth_understanding_string_intern.html)
@@ -583,19 +583,7 @@ protected void finalize() throws Throwable {}
 
 ## equals()
 
-**1. equals() 与 == 的区别** 
-
-- 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
-- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用的对象的 equals() 方法的具体实现来进行比较。
-
-```java
-Integer x = new Integer(1);
-Integer y = new Integer(1);
-System.out.println(x.equals(y)); // true
-System.out.println(x == y);      // false
-```
-
-**2. 等价关系** 
+**1. 等价关系** 
 
 （一）自反性
 
@@ -632,6 +620,18 @@ x.equals(y) == x.equals(y); // true
 x.equals(null); // false;
 ```
 
+**2. equals() 与 ==** 
+
+- 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
+- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用对象 equals() 方法的具体实现来进行比较。
+
+```java
+Integer x = new Integer(1);
+Integer y = new Integer(1);
+System.out.println(x.equals(y)); // true
+System.out.println(x == y);      // false
+```
+
 **3. 实现** 
 
 - 检查是否为同一个对象的引用，如果是直接返回 true；
@@ -1005,7 +1005,7 @@ public class A {
 
 **4. 静态内部类** 
 
-非静态内部类依赖于需要外部类的实例，而静态内部类不需要。
+非静态内部类依赖于外部类的实例，而静态内部类不需要。
 
 ```java
 public class OuterClass {
diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index c6fc88f4..27a783f1 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -14,6 +14,7 @@
     * [ConcurrentHashMap](#concurrenthashmap)
     * [LinkedHashMap](#linkedhashmap)
     * [WeekHashMap](#weekhashmap)
+* [附录](#附录)
 * [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
@@ -24,15 +25,15 @@
 
 ## Collection
 
-<div align="center"> <img src="../pics//java-collections.png"/> </div><br>
+<div align="center"> <img src="../pics//NP4z3i8m38Ntd28NQ4_0KCJ2q044Oez.png"/> </div><br>
 
 ### 1. Set
 
-- HashSet：基于哈希实现，支持快速查找，但不支持有序性操作，例如根据一个范围查找元素的操作。并且失去了元素的插入顺序信息，也就是说使用 Iterator 遍历 HashSet 得到的结果是不确定的。
+- HashSet：基于哈希表实现，支持快速查找。但不支持有序性操作，例如根据一个范围查找元素的操作。并且失去了元素的插入顺序信息，也就是说使用 Iterator 遍历 HashSet 得到的结果是不确定的。
 
 - TreeSet：基于红黑树实现，支持有序性操作，但是查找效率不如 HashSet，HashSet 查找时间复杂度为 O(1)，TreeSet 则为 O(logN)。
 
-- LinkedHashSet：具有 HashSet 的查找效率，且内部使用链表维护元素的插入顺序。
+- LinkedHashSet：具有 HashSet 的查找效率，且内部使用双向链表维护元素的插入顺序。
 
 ### 2. List
 
@@ -50,13 +51,13 @@
 
 ## Map
 
-<div align="center"> <img src="../pics//java-collections1.png"/> </div><br>
+<div align="center"> <img src="../pics//SoWkIImgAStDuUBAp2j9BKfBJ4vLy4q.png"/> </div><br>
 
-- HashMap：基于哈希实现；
+- HashMap：基于哈希表实现；
 
 - HashTable：和 HashMap 类似，但它是线程安全的，这意味着同一时刻多个线程可以同时写入 HashTable 并且不会导致数据不一致。它是遗留类，不应该去使用它。现在可以使用 ConcurrentHashMap 来支持线程安全，并且 ConcurrentHashMap 的效率会更高，因为 ConcurrentHashMap 引入了分段锁。
 
-- LinkedHashMap：使用链表来维护元素的顺序，顺序为插入顺序或者最近最少使用（LRU）顺序。
+- LinkedHashMap：使用双向链表来维护元素的顺序，顺序为插入顺序或者最近最少使用（LRU）顺序。
 
 - TreeMap：基于红黑树实现。
 
@@ -64,7 +65,7 @@
 
 ## 迭代器模式
 
-<div align="center"> <img src="../pics//Iterator-1.jpg"/> </div><br>
+<div align="center"> <img src="../pics//SoWkIImgAStDuUBAp2j9BKfBJ4vLy0G.png"/> </div><br>
 
 Collection 实现了 Iterable 接口，其中的 iterator() 方法能够产生一个 Iterator 对象，通过这个对象就可以迭代遍历 Collection 中的元素。
 
@@ -95,7 +96,7 @@ Integer[] arr = {1, 2, 3};
 List list = Arrays.asList(arr);
 ```
 
-也可以使用以下方式使用 asList()：
+也可以使用以下方式调用 asList()：
 
 ```java
 List list = Arrays.asList(1,2,3);
@@ -111,7 +112,7 @@ List list = Arrays.asList(1,2,3);
 
 ### 1. 概览
 
-实现了 RandomAccess 接口，因此支持随机访问，这是理所当然的，因为 ArrayList 是基于数组实现的。
+实现了 RandomAccess 接口，因此支持随机访问。这是理所当然的，因为 ArrayList 是基于数组实现的。
 
 ```java
 public class ArrayList<E> extends AbstractList<E>
@@ -126,7 +127,9 @@ private static final int DEFAULT_CAPACITY = 10;
 
 ### 2. 序列化
 
-基于数组实现，保存元素的数组使用 transient 修饰，该关键字声明数组默认不会被序列化。ArrayList 具有动态扩容特性，因此保存元素的数组不一定都会被使用，那么就没必要全部进行序列化。ArrayList 重写了 writeObject() 和 readObject() 来控制只序列化数组中有元素填充那部分内容。
+ArrayList 基于数组实现，并且具有动态扩容特性，因此保存元素的数组不一定都会被使用，那么就没必要全部进行序列化。
+
+保存元素的数组 elementData 使用 transient 修饰，该关键字声明数组默认不会被序列化。ArrayList 重写了 writeObject() 和 readObject() 来控制只序列化数组中有元素填充那部分内容。
 
 ```java
 transient Object[] elementData; // non-private to simplify nested class access
@@ -174,7 +177,7 @@ private void grow(int minCapacity) {
 
 ### 4. 删除元素
 
-需要调用 System.arraycopy() 将 index+1 后面的元素都复制到 index 位置上。
+需要调用 System.arraycopy() 将 index+1 后面的元素都复制到 index 位置上，该操作的时间复杂度为 O(N)，可以看出 ArrayList 删除元素的代价是非常高的。
 
 ```java
 public E remove(int index) {
@@ -311,7 +314,7 @@ CopyOnWriteArrayList 在写操作的同时允许读操作，大大提高了读
 
 ### 1. 概览
 
-基于双向链表实现，内部使用 Node 来存储链表节点信息。
+基于双向链表实现，使用 Node 存储链表节点信息。
 
 ```java
 private static class Node<E> {
@@ -328,7 +331,7 @@ transient Node<E> first;
 transient Node<E> last;
 ```
 
-<div align="center"> <img src="../pics//5158bc2f-83a6-4351-817e-c9b07f955d76.png"/> </div><br>
+<div align="center"> <img src="../pics//49495c95-52e5-4c9a-b27b-92cf235ff5ec.png"/> </div><br>
 
 ### 2. ArrayList 与 LinkedList
 
@@ -474,7 +477,7 @@ public V put(K key, V value) {
 }
 ```
 
-HashMap 允许插入键为 null 的键值对。因为无法调用 null 的 hashCode()，也就无法确定该键值对的桶下标，只能通过强制指定一个桶下标来存放。HashMap 使用第 0 个桶存放键为 null 的键值对。
+HashMap 允许插入键为 null 的键值对。但是因为无法调用 null 的 hashCode() 方法，也就无法确定该键值对的桶下标，只能通过强制指定一个桶下标来存放。HashMap 使用第 0 个桶存放键为 null 的键值对。
 
 ```java
 private V putForNullKey(V value) {
@@ -601,10 +604,10 @@ static int indexFor(int h, int length) {
 
 | 参数 | 含义 |
 | :--: | :-- |
-| capacity | table 的容量大小，默认为 16，需要注意的是 capacity 必须保证为 2 的 n 次方。|
+| capacity | table 的容量大小，默认为 16。需要注意的是 capacity 必须保证为 2 的 n 次方。|
 | size | table 的实际使用量。 |
 | threshold | size 的临界值，size 必须小于 threshold，如果大于等于，就必须进行扩容操作。 |
-| load_factor | 装载因子，table 能够使用的比例，threshold = capacity * load_factor。|
+| loadFactor | 装载因子，table 能够使用的比例，threshold = capacity * loadFactor。|
 
 ```java
 static final int DEFAULT_INITIAL_CAPACITY = 16;
@@ -851,7 +854,7 @@ public int size() {
 
 ### 3. JDK 1.8 的改动
 
-JDK 1.7 使用分段锁机制来实现并发更新操作，核心类为 Segment，它继承自重入锁 ReentrantLock，并发程度与 Segment 数量相等。
+JDK 1.7 使用分段锁机制来实现并发更新操作，核心类为 Segment，它继承自重入锁 ReentrantLock，并发度与 Segment 数量相等。
 
 JDK 1.8 使用了 CAS 操作来支持更高的并发度，在 CAS 操作失败时使用内置锁 synchronized。
 
@@ -867,7 +870,7 @@ JDK 1.8 使用了 CAS 操作来支持更高的并发度，在 CAS 操作失败
 public class LinkedHashMap<K,V> extends HashMap<K,V> implements Map<K,V>
 ```
 
-内存维护了一个双向循环链表，用来维护插入顺序或者 LRU 顺序。
+内存维护了一个双向链表，用来维护插入顺序或者 LRU 顺序。
 
 ```java
 /**
@@ -881,13 +884,13 @@ transient LinkedHashMap.Entry<K,V> head;
 transient LinkedHashMap.Entry<K,V> tail;
 ```
 
-顺序使用 accessOrder 来决定，默认为 false，此时使用的是插入顺序。
+accessOrder 决定了顺序，默认为 false，此时使用的是插入顺序。
 
 ```java
 final boolean accessOrder;
 ```
 
-LinkedHashMap 最重要的是以下用于记录顺序的函数，它们会在 put、get 等方法中调用。
+LinkedHashMap 最重要的是以下用于维护顺序的函数，它们会在 put、get 等方法中调用。
 
 ```java
 void afterNodeAccess(Node<K,V> p) { }
@@ -896,7 +899,7 @@ void afterNodeInsertion(boolean evict) { }
 
 ### afterNodeAccess()
 
-当一个 Node 被访问时，如果 accessOrder 为 true，会将它移到链表尾部。也就是说指定为 LRU 顺序之后，在每次访问一个节点时，会将这个节点移到链表尾部，保证链表尾部是最近访问的节点，那么链表首部就是最近最久未使用的节点。
+当一个节点被访问时，如果 accessOrder 为 true，则会将 该节点移到链表尾部。也就是说指定为 LRU 顺序之后，在每次访问一个节点时，会将这个节点移到链表尾部，保证链表尾部是最近访问的节点，那么链表首部就是最近最久未使用的节点。
 
 ```java
 void afterNodeAccess(Node<K,V> e) { // move node to last
@@ -951,7 +954,11 @@ protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
 
 ### LRU 缓存
 
-以下是使用 LinkedHashMap 实现的一个 LRU 缓存，设定最大缓存空间 MAX_ENTRIES  为 3。使用 LinkedHashMap 的构造函数将 accessOrder 设置为 true，开启 LUR 顺序。覆盖 removeEldestEntry() 方法实现，在节点多于 MAX_ENTRIES 就会将最近最久未使用的数据移除。
+以下是使用 LinkedHashMap 实现的一个 LRU 缓存：
+
+- 设定最大缓存空间 MAX_ENTRIES  为 3；
+- 使用 LinkedHashMap 的构造函数将 accessOrder 设置为 true，开启 LUR 顺序；
+- 覆盖 removeEldestEntry() 方法实现，在节点多于 MAX_ENTRIES 就会将最近最久未使用的数据移除。
 
 ```java
 class LRUCache<K, V> extends LinkedHashMap<K, V> {
@@ -1002,7 +1009,9 @@ Tomcat 中的 ConcurrentCache 就使用了 WeakHashMap 来实现缓存功能。
 ConcurrentCache 采取的是分代缓存：
 
 - 经常使用的对象放入 eden 中，eden 使用 ConcurrentHashMap 实现，不用担心会被回收（伊甸园）；
-- 不常用的对象放入 longterm，longterm 使用 WeakHashMap 实现，用来存放比较老的对象，这些老对象会被垃圾收集器回收。
+- 不常用的对象放入 longterm，longterm 使用 WeakHashMap 实现，这些老对象会被垃圾收集器回收。
+- 当调用  get() 方法时，会先从 eden 区获取，如果没有找到的话再到 longterm 获取，当从 longterm 获取到就把对象放入 eden 中，保证频繁被访问的节点不容易被回收。
+- 当调用 put() 方法时，如果缓存当前容量大小超过了 size，那么就将 eden 中的所有对象都放入 longterm 中，利用虚拟机回收掉一部分不经常使用的对象。
 
 ```java
 public final class ConcurrentCache<K, V> {
@@ -1039,6 +1048,90 @@ public final class ConcurrentCache<K, V> {
 }
 ```
 
+# 附录
+
+Collection 绘图源码：
+
+```
+@startuml
+
+interface Collection
+interface Set
+interface List
+interface Queue
+interface SortSet
+
+class HashSet
+class LinkedHashSet
+class TreeSet
+class ArrayList
+class Vector
+class LinkedList
+class PriorityQueue
+
+
+Collection <|-- Set
+Collection <|-- List
+Collection <|-- Queue
+Set <|-- SortSet
+
+Set <|.. HashSet
+Set <|.. LinkedHashSet
+SortSet <|.. TreeSet
+List <|.. ArrayList
+List <|.. Vector
+List <|.. LinkeList
+Queue <|.. LinkedList
+Queue <|.. PriorityQueue
+
+@enduml
+```
+
+Map 绘图源码
+
+```
+@startuml
+
+interface Map
+interface SortMap
+
+class HashTable
+class LinkedHashMap
+class HashMap
+class TreeMap
+
+Map <|.. HashTable
+Map <|.. LinkedHashMap
+Map <|.. HashMap
+Map <|-- SortMap
+SortMap <|.. TreeMap
+
+@enduml
+```
+
+迭代器类图
+
+```
+@startuml
+
+interface Iterable
+interface Collection
+interface List
+interface Set
+interface Queue
+interface Iterator
+interface ListIterator
+
+Iterable <|-- Collection
+Collection <|.. List
+Collection <|.. Set
+Collection <|-- Queue
+Iterator <-- Iterable
+Iterator <|.. ListIterator
+ListIterator <-- List
+
+@enduml
+```
 
 # 参考资料
 
diff --git a/notes/算法.md b/notes/算法.md
index f440111c..fb878742 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -2050,7 +2050,7 @@ public class SparseVector {
 
 如果只有一个圆盘，那么只需要进行一次移动操作。
 
-从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然a<sub>n</sub> = 2<sup>n</sup> - 1，n 圆盘需要移动 2<sup>n</sup> - 1 次。
+从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然 a<sub>n</sub> = 2<sup>n</sup> - 1，n 个圆盘需要移动 2<sup>n</sup> - 1 次。
 
 <div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
 
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literal 0
HcmV?d00001


From cc5277542918f37e8cf415c9952b6bd71a3cf93d Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Sun, 5 Aug 2018 15:30:38 +0800
Subject: [PATCH 054/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 39b43f30..9bdb297b 100644
--- a/README.md
+++ b/README.md
@@ -89,7 +89,7 @@ Leetcode 上数据库题目的解题记录。
 
 > [Java 容器](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20容器.md)
 
-源码分析：ArrayListVector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
+源码分析：ArrayList、Vector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
 
 > [Java 并发](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20并发.md)
 

From eace8369dd11d548c5a57491bbeee4d439ef18bb Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 5 Aug 2018 16:35:58 +0800
Subject: [PATCH 055/172] auto commit

---
 notes/Java IO.md | 123 ++++++++++++++++++++++++++++++-----------------
 1 file changed, 80 insertions(+), 43 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index 08dc80f4..d93ff47c 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -40,8 +40,7 @@ File 类可以用于表示文件和目录的信息，但是它不表示文件的
 递归地输出一个目录下所有文件：
 
 ```java
-public static void listAllFiles(File dir)
-{
+public static void listAllFiles(File dir) {
     if (dir == null || !dir.exists()) {
         return;
     }
@@ -60,8 +59,7 @@ public static void listAllFiles(File dir)
 使用字节流操作进行文件复制：
 
 ```java
-public static void copyFile(String src, String dist) throws IOException
-{
+public static void copyFile(String src, String dist) throws IOException {
     FileInputStream in = new FileInputStream(src);
     FileOutputStream out = new FileOutputStream(dist);
     byte[] buffer = new byte[20 * 1024];
@@ -78,7 +76,11 @@ public static void copyFile(String src, String dist) throws IOException
 
 <div align="center"> <img src="../pics//DP-Decorator-java.io.png" width="500"/> </div><br>
 
-Java I/O 使用了装饰者模式来实现。以 InputStream 为例，InputStream 是抽象组件，FileInputStream 是 InputStream 的子类，属于具体组件，提供了字节流的输入操作。FilterInputStream 属于抽象装饰者，装饰者用于装饰组件，为组件提供额外的功能，例如 BufferedInputStream 为 FileInputStream 提供缓存的功能。
+Java I/O 使用了装饰者模式来实现。以 InputStream 为例，
+
+- InputStream 是抽象组件；
+- FileInputStream 是 InputStream 的子类，属于具体组件，提供了字节流的输入操作；
+- FilterInputStream 属于抽象装饰者，装饰者用于装饰组件，为组件提供额外的功能，例如 BufferedInputStream 为 FileInputStream 提供缓存的功能。
 
 实例化一个具有缓存功能的字节流对象时，只需要在 FileInputStream 对象上再套一层 BufferedInputStream 对象即可。
 
@@ -99,8 +101,7 @@ DataInputStream 装饰者提供了对更多数据类型进行输入的操作，
 逐行输出文本文件的内容：
 
 ```java
-public static void readFileContent(String filePath) throws IOException
-{
+public static void readFileContent(String filePath) throws IOException {
     FileReader fileReader = new FileReader(filePath);
     BufferedReader bufferedReader = new BufferedReader(fileReader);
     String line;
@@ -126,7 +127,7 @@ UTF-16be 中的 be 指的是 Big Endian，也就是大端。相应地也有 UTF-
 
 Java 使用双字节编码 UTF-16be，这不是指 Java 只支持这一种编码方式，而是说 char 这种类型使用 UTF-16be 进行编码。char 类型占 16 位，也就是两个字节，Java 使用这种双字节编码是为了让一个中文或者一个英文都能使用一个 char 来存储。
 
-String 可以看成一个字符序列，可以指定一个编码方式将它转换为字节序列，也可以指定一个编码方式将一个字节序列转换为 String。
+String 可以看成一个字符序列，可以指定一个编码方式将它编码为字节序列，也可以指定一个编码方式将一个字节序列解码为 String。
 
 ```java
 String str1 = "中文";
@@ -151,8 +152,7 @@ byte[] bytes = str1.getBytes();
 序列化的类需要实现 Serializable 接口，它只是一个标准，没有任何方法需要实现，但是如果不去实现它的话而进行序列化，会抛出异常。
 
 ```java
-public static void main(String[] args) throws IOException, ClassNotFoundException
-{
+public static void main(String[] args) throws IOException, ClassNotFoundException {
     A a1 = new A(123, "abc");
     String objectFile = "file/a1";
     ObjectOutputStream objectOutputStream = new ObjectOutputStream(new FileOutputStream(objectFile));
@@ -188,7 +188,7 @@ private static class A implements Serializable
 
 transient 关键字可以使一些属性不会被序列化。
 
-**ArrayList 序列化和反序列化的实现** ：ArrayList 中存储数据的数组是用 transient 修饰的，因为这个数组是动态扩展的，并不是所有的空间都被使用，因此就不需要所有的内容都被序列化。通过重写序列化和反序列化方法，使得可以只序列化数组中有内容的那部分数据。
+ArrayList 中存储数据的数组是用 transient 修饰的，因为这个数组是动态扩展的，并不是所有的空间都被使用，因此就不需要所有的内容都被序列化。通过重写序列化和反序列化方法，使得可以只序列化数组中有内容的那部分数据。
 
 ```java
 private transient Object[] elementData;
@@ -205,7 +205,7 @@ Java 中的网络支持：
 
 ## InetAddress
 
-没有公有构造函数，只能通过静态方法来创建实例。
+没有公有的构造函数，只能通过静态方法来创建实例。
 
 ```java
 InetAddress.getByName(String host);
@@ -217,19 +217,24 @@ InetAddress.getByAddress(byte[] address);
 可以直接从 URL 中读取字节流数据。
 
 ```java
-public static void main(String[] args) throws IOException
-{
+public static void main(String[] args) throws IOException {
+
     URL url = new URL("http://www.baidu.com");
-    // 字节流
+
+    /* 字节流 */
     InputStream is = url.openStream();
-    // 字符流
+
+    /* 字符流 */
     InputStreamReader isr = new InputStreamReader(is, "utf-8");
+
+    /* 提供缓存功能 */
     BufferedReader br = new BufferedReader(isr);
-    String line = br.readLine();
-    while (line != null) {
+
+    String line;
+    while ((line = br.readLine()) != null) {
         System.out.println(line);
-        line = br.readLine();
     }
+
     br.close();
 }
 ```
@@ -253,7 +258,7 @@ public static void main(String[] args) throws IOException
 - [Java NIO 浅析](https://tech.meituan.com/nio.html)
 - [IBM: NIO 入门](https://www.ibm.com/developerworks/cn/education/java/j-nio/j-nio.html)
 
-新的输入/输出 (NIO) 库是在 JDK 1.4 中引入的。NIO 弥补了原来的 I/O 的不足，提供了高速的、面向块的 I/O。
+新的输入/输出 (NIO) 库是在 JDK 1.4 中引入的，弥补了原来的 I/O 的不足，提供了高速的、面向块的 I/O。
 
 ## 流与块
 
@@ -271,7 +276,7 @@ I/O 包和 NIO 已经很好地集成了，java.io.\* 已经以 NIO 为基础重
 
 通道 Channel 是对原 I/O 包中的流的模拟，可以通过它读取和写入数据。
 
-通道与流的不同之处在于，流只能在一个方向上移动，(一个流必须是 InputStream 或者 OutputStream 的子类)，而通道是双向的，可以用于读、写或者同时用于读写。
+通道与流的不同之处在于，流只能在一个方向上移动(一个流必须是 InputStream 或者 OutputStream 的子类)，而通道是双向的，可以用于读、写或者同时用于读写。
 
 通道包括以下类型：
 
@@ -329,21 +334,41 @@ I/O 包和 NIO 已经很好地集成了，java.io.\* 已经以 NIO 为基础重
 以下展示了使用 NIO 快速复制文件的实例：
 
 ```java
-public static void fastCopy(String src, String dist) throws IOException
-{
-    FileInputStream fin = new FileInputStream(src);      /* 获取源文件的输入字节流 */
-    FileChannel fcin = fin.getChannel();                 /* 获取输入字节流的文件通道 */
-    FileOutputStream fout = new FileOutputStream(dist);  /* 获取目标文件的输出字节流 */
-    FileChannel fcout = fout.getChannel();               /* 获取输出字节流的通道 */
-    ByteBuffer buffer = ByteBuffer.allocateDirect(1024); /* 为缓冲区分配 1024 个字节 */
+public static void fastCopy(String src, String dist) throws IOException {
+
+    /* 获得源文件的输入字节流 */
+    FileInputStream fin = new FileInputStream(src);
+
+    /* 获取输入字节流的文件通道 */
+    FileChannel fcin = fin.getChannel();
+
+    /* 获取目标文件的输出字节流 */
+    FileOutputStream fout = new FileOutputStream(dist);
+
+    /* 获取输出字节流的通道 */
+    FileChannel fcout = fout.getChannel();
+
+    /* 为缓冲区分配 1024 个字节 */
+    ByteBuffer buffer = ByteBuffer.allocateDirect(1024);
+
     while (true) {
-        int r = fcin.read(buffer);                       /* 从输入通道中读取数据到缓冲区中 */
-        if (r == -1) {                                   /* read() 返回 -1 表示 EOF */
+
+        /* 从输入通道中读取数据到缓冲区中 */
+        int r = fcin.read(buffer);
+
+        /* read() 返回 -1 表示 EOF */
+        if (r == -1) {
             break;
         }
-        buffer.flip();                                   /* 切换读写 */
-        fcout.write(buffer);                             /* 把缓冲区的内容写入输出文件中 */
-        buffer.clear();                                  /* 清空缓冲区 */
+
+        /* 切换读写 */
+        buffer.flip();
+
+        /* 把缓冲区的内容写入输出文件中 */
+        fcout.write(buffer);
+        
+        /* 清空缓冲区 */
+        buffer.clear();
     }
 }
 ```
@@ -448,10 +473,10 @@ while (true) {
 ## 套接字 NIO 实例
 
 ```java
-public class NIOServer
-{
-    public static void main(String[] args) throws IOException
-    {
+public class NIOServer {
+
+    public static void main(String[] args) throws IOException {
+
         Selector selector = Selector.open();
 
         ServerSocketChannel ssChannel = ServerSocketChannel.open();
@@ -463,32 +488,45 @@ public class NIOServer
         serverSocket.bind(address);
 
         while (true) {
+
             selector.select();
             Set<SelectionKey> keys = selector.selectedKeys();
             Iterator<SelectionKey> keyIterator = keys.iterator();
+
             while (keyIterator.hasNext()) {
+
                 SelectionKey key = keyIterator.next();
+
                 if (key.isAcceptable()) {
+
                     ServerSocketChannel ssChannel1 = (ServerSocketChannel) key.channel();
+
                     // 服务器会为每个新连接创建一个 SocketChannel
                     SocketChannel sChannel = ssChannel1.accept();
                     sChannel.configureBlocking(false);
+
                     // 这个新连接主要用于从客户端读取数据
                     sChannel.register(selector, SelectionKey.OP_READ);
+
                 } else if (key.isReadable()) {
+
                     SocketChannel sChannel = (SocketChannel) key.channel();
                     System.out.println(readDataFromSocketChannel(sChannel));
                     sChannel.close();
                 }
+
                 keyIterator.remove();
             }
         }
     }
 
     private static String readDataFromSocketChannel(SocketChannel sChannel) throws IOException {
+
         ByteBuffer buffer = ByteBuffer.allocate(1024);
         StringBuilder data = new StringBuilder();
+
         while (true) {
+
             buffer.clear();
             int n = sChannel.read(buffer);
             if (n == -1) {
@@ -509,10 +547,9 @@ public class NIOServer
 ```
 
 ```java
-public class NIOClient
-{
-    public static void main(String[] args) throws IOException 
-    {
+public class NIOClient {
+
+    public static void main(String[] args) throws IOException {
         Socket socket = new Socket("127.0.0.1", 8888);
         OutputStream out = socket.getOutputStream();
         String s = "hello world";
@@ -526,9 +563,9 @@ public class NIOClient
 
 内存映射文件 I/O 是一种读和写文件数据的方法，它可以比常规的基于流或者基于通道的 I/O 快得多。
 
-向内存映射文件写入可能是危险的，仅只是改变数组的单个元素这样的简单操作，就可能会直接修改磁盘上的文件。修改数据与将数据保存到磁盘是没有分开的。
+向内存映射文件写入可能是危险的，只是改变数组的单个元素这样的简单操作，就可能会直接修改磁盘上的文件。修改数据与将数据保存到磁盘是没有分开的。
 
-下面代码行将文件的前 1024 个字节映射到内存中，map() 方法返回一个 MappedByteBuffer，它是 ByteBuffer 的子类。因此，您可以像使用其他任何 ByteBuffer 一样使用新映射的缓冲区，操作系统会在需要时负责执行映射。
+下面代码行将文件的前 1024 个字节映射到内存中，map() 方法返回一个 MappedByteBuffer，它是 ByteBuffer 的子类。因此，可以像使用其他任何 ByteBuffer 一样使用新映射的缓冲区，操作系统会在需要时负责执行映射。
 
 ```java
 MappedByteBuffer mbb = fc.map(FileChannel.MapMode.READ_WRITE, 0, 1024);

From bd3f5ca2437cc75f6dc1634255e72fd7a1e55091 Mon Sep 17 00:00:00 2001
From: aikin <1@aikin.me>
Date: Sun, 5 Aug 2018 17:36:53 +0800
Subject: [PATCH 056/172] =?UTF-8?q?add=20refactoring=20kata=20into=20?=
 =?UTF-8?q?=E9=87=8D=E6=9E=84.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/重构.md | 5 +++++
 1 file changed, 5 insertions(+)

diff --git a/notes/重构.md b/notes/重构.md
index 9a3242ba..72021355 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -107,6 +107,7 @@
     * [10. 塑造模板函数](#10-塑造模板函数)
     * [11. 以委托取代继承](#11-以委托取代继承)
     * [12. 以继承取代委托](#12-以继承取代委托)
+* [练习重构的栗子](#练习重构的栗子)    
 * [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
@@ -1415,6 +1416,10 @@ public Manager(String name, String id, int grade) {
 
 让委托类继承受托类。
 
+
+# 练习重构的栗子
+* [Refactoring Kata](https://github.com/aikin/refactoring-kata)
+
 # 参考资料
 
 - MartinFowler, 福勒, 贝克, 等. 重构: 改善既有代码的设计 [M]. 电子工业出版社, 2011.

From 1366a2e149d4a2cc9b4af30fea91a7af40d3f497 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 5 Aug 2018 17:37:04 +0800
Subject: [PATCH 057/172] auto commit

---
 notes/Java IO.md    | 12 ++++++------
 notes/计算机网络.md | 26 +++++++++++++++-----------
 2 files changed, 21 insertions(+), 17 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index d93ff47c..1e3aedcd 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -60,15 +60,18 @@ public static void listAllFiles(File dir) {
 
 ```java
 public static void copyFile(String src, String dist) throws IOException {
+
     FileInputStream in = new FileInputStream(src);
     FileOutputStream out = new FileOutputStream(dist);
     byte[] buffer = new byte[20 * 1024];
+
     // read() 最多读取 buffer.length 个字节
     // 返回的是实际读取的个数
     // 返回 -1 的时候表示读到 eof，即文件尾
     while (in.read(buffer, 0, buffer.length) != -1) {
         out.write(buffer);
     }
+
     in.close();
     out.close();
 }
@@ -165,20 +168,17 @@ public static void main(String[] args) throws IOException, ClassNotFoundExceptio
     System.out.println(a2);
 }
 
-private static class A implements Serializable
-{
+private static class A implements Serializable {
     private int x;
     private String y;
 
-    A(int x, String y)
-    {
+    A(int x, String y) {
         this.x = x;
         this.y = y;
     }
 
     @Override
-    public String toString()
-    {
+    public String toString() {
         return "x = " + x + "  " + "y = " + y;
     }
 }
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index f1460844..9b213d74 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -91,7 +91,7 @@
 
 每个分组都有首部和尾部，包含了源地址和目的地址等控制信息，在同一个传输线路上同时传输多个分组互相不会影响，因此在同一条传输线路上允许同时传输多个分组，也就是说分组交换不需要占用传输线路。
 
-考虑在一个邮局通信系统中，邮局收到一份邮件之后，先存储下来，然后把相同目的地的邮件一起转发到下一个目的地，这个过程就是存储转发过程，分组交换也使用了存储转发过程。
+在一个邮局通信系统中，邮局收到一份邮件之后，先存储下来，然后把相同目的地的邮件一起转发到下一个目的地，这个过程就是存储转发过程，分组交换也使用了存储转发过程。
 
 ## 时延
 
@@ -109,11 +109,11 @@
 
 ### 2. 传播时延
 
-电磁波在信道中传播一定的距离需要花费的时间，电磁波传播速度接近光速。
+电磁波在信道中传播所需要花费的时间，电磁波传播的速度接近光速。
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?delay=\frac{l(m)}{v(m/s)}"/></div> <br>
 
-其中 l 表示信道长度，v 表示电磁波在信道上的传播速率。
+其中 l 表示信道长度，v 表示电磁波在信道上的传播速度。
 
 ### 3. 处理时延
 
@@ -135,7 +135,7 @@
 
 -  **网络层** ：为主机间提供数据传输服务，而运输层协议是为主机中的进程提供服务。网络层把运输层传递下来的报文段或者用户数据报封装成分组。
 
--  **数据链路层** ：网络层针对的还是主机之间的数据传输服务，而主机之间可以有很多链路，链路层协议就是为同一链路的节点提供服务。数据链路层把网络层传来的分组封装成帧。
+-  **数据链路层** ：网络层针对的还是主机之间的数据传输服务，而主机之间可以有很多链路，链路层协议就是为同一链路的主机提供服务。数据链路层把网络层传下来的分组封装成帧。
 
 -  **物理层** ：考虑的是怎样在传输媒体上传输数据比特流，而不是指具体的传输媒体。物理层的作用是尽可能屏蔽传输媒体和通信手段的差异，使数据链路层感觉不到这些差异。
 
@@ -153,7 +153,7 @@
 
 它只有四层，相当于五层协议中数据链路层和物理层合并为网络接口层。
 
-现在的 TCP/IP 体系结构不严格遵循 OSI 分层概念，应用层可能会直接使用 IP 层或者网络接口层。
+TCP/IP 体系结构不严格遵循 OSI 分层概念，应用层可能会直接使用 IP 层或者网络接口层。
 
 <div align="center"> <img src="../pics//45e0e0bf-386d-4280-a341-a0b9496c7674.png" width="400"/> </div><br>
 
@@ -314,7 +314,7 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 主要有以太网、令牌环网、FDDI 和 ATM 等局域网技术，目前以太网占领着有线局域网市场。
 
-可以按照网络拓结构扑对局域网进行分类：
+可以按照网络拓扑结构对局域网进行分类：
 
 <div align="center"> <img src="../pics//a6026bb4-3daf-439f-b1ec-a5a24e19d2fb.jpg" width="600"/> </div><br>
 
@@ -322,9 +322,9 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 以太网是一种星型拓扑结构局域网。
 
-早期使用集线器进行连接。集线器是一种物理层设备，作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离。之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
+早期使用集线器进行连接，集线器是一种物理层设备，作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离，之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
 
-目前以太网使用交换机替代了集线器。交换机是一种链路层设备，它不会发生碰撞，能根据 MAC 地址进行存储转发。
+目前以太网使用交换机替代了集线器，交换机是一种链路层设备，它不会发生碰撞，能根据 MAC 地址进行存储转发。
 
 以太网帧格式：
 
@@ -337,7 +337,9 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 ## 交换机*
 
-交换机具有自学习能力，学习的是交换表的内容，交换表中存储着 MAC 地址到接口的映射。正是由于这种自学习能力，因此交换机是一种即插即可即用设备，不需要网络管理员手动配置交换表内容。
+交换机具有自学习能力，学习的是交换表的内容，交换表中存储着 MAC 地址到接口的映射。
+
+正是由于这种自学习能力，因此交换机是一种即插即用设备，不需要网络管理员手动配置交换表内容。
 
 下图中，交换机有 4 个接口，主机 A 向主机 B 发送数据帧时，交换机把主机 A 到接口 1 的映射写入交换表中。为了发送数据帧到 B，先查交换表，此时没有主机 B 的表项，那么主机 A 就发送广播帧，主机 C 和主机 D 会丢弃该帧。主机 B 收下之后，查找交换表得到主机 A 映射的接口为 1，就发送数据帧到接口 1，同时交换机添加主机 B 到接口 3 的映射。
 
@@ -345,9 +347,11 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 ## 虚拟局域网
 
-虚拟局域网可以建立与物理位置无关的逻辑组，只有在同一个虚拟局域网中的成员才会收到链路层广播信息。例如下图中 (A1, A2, A3, A4) 属于一个虚拟局域网，A1 发送的广播会被 A2、A3、A4 收到，而其它站点收不到。
+虚拟局域网可以建立与物理位置无关的逻辑组，只有在同一个虚拟局域网中的成员才会收到链路层广播信息。
 
-使用 VLAN 干线连接来建立虚拟局域网，每台交换机上的一个特殊端口被设置为干线端口，以互连 VLAN 交换机。IEEE 定义了一种扩展的以太网帧格式 802.1Q，它在标准以太网帧上加进了 4 字节首部 VLAN 标签，用于表示该帧属于哪一个虚拟局域网。
+例如下图中 (A1, A2, A3, A4) 属于一个虚拟局域网，A1 发送的广播会被 A2、A3、A4 收到，而其它站点收不到。
+
+使用 VLAN 干线连接来建立虚拟局域网，每台交换机上的一个特殊接口被设置为干线接口，以互连 VLAN 交换机。IEEE 定义了一种扩展的以太网帧格式 802.1Q，它在标准以太网帧上加进了 4 字节首部 VLAN 标签，用于表示该帧属于哪一个虚拟局域网。
 
 <div align="center"> <img src="../pics//a74b70ac-323a-4b31-b4d5-90569b8a944b.png" width="500"/> </div><br>
 

From 6e9e1d675e8eb961516637b052164f4841995ec4 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 6 Aug 2018 00:13:34 +0800
Subject: [PATCH 058/172] auto commit

---
 notes/分布式.md       | 10 ++++------
 notes/系统设计基础.md |  4 ++--
 notes/重构.md         |  5 +++++
 3 files changed, 11 insertions(+), 8 deletions(-)

diff --git a/notes/分布式.md b/notes/分布式.md
index 0f136855..c048b43e 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -61,7 +61,7 @@ EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了
 
 - 尝试从 N 个相互独立 Redis 实例获取锁，如果一个实例不可用，应该尽快尝试下一个；
 - 计算获取锁消耗的时间，只有当这个时间小于锁的过期时间，并且从大多数（N / 2 + 1）实例上获取了锁，那么就认为锁获取成功了；
-- 如果锁获取失败，会到每个实例上释放锁。
+- 如果锁获取失败，就到每个实例上释放锁。
 
 ## Zookeeper 的有序节点
 
@@ -124,9 +124,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 ## 2PC
 
-两阶段提交（Two-phase Commit，2PC）
-
-通过引入协调者（Coordinator）来协调参与者的行为，并最终决定这些参与者是否要真正执行事务。
+两阶段提交（Two-phase Commit，2PC），通过引入协调者（Coordinator）来协调参与者的行为，并最终决定这些参与者是否要真正执行事务。
 
 ### 1. 运行过程
 
@@ -152,7 +150,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 （二）单点问题
 
-协调者在 2PC 中起到非常大的作用，发生故障将会造成很大影响，特别是在阶段二发生故障，所有参与者会一直等待状态，无法完成其它操作。
+协调者在 2PC 中起到非常大的作用，发生故障将会造成很大影响。特别是在阶段二发生故障，所有参与者会一直等待状态，无法完成其它操作。
 
 （三）数据不一致
 
@@ -214,7 +212,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 BASE 是基本可用（Basically Available）、软状态（Soft State）和最终一致性（Eventually Consistent）三个短语的缩写。
 
-BASE 理论是对 CAP 中一致性和可用性权衡的结果，它的理论的核心思想是：即使无法做到强一致性，但每个应用都可以根据自身业务特点，采用适当的方式来使系统达到最终一致性。
+BASE 理论是对 CAP 中一致性和可用性权衡的结果，它的核心思想是：即使无法做到强一致性，但每个应用都可以根据自身业务特点，采用适当的方式来使系统达到最终一致性。
 
 <div align="center"> <img src="../pics//bc603930-d74d-4499-a3e7-2d740fc07f33.png"/> </div><br>
 
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index e132aa2c..40f0ca9d 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -13,7 +13,7 @@
 
 ### 1. 响应时间
 
-指从某个请求发出到接收到响应消耗的时间。
+指从某个请求从发出到接收到响应消耗的时间。
 
 在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
 
@@ -79,7 +79,7 @@
 实现可扩展主要有两种方式：
 
 - 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
-- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务器框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
+- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
 
 # 四、可用性
 
diff --git a/notes/重构.md b/notes/重构.md
index 9a3242ba..b5a834e0 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -107,6 +107,7 @@
     * [10. 塑造模板函数](#10-塑造模板函数)
     * [11. 以委托取代继承](#11-以委托取代继承)
     * [12. 以继承取代委托](#12-以继承取代委托)
+* [重构练习](#重构练习)
 * [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
@@ -1415,6 +1416,10 @@ public Manager(String name, String id, int grade) {
 
 让委托类继承受托类。
 
+# 重构练习
+
+- [Refactoring Kata](https://github.com/aikin/refactoring-kata)
+
 # 参考资料
 
 - MartinFowler, 福勒, 贝克, 等. 重构: 改善既有代码的设计 [M]. 电子工业出版社, 2011.

From 4bee3f40f5650a356719a6b3de9e471df5a092d1 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 6 Aug 2018 00:17:57 +0800
Subject: [PATCH 059/172] auto commit

---
 notes/重构.md | 10 ----------
 1 file changed, 10 deletions(-)

diff --git a/notes/重构.md b/notes/重构.md
index abdd99bd..b5a834e0 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -107,11 +107,7 @@
     * [10. 塑造模板函数](#10-塑造模板函数)
     * [11. 以委托取代继承](#11-以委托取代继承)
     * [12. 以继承取代委托](#12-以继承取代委托)
-<<<<<<< HEAD
 * [重构练习](#重构练习)
-=======
-* [练习重构的栗子](#练习重构的栗子)    
->>>>>>> e6475eb3bca6dc610b1db681f91da9d266151e0a
 * [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
@@ -1420,15 +1416,9 @@ public Manager(String name, String id, int grade) {
 
 让委托类继承受托类。
 
-<<<<<<< HEAD
 # 重构练习
 
 - [Refactoring Kata](https://github.com/aikin/refactoring-kata)
-=======
-
-# 练习重构的栗子
-* [Refactoring Kata](https://github.com/aikin/refactoring-kata)
->>>>>>> e6475eb3bca6dc610b1db681f91da9d266151e0a
 
 # 参考资料
 

From ea1c200cde84c393a5d1502bc38ce97f295fb4aa Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 6 Aug 2018 00:29:53 +0800
Subject: [PATCH 060/172] auto commit

---
 notes/集群.md | 40 ++++++++++++++++++++++++++--------------
 1 file changed, 26 insertions(+), 14 deletions(-)

diff --git a/notes/集群.md b/notes/集群.md
index b36a0bfe..717ae637 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -3,7 +3,7 @@
     * [算法实现](#算法实现)
     * [转发实现](#转发实现)
 * [二、集群下的 Session 管理](#二集群下的-session-管理)
-    * [Sticky Sessions](#sticky-sessions)
+    * [Sticky Session](#sticky-session)
     * [Session Replication](#session-replication)
     * [Session Server](#session-server)
 <!-- GFM-TOC -->
@@ -22,14 +22,16 @@
 
 负载均衡运行过程包含两个部分：
 
-1. 根据负载均衡算法得到应该将请求转发到哪个节点上；
+1. 根据负载均衡算法得到请求转发的节点；
 2. 将请求进行转发；
 
 ## 算法实现
 
 ### 1. 轮询（Round Robin）
 
-轮询算法把每个请求轮流发送到每个服务器上。下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
+轮询算法把每个请求轮流发送到每个服务器上。
+
+下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
 
 <div align="center"> <img src="../pics//2766d04f-7dad-42e4-99d1-60682c9d5c61.jpg"/> </div><br>
 
@@ -39,17 +41,23 @@
 
 ### 2. 加权轮询（Weighted Round Robbin）
 
-加权轮询是在轮询的基础上，根据服务器的性能差异，为服务器赋予一定的权值，性能高的服务器分配更高的权值。例如下图中，服务器 1 被赋予的权值为 5，服务器 2 被赋予的权值为 1，那么 (1, 2, 3, 4, 5) 请求会被发送到服务器 1，(6) 请求会被发送到服务器 2。
+加权轮询是在轮询的基础上，根据服务器的性能差异，为服务器赋予一定的权值，性能高的服务器分配更高的权值。
+
+例如下图中，服务器 1 被赋予的权值为 5，服务器 2 被赋予的权值为 1，那么 (1, 2, 3, 4, 5) 请求会被发送到服务器 1，(6) 请求会被发送到服务器 2。
 
 <div align="center"> <img src="../pics//211c60d4-75ca-4acd-8a4f-171458ed58b4.jpg"/> </div><br>
 
 ### 3. 最少连接（least Connections）
 
-由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
+由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。
+
+例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
 
 <div align="center"> <img src="../pics//3b0d1aa8-d0e0-46c2-8fd1-736bf08a11aa.jpg"/> </div><br>
 
-最少连接算法就是将请求发送给当前最少连接数的服务器上。例如下图中，服务器 1 当前连接数最小，那么新到来的请求 6 就会被发送到服务器 1 上。
+最少连接算法就是将请求发送给当前最少连接数的服务器上。
+
+例如下图中，服务器 1 当前连接数最小，那么新到来的请求 6 就会被发送到服务器 1 上。
 
 <div align="center"> <img src="../pics//1f4a7f10-52b2-4bd7-a67d-a9581d66dc62.jpg"/> </div><br>
 
@@ -61,13 +69,17 @@
 
 ### 5. 随机算法（Random）
 
-把请求随机发送到服务器上。和轮询算法类似，该算法比较适合服务器性能差不多的场景。
+把请求随机发送到服务器上。
+
+和轮询算法类似，该算法比较适合服务器性能差不多的场景。
 
 <div align="center"> <img src="../pics//0ee0f61b-c782-441e-bf34-665650198ae0.jpg"/> </div><br>
 
 ### 6. 源地址哈希法 (IP Hash)
 
-源地址哈希通过对客户端 IP 计算哈希值之后，再对服务器数量进行取模运算得到目标服务器的序号。可以保证同一 IP 的客户端的请求会转发到同一台服务器上，可以用来实现会话粘滞（Sticky Session）
+源地址哈希通过对客户端 IP 计算哈希值之后，再对服务器数量取模得到目标服务器的序号。
+
+可以保证同一 IP 的客户端的请求会转发到同一台服务器上，用来实现会话粘滞（Sticky Session）
 
 <div align="center"> <img src="../pics//2018040302.jpg"/> </div><br>
 
@@ -96,7 +108,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 缺点：
 
-- 由于 DNS 具有多级结构，每一级的域名记录都可能被缓存，当下线一台服务器需要修改 DNS 记录时，需要过很长一段时间才能生效；
+- 由于 DNS 具有多级结构，每一级的域名记录都可能被缓存，当下线一台服务器需要修改 DNS 记录时，需要过很长一段时间才能生效。
 
 大型网站基本使用了 DNS 做为第一级负载均衡手段，然后在内部使用其它方式做第二级负载均衡。也就是说，域名解析的结果为内部的负载均衡服务器 IP 地址。
 
@@ -123,7 +135,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 ### 4. 网络层
 
-负载均衡服务器在操作系统内核进程获取网络数据包，根据负载均衡算法计算源服务器的 IP 地址，并修改请求数据包的目的 IP 地址，最后进行转发。
+在操作系统内核进程获取网络数据包，根据负载均衡算法计算源服务器的 IP 地址，并修改请求数据包的目的 IP 地址，最后进行转发。
 
 源服务器返回的响应也需要经过负载均衡服务器，通常是让负载均衡服务器同时作为集群的网关服务器来实现。
 
@@ -133,15 +145,15 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 缺点：
 
-- 和反向代理一样，所有的请求和相应都经过负载均衡服务器，会成为性能瓶颈。
+- 和反向代理一样，所有的请求和响应都经过负载均衡服务器，会成为性能瓶颈。
 
 ### 5. 链路层
 
 在链路层根据负载均衡算法计算源服务器的 MAC 地址，并修改请求数据包的目的 MAC 地址，并进行转发。
 
-通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要需要 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
+通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
 
-这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用 LVS（Linux Virtual Server）。
+这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
 
 参考：
 
@@ -152,7 +164,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 一个用户的 Session 信息如果存储在一个服务器上，那么当负载均衡器把用户的下一个请求转发到另一个服务器，由于服务器没有用户的 Session 信息，那么该用户就需要重新进行登录等操作。
 
-## Sticky Sessions
+## Sticky Session
 
 需要配置负载均衡器，使得一个用户的所有请求都路由到同一个服务器，这样就可以把用户的 Session 存放在该服务器中。
 

From 64acfecb2933004474c510c80da9ce8ac74f6af8 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Mon, 6 Aug 2018 16:35:34 +0800
Subject: [PATCH 061/172] Add files via upload

---
 other/group.jpg | Bin 114283 -> 84645 bytes
 1 file changed, 0 insertions(+), 0 deletions(-)

diff --git a/other/group.jpg b/other/group.jpg
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From 7aec4940558e381f09b4c2e7a5562c28ab3d2b74 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Mon, 6 Aug 2018 16:42:08 +0800
Subject: [PATCH 062/172] Add files via upload

---
 other/group.jpg | Bin 84645 -> 68267 bytes
 1 file changed, 0 insertions(+), 0 deletions(-)

diff --git a/other/group.jpg b/other/group.jpg
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From 1e02771555f8dcafabd977493fdabd2b9294954f Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Mon, 6 Aug 2018 16:43:58 +0800
Subject: [PATCH 063/172] Add files via upload

---
 other/group.jpg | Bin 68267 -> 84645 bytes
 1 file changed, 0 insertions(+), 0 deletions(-)

diff --git a/other/group.jpg b/other/group.jpg
index c76902c6e40c28867bab121158cf186ad687b2c3..af2128418e53b5f858ccadb4e289ea7691e5a949 100644
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zgDIgv3AHR0f)t~q`Y;Y&5mo2^kN<1k|3$B!PbB8toQ08p^W4C}|6jgjhO_e+zux>`
DhKlG(


From ccf76435f8352c85bc6834241210a6107f34ebf7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Mon, 6 Aug 2018 21:26:28 +0800
Subject: [PATCH 065/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 9bdb297b..2b4e688d 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-wechat-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->

From 86c26a7cf6fecd5a5f41b6a82c7aa372f3a50556 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 6 Aug 2018 22:29:40 +0800
Subject: [PATCH 066/172] auto commit

---
 notes/Redis.md          | 15 +++++++++------
 notes/SQL.md            | 24 ++++++++++++------------
 notes/数据库系统原理.md | 27 ++++++++++++++++++---------
 notes/正则表达式.md     | 39 +++++++++++++++++++--------------------
 4 files changed, 58 insertions(+), 47 deletions(-)

diff --git a/notes/Redis.md b/notes/Redis.md
index c54287c8..0981abf0 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -416,7 +416,7 @@ Reids 具体有 6 种淘汰策略：
 | allkeys-random | 从所有数据集中任意选择数据进行淘汰 |
 | noeviction | 禁止驱逐数据 |
 
-作为内存数据库，出于对性能和内存消耗的考虑，Redis 的淘汰算法实际实现上并非针对所有 key，而是抽样一小部分 key 从中选出被淘汰 key。
+作为内存数据库，出于对性能和内存消耗的考虑，Redis 的淘汰算法实际实现上并非针对所有 key，而是抽样一小部分并且从中选出被淘汰的 key。
 
 使用 Redis 缓存数据时，为了提高缓存命中率，需要保证缓存数据都是热点数据。可以将内存最大使用量设置为热点数据占用的内存量，然后启用 allkeys-lru 淘汰策略，将最近最少使用的数据淘汰。
 
@@ -440,7 +440,7 @@ Redis 是内存型数据库，为了保证数据在断电后不会丢失，需
 
 将写命令添加到 AOF 文件（Append Only File）的末尾。
 
-使用 AOF 持久化需要设置同步选项，从而确保写命令什么时候会同步到磁盘文件上。这是因为对硬盘的文件进行写入并不会马上将内容同步到磁盘文件上，而是先存储到缓冲区，然后由操作系统决定什么时候同步到硬盘。有以下同步选项：
+使用 AOF 持久化需要设置同步选项，从而确保写命令什么时候会同步到磁盘文件上。这是因为对文件进行写入并不会马上将内容同步到磁盘上，而是先存储到缓冲区，然后由操作系统决定什么时候同步到磁盘。有以下同步选项：
 
 | 选项 | 同步频率 |
 | :--: | :--: |
@@ -449,8 +449,8 @@ Redis 是内存型数据库，为了保证数据在断电后不会丢失，需
 | no | 让操作系统来决定何时同步 |
 
 - always 选项会严重减低服务器的性能；
-- everysec 选项比较合适，可以保证系统奔溃时只会丢失一秒左右的数据，并且 Redis 每秒执行一次同步对服务器性能几乎没有任何影响；
-- no 选项并不能给服务器性能带来多大的提升，而且也会增加系统奔溃时数据丢失的数量。
+- everysec 选项比较合适，可以保证系统崩溃时只会丢失一秒左右的数据，并且 Redis 每秒执行一次同步对服务器性能几乎没有任何影响；
+- no 选项并不能给服务器性能带来多大的提升，而且也会增加系统崩溃时数据丢失的数量。
 
 随着服务器写请求的增多，AOF 文件会越来越大。Redis 提供了一种将 AOF 重写的特性，能够去除 AOF 文件中的冗余写命令。
 
@@ -553,9 +553,12 @@ Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状
 
 # 十三、分片
 
-分片是将数据划分为多个部分的方法，可以将数据存储到多台机器里面，也可以从多台机器里面获取数据，这种方法在解决某些问题时可以获得线性级别的性能提升。
+分片是将数据划分为多个部分的方法，可以将数据存储到多台机器里面，这种方法在解决某些问题时可以获得线性级别的性能提升。
 
-假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... 等等，有不同的方式来选择一个指定的键存储在哪个实例中。最简单的方式是范围分片，例如用户 id 从 0\~1000 的存储到实例 R0 中，用户 id 从 1001\~2000 的存储到实例 R1 中，等等。但是这样需要维护一张映射范围表，维护操作代价很高。还有一种方式是哈希分片，使用 CRC32 哈希函数将键转换为一个数字，再对实例数量求模就能知道应该存储的实例。
+假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... 等等，有不同的方式来选择一个指定的键存储在哪个实例中。
+
+- 最简单的方式是范围分片，例如用户 id 从 0\~1000 的存储到实例 R0 中，用户 id 从 1001\~2000 的存储到实例 R1 中，等等。但是这样需要维护一张映射范围表，维护操作代价很高。
+- 还有一种方式是哈希分片，使用 CRC32 哈希函数将键转换为一个数字，再对实例数量求模就能知道应该存储的实例。
 
 根据执行分片的位置，可以分为三种分片方式：
 
diff --git a/notes/SQL.md b/notes/SQL.md
index ca9b7b29..4a44aee8 100644
--- a/notes/SQL.md
+++ b/notes/SQL.md
@@ -19,7 +19,7 @@
 * [十八、存储过程](#十八存储过程)
 * [十九、游标](#十九游标)
 * [二十、触发器](#二十触发器)
-* [二十一、事务处理](#二十一事务处理)
+* [二十一、事务管理](#二十一事务管理)
 * [二十二、字符集](#二十二字符集)
 * [二十三、权限管理](#二十三权限管理)
 * [参考资料](#参考资料)
@@ -553,7 +553,7 @@ WHERE col5 = val;
 
 # 十八、存储过程
 
-存储过程可以看成是对一系列 SQL 操作的批处理；
+存储过程可以看成是对一系列 SQL 操作的批处理。
 
 使用存储过程的好处：
 
@@ -642,11 +642,11 @@ SELECT @result; -- 获取结果
 
 DELETE 触发器包含一个名为 OLD 的虚拟表，并且是只读的。
 
-UPDATE 触发器包含一个名为 NEW 和一个名为 OLD 的虚拟表，其中 NEW 是可以被修改地，而 OLD 是只读的。
+UPDATE 触发器包含一个名为 NEW 和一个名为 OLD 的虚拟表，其中 NEW 是可以被修改的，而 OLD 是只读的。
 
 MySQL 不允许在触发器中使用 CALL 语句，也就是不能调用存储过程。
 
-# 二十一、事务处理
+# 二十一、事务管理
 
 基本术语：
 
@@ -708,12 +708,12 @@ SELECT user FROM user;
 
 **创建账户** 
 
+新创建的账户没有任何权限。
+
 ```sql
 CREATE USER myuser IDENTIFIED BY 'mypassword';
 ```
 
-新创建的账户没有任何权限。
-
 **修改账户名** 
 
 ```sql
@@ -734,18 +734,14 @@ SHOW GRANTS FOR myuser;
 
 **授予权限** 
 
+账户用 username@host 的形式定义，username@% 使用的是默认主机名。
+
 ```sql
 GRANT SELECT, INSERT ON mydatabase.* TO myuser;
 ```
 
-账户用 username@host 的形式定义，username@% 使用的是默认主机名。
-
 **删除权限** 
 
-```sql
-REVOKE SELECT, INSERT ON mydatabase.* FROM myuser;
-```
-
 GRANT 和 REVOKE 可在几个层次上控制访问权限：
 
 - 整个服务器，使用 GRANT ALL 和 REVOKE ALL；
@@ -754,6 +750,10 @@ GRANT 和 REVOKE 可在几个层次上控制访问权限：
 - 特定的列；
 - 特定的存储过程。
 
+```sql
+REVOKE SELECT, INSERT ON mydatabase.* FROM myuser;
+```
+
 **更改密码** 
 
 必须使用 Password() 函数
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index cabc2f97..151ffc0f 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -44,10 +44,10 @@
 
 ## 概念
 
-<div align="center"> <img src="../pics//185b9c49-4c13-4241-a848-fbff85c03a64.png" width="400"/> </div><br>
-
 事务指的是满足 ACID 特性的一组操作，可以通过 Commit 提交一个事务，也可以使用 Rollback 进行回滚。
 
+<div align="center"> <img src="../pics//185b9c49-4c13-4241-a848-fbff85c03a64.png" width="400"/> </div><br>
+
 ## ACID
 
 ### 1. 原子性（Atomicity）
@@ -77,7 +77,7 @@
 事务的 ACID 特性概念简单，但不是很好理解，主要是因为这几个特性不是一种平级关系：
 
 - 只有满足一致性，事务的执行结果才是正确的。
-- 在无并发的情况下，事务串行执行，隔离性一定能够满足。此时要只要能满足原子性，就一定能满足一致性。
+- 在无并发的情况下，事务串行执行，隔离性一定能够满足。此时只要能满足原子性，就一定能满足一致性。
 - 在并发的情况下，多个事务并发执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
 - 事务满足持久化是为了能应对数据库崩溃的情况。
 
@@ -123,7 +123,6 @@ T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插
 
 ## 封锁粒度
 
-<div align="center"> <img src="../pics//1a851e90-0d5c-4d4f-ac54-34c20ecfb903.jpg" width="300"/> </div><br>
 
 MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
@@ -133,6 +132,8 @@ MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 在选择封锁粒度时，需要在锁开销和并发程度之间做一个权衡。
 
+<div align="center"> <img src="../pics//1a851e90-0d5c-4d4f-ac54-34c20ecfb903.jpg" width="300"/> </div><br>
+
 ## 封锁类型
 
 ### 1. 读写锁
@@ -304,7 +305,11 @@ SELECT ... FOR UPDATE;
 
 # 五、多版本并发控制
 
-多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC；可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
+多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。
+
+而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC。
+
+可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
 
 ## 版本号
 
@@ -372,11 +377,15 @@ delete;
 
 # 六、Next-Key Locks
 
-Next-Key Locks 也是 MySQL 的 InnoDB 存储引擎的一种锁实现。MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题而存在的。在可重复读（REPEATABLE READ）隔离级别下，使用 MVCC + Next-Key Locks 可以解决幻读问题。
+Next-Key Locks 是 MySQL 的 InnoDB 存储引擎的一种锁实现。
+
+MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题而存在的。在可重复读（REPEATABLE READ）隔离级别下，使用 MVCC + Next-Key Locks 可以解决幻读问题。
 
 ## Record Locks
 
-锁定整个记录（行）。锁定的对象是记录的索引，而不是记录本身。如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
+锁定的对象是记录的索引，而不是记录本身。
+
+如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
 
 ## Gap Locks
 
@@ -420,7 +429,7 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 如果 {A1，A2，... ，An} 是关系的一个或多个属性的集合，该集合函数决定了关系的其它所有属性并且是最小的，那么该集合就称为键码。
 
-对于 A->B，如果能找到 A 的真子集 A'，使得 A'-> B，那么 A->B 就是部分函数依赖，否则就是完全函数依赖；
+对于 A->B，如果能找到 A 的真子集 A'，使得 A'-> B，那么 A->B 就是部分函数依赖，否则就是完全函数依赖。
 
 对于 A->B，B->C，则 A->C 是一个传递函数依赖。
 
@@ -452,7 +461,7 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 ### 1. 第一范式 (1NF)
 
-属性不可分；
+属性不可分。
 
 ### 2. 第二范式 (2NF)
 
diff --git a/notes/正则表达式.md b/notes/正则表达式.md
index c6c21f2c..c713da63 100644
--- a/notes/正则表达式.md
+++ b/notes/正则表达式.md
@@ -23,11 +23,11 @@
 
 # 二、匹配单个字符
 
-正则表达式一般是区分大小写的，但是也有些实现是不区分。
-
 **.**  可以用来匹配任何的单个字符，但是在绝大多数实现里面，不能匹配换行符；
 
-**\\**  是元字符，表示它有特殊的含义，而不是字符本身的含义。如果需要匹配 . ，那么要用 \ 进行转义，即在 . 前面加上 \ 。
+**.**  是元字符，表示它有特殊的含义，而不是字符本身的含义。如果需要匹配 . ，那么要用 \ 进行转义，即在 . 前面加上 \ 。
+
+正则表达式一般是区分大小写的，但是也有些实现是不区分。
 
 **正则表达式** 
 
@@ -43,11 +43,11 @@ My  **name**  is Zheng.
 
 **[ ]**  定义一个字符集合；
 
-0-9、a-z 定义了一个字符区间，区间使用 ASCII 码来确定，字符区间只能用在 [ ] 之间。
+0-9、a-z 定义了一个字符区间，区间使用 ASCII 码来确定，字符区间在 [ ] 中使用。
 
-**-**  元字符只有在 [ ] 之间才是元字符，在 [ ] 之外就是一个普通字符；
+**-**  只有在 [ ] 之间才是元字符，在 [ ] 之外就是一个普通字符；
 
-**^**  在 [ ] 字符集合中是取非操作。
+**^**  在 [ ] 中是取非操作。
 
 **应用** 
 
@@ -78,9 +78,9 @@ abc[^0-9]
 |  \t |  制表符 |
 |  \v |  垂直制表符 |
 
-\r\n 是 Windows 中的文本行结束标签，在 Unix/Linux 则是 \n ；\r\n\r\n 可以匹配 Windows 下的空白行，因为它将匹配两个连续的行尾标签，而这正是两条记录之间的空白行；
+\r\n 是 Windows 中的文本行结束标签，在 Unix/Linux 则是 \n。
 
-. 是元字符，前提是没有对它们进行转义；f 和 n 也是元字符，但是前提是对它们进行了转义。
+\r\n\r\n 可以匹配 Windows 下的空白行，因为它将匹配两个连续的行尾标签，而这正是两条记录之间的空白行；
 
 ## 匹配特定的字符类别
 
@@ -105,11 +105,13 @@ abc[^0-9]
 |  \s | 任何一个空白字符，等价于 [\f\n\r\t\v]  |
 | \S  |  对 \s 取非  |
 
-\x 匹配十六进制字符，\0 匹配八进制，例如 \x0A 对应 ASCII 字符 10 ，等价于 \n，也就是它会匹配 \n 。
+\x 匹配十六进制字符，\0 匹配八进制，例如 \x0A 对应 ASCII 字符 10，等价于 \n。
 
 # 五、重复匹配
 
-**\+**  匹配 1 个或者多个字符， **\*** 匹配 0 个或者多个，**?** 匹配 0 个或者 1 个。
+-  **\+**  匹配 1 个或者多个字符
+-  **\** * 匹配 0 个或者多个
+-  **?**  匹配 0 个或者 1 个
 
 **应用** 
 
@@ -127,16 +129,11 @@ abc[^0-9]
 
 **abc.def<span>@</span>qq.com** 
 
-为了可读性，常常把转义的字符放到字符集合 [ ] 中，但是含义是相同的。
+-  **{n}**  匹配 n 个字符
+-  **{m, n}**  匹配 m\~n 个字符
+-  **{m,}**  至少匹配 m 个字符
 
-```
-[\w.]+@\w+\.\w+
-[\w.]+@[\w]+[\.][\w]+
-```
-
-**{n}**  匹配 n 个字符，**{m, n}** 匹配 m\~n 个字符，**{m,}** 至少匹配 m 个字符；
-
-\* 和 + 都是贪婪型元字符，会匹配最多的内容，在元字符后面加 ? 可以转换为懒惰型元字符，例如 \*?、+? 和 {m, n}? 。
+\* 和 + 都是贪婪型元字符，会匹配最多的内容。在后面加 ? 可以转换为懒惰型元字符，例如 \*?、+? 和 {m, n}? 。
 
 **正则表达式** 
 
@@ -220,7 +217,9 @@ a.+c
 
 **应用** 
 
-匹配 IP 地址。IP 地址中每部分都是 0-255 的数字，用正则表达式匹配时以下情况是合法的：
+匹配 IP 地址。
+
+IP 地址中每部分都是 0-255 的数字，用正则表达式匹配时以下情况是合法的：
 
 - 一位数字
 - 不以 0 开头的两位数字

From f75f18722b8fcf3ca3f969375e18ea2d90db73fa Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Mon, 6 Aug 2018 23:15:46 +0800
Subject: [PATCH 067/172] =?UTF-8?q?Create=20leetcode=E6=80=BB=E7=BB=93.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 other/leetcode总结.md | 37 +++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 other/leetcode总结.md

diff --git a/other/leetcode总结.md b/other/leetcode总结.md
new file mode 100644
index 00000000..a864b791
--- /dev/null
+++ b/other/leetcode总结.md
@@ -0,0 +1,37 @@
+# LeetCode 面试必备
+ - 💪 就是干！如果你觉得有帮助请点个star，谢谢！
+
+> **欢迎任何人参与和完善：一个人可以走的很快，但是一群人却可以走的更远**
+
+## LeetCode 习题集合
+
+* [LeetCode 解题集合](https://github.com/apachecn/leetcode/docs/Leetcode_Solutions)
+
+
+## 模版要求
+
+> 提交PR基本要求（满足任意一种即可）
+
+* 1. 不一样的思路
+* 2. 优化时间复杂度和空间复杂度，或者解决题目的Follow up
+* 3. 有意义的简化代码
+* 4. 未提交过的题目
+
+> **案例模版**
+
+[模版：007. Reverse Integer 反转整数](https://github.com/apachecn/leetcode/docs/Leetcode_Solutions/007._Reverse_Integer.md)
+
+
+## 项目贡献者
+
+> 项目发起人
+
+* [@Lisanaaa](https://github.com/Lisanaaa)
+* [@片刻](https://github.com/jiangzhonglian)
+
+> 贡献者（欢迎大家来追加）
+
+* [@Lisanaaa](https://github.com/Lisanaaa)
+* [@片刻](https://github.com/jiangzhonglian)
+* [@小瑶](https://github.com/chenyyx)
+

From 963b1fccdf92f44ca249d57d16ca8ffc6ba083e0 Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Mon, 6 Aug 2018 23:17:13 +0800
Subject: [PATCH 068/172] =?UTF-8?q?Rename=20leetcode=E6=80=BB=E7=BB=93.md?=
 =?UTF-8?q?=20to=20leetcode=20=E6=80=BB=E7=BB=93.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 other/{leetcode总结.md => leetcode 总结.md} | 0
 1 file changed, 0 insertions(+), 0 deletions(-)
 rename other/{leetcode总结.md => leetcode 总结.md} (100%)

diff --git a/other/leetcode总结.md b/other/leetcode 总结.md
similarity index 100%
rename from other/leetcode总结.md
rename to other/leetcode 总结.md

From b98c2a709474a637752a28eab02f5b7804390c62 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 6 Aug 2018 23:23:25 +0800
Subject: [PATCH 069/172] auto commit

---
 notes/计算机网络.md | 14 ++++++++++----
 1 file changed, 10 insertions(+), 4 deletions(-)

diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index 9b213d74..91a4d84c 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -415,12 +415,14 @@ IP 地址 ::= {< 网络号 >, < 主机号 >}
 
 ### 2. 子网划分
 
-通过在主机号字段中拿一部分作为子网号，把两级 IP 地址划分为三级 IP 地址。注意，外部网络看不到子网的存在。
+通过在主机号字段中拿一部分作为子网号，把两级 IP 地址划分为三级 IP 地址。
 
 IP 地址 ::= {< 网络号 >, < 子网号 >, < 主机号 >}
 
 要使用子网，必须配置子网掩码。一个 B 类地址的默认子网掩码为 255.255.0.0，如果 B 类地址的子网占两个比特，那么子网掩码为 11111111 11111111 11000000 00000000，也就是 255.255.192.0。
 
+注意，外部网络看不到子网的存在。
+
 ### 3. 无分类
 
 无分类编址 CIDR 消除了传统 A 类、B 类和 C 类地址以及划分子网的概念，使用网络前缀和主机号来对 IP 地址进行编码，网络前缀的长度可以根据需要变化。
@@ -445,7 +447,7 @@ ARP 实现由 IP 地址得到 MAC 地址。
 
 <div align="center"> <img src="../pics//b9d79a5a-e7af-499b-b989-f10483e71b8b.jpg" width="500"/> </div><br>
 
-每个主机都有一个 ARP 高速缓存，里面有本局域网上的各主机和路由器的 IP 地址到硬件地址的映射表。
+每个主机都有一个 ARP 高速缓存，里面有本局域网上的各主机和路由器的 IP 地址到 MAC 地址的映射表。
 
 如果主机 A 知道主机 B 的 IP 地址，但是 ARP 高速缓存中没有该 IP 地址到 MAC 地址的映射，此时主机 A 通过广播的方式发送 ARP 请求分组，主机 B 收到该请求后会发送 ARP 响应分组给主机 A 告知其 MAC 地址，随后主机 A 向其高速缓存中写入主机 B 的 IP 地址到 MAC 地址的映射。
 
@@ -465,12 +467,14 @@ ICMP 报文分为差错报告报文和询问报文。
 
 Ping 是 ICMP 的一个重要应用，主要用来测试两台主机之间的连通性。
 
-Ping 发送的 IP 数据报封装的是无法交付的 UDP 用户数据报。
+Ping 的原理是通过向目的主机发送 ICMP Echo 请求报文，目的主机收到之后会发送 Echo 回答报文。Ping 会根据时间和成功响应的次数估算出数据包往返时间以及丢包率。
 
 ### 2. Traceroute
 
 Traceroute 是 ICMP 的另一个应用，用来跟踪一个分组从源点到终点的路径。
 
+Traceroute 发送的 IP 数据报封装的是无法交付的 UDP 用户数据报，并由目的主机发送终点不可达差错报告报文。
+
 - 源主机向目的主机发送一连串的 IP 数据报。第一个数据报 P1 的生存时间 TTL 设置为 1，当 P1 到达路径上的第一个路由器 R1 时，R1 收下它并把 TTL 减 1，此时 TTL 等于 0，R1 就把 P1 丢弃，并向源主机发送一个 ICMP 时间超过差错报告报文；
 - 源主机接着发送第二个数据报 P2，并把 TTL 设置为 2。P2 先到达 R1，R1 收下后把 TTL 减 1 再转发给 R2，R2 收下后也把 TTL 减 1，由于此时 TTL 等于 0，R2 就丢弃 P2，并向源主机发送一个 ICMP 时间超过差错报文。
 - 不断执行这样的步骤，直到最后一个数据报刚刚到达目的主机，主机不转发数据报，也不把 TTL 值减 1。但是因为数据报封装的是无法交付的 UDP，因此目的主机要向源主机发送 ICMP 终点不可达差错报告报文。
@@ -883,7 +887,9 @@ DHCP 工作过程如下：
 - JamesF.Kurose, KeithW.Ross, 库罗斯, 等. 计算机网络: 自顶向下方法 [M]. 机械工业出版社, 2014.
 - W.RichardStevens. TCP/IP 详解. 卷 1, 协议 [M]. 机械工业出版社, 2006.
 - [Active vs Passive FTP Mode: Which One is More Secure?](https://securitywing.com/active-vs-passive-ftp-mode/)
-- [Active and Passive FTP Transfers Defined  -  KB Article #1138](http://www.serv-u.com/kb/1138/active-and-passive-ftp-transfers-defined)
+- [Active and Passive FTP Transfers Defined - KB Article #1138](http://www.serv-u.com/kb/1138/active-and-passive-ftp-transfers-defined)
+- [Traceroute](https://zh.wikipedia.org/wiki/Traceroute)
+- [ping](https://zh.wikipedia.org/wiki/Ping)
 - [How DHCP works and DHCP Interview Questions and Answers](http://webcache.googleusercontent.com/search?q=cache:http://anandgiria.blogspot.com/2013/09/windows-dhcp-interview-questions-and.html)
 - [What is process of DORA in DHCP?](https://www.quora.com/What-is-process-of-DORA-in-DHCP)
 - [What is DHCP Server ?](https://tecadmin.net/what-is-dhcp-server/)

From a06245c01c7391b36f71159e7a4ec49b383050f2 Mon Sep 17 00:00:00 2001
From: Elong <peielcn@gmail.com>
Date: Tue, 7 Aug 2018 10:39:08 +0800
Subject: [PATCH 070/172] =?UTF-8?q?Update=20Java=20=E5=AE=B9=E5=99=A8.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

8和16的二进制表示问题
---
 notes/Java 容器.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 27a783f1..24ce2405 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -680,8 +680,8 @@ void transfer(Entry[] newTable) {
 假设原数组长度 capacity 为 8，扩容之后 new capacity 为 16：
 
 ```html
-capacity     : 00010000
-new capacity : 00100000
+capacity     : 00001000
+new capacity : 00010000
 ```
 
 对于一个 Key，它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；如果为 1，那么得到的结果为原来的结果 + 8。

From 8c67483fe9ac599aa6f6358fb32a541d7e279666 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 7 Aug 2018 11:12:35 +0800
Subject: [PATCH 071/172] auto commit

---
 notes/Java 容器.md  |  8 ++--
 notes/计算机网络.md | 94 +++++++++++++++++++++++----------------------
 2 files changed, 52 insertions(+), 50 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 24ce2405..bf1921fc 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -677,14 +677,14 @@ void transfer(Entry[] newTable) {
 
 在进行扩容时，需要把键值对重新放到对应的桶上。HashMap 使用了一个特殊的机制，可以降低重新计算桶下标的操作。
 
-假设原数组长度 capacity 为 8，扩容之后 new capacity 为 16：
+假设原数组长度 capacity 为 16，扩容之后 new capacity 为 32：
 
 ```html
-capacity     : 00001000
-new capacity : 00010000
+capacity     : 00010000
+new capacity : 00100000
 ```
 
-对于一个 Key，它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；如果为 1，那么得到的结果为原来的结果 + 8。
+对于一个 Key，它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；如果为 1，那么得到的结果为原来的结果 +16。
 
 ### 7. 扩容-计算数组容量
 
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index 91a4d84c..e80ba7f8 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -37,16 +37,16 @@
     * [TCP 首部格式](#tcp-首部格式)
     * [TCP 的三次握手](#tcp-的三次握手)
     * [TCP 的四次挥手](#tcp-的四次挥手)
-    * [TCP 滑动窗口](#tcp-滑动窗口)
     * [TCP 可靠传输](#tcp-可靠传输)
+    * [TCP 滑动窗口](#tcp-滑动窗口)
     * [TCP 流量控制](#tcp-流量控制)
     * [TCP 拥塞控制](#tcp-拥塞控制)
 * [六、应用层](#六应用层)
     * [域名系统](#域名系统)
     * [文件传送协议](#文件传送协议)
+    * [动态主机配置协议](#动态主机配置协议)
     * [远程登录协议](#远程登录协议)
     * [电子邮件协议](#电子邮件协议)
-    * [动态主机配置协议](#动态主机配置协议)
     * [常用端口](#常用端口)
     * [Web 页面请求过程](#web-页面请求过程)
 * [参考资料](#参考资料)
@@ -622,11 +622,11 @@ BGP 只能寻找一条比较好的路由，而不是最佳路由。
 
 - 首先 B 处于 LISTEN（监听）状态，等待客户的连接请求。
 
-- A 向 B 发送连接请求报文段，SYN=1，ACK=0，选择一个初始的序号 x。
+- A 向 B 发送连接请求报文，SYN=1，ACK=0，选择一个初始的序号 x。
 
-- B 收到连接请求报文段，如果同意建立连接，则向 A 发送连接确认报文段，SYN=1，ACK=1，确认号为 x+1，同时也选择一个初始的序号 y。
+- B 收到连接请求报文，如果同意建立连接，则向 A 发送连接确认报文，SYN=1，ACK=1，确认号为 x+1，同时也选择一个初始的序号 y。
 
-- A 收到 B 的连接确认报文段后，还要向 B 发出确认，确认号为 y+1，序号为 x+1。
+- A 收到 B 的连接确认报文后，还要向 B 发出确认，确认号为 y+1，序号为 x+1。
 
 - B 收到 A 的确认后，连接建立。
 
@@ -642,11 +642,11 @@ BGP 只能寻找一条比较好的路由，而不是最佳路由。
 
 以下描述不讨论序号和确认号，因为序号和确认号的规则比较简单。并且不讨论 ACK，因为 ACK 在连接建立之后都为 1。
 
-- A 发送连接释放报文段，FIN=1。
+- A 发送连接释放报文，FIN=1。
 
 - B 收到之后发出确认，此时 TCP 属于半关闭状态，B 能向 A 发送数据但是 A 不能向 B 发送数据。
 
-- 当 B 不再需要连接时，发送连接释放请求报文段，FIN=1。
+- 当 B 不再需要连接时，发送连接释放报文，FIN=1。
 
 - A 收到后发出确认，进入 TIME-WAIT 状态，等待 2 MSL（最大报文存活时间）后释放连接。
 
@@ -660,19 +660,9 @@ BGP 只能寻找一条比较好的路由，而不是最佳路由。
 
 客户端接收到服务器端的 FIN 报文后进入此状态，此时并不是直接进入 CLOSED 状态，还需要等待一个时间计时器设置的时间 2MSL。这么做有两个理由：
 
-- 确保最后一个确认报文段能够到达。如果 B 没收到 A 发送来的确认报文段，那么就会重新发送连接释放请求报文段，A 等待一段时间就是为了处理这种情况的发生。
+- 确保最后一个确认报文能够到达。如果 B 没收到 A 发送来的确认报文，那么就会重新发送连接释放请求报文，A 等待一段时间就是为了处理这种情况的发生。
 
-- 等待一段时间是为了让本连接持续时间内所产生的所有报文段都从网络中消失，使得下一个新的连接不会出现旧的连接请求报文段。
-
-## TCP 滑动窗口
-
-<div align="center"> <img src="../pics//a3253deb-8d21-40a1-aae4-7d178e4aa319.jpg" width="800"/> </div><br>
-
-窗口是缓存的一部分，用来暂时存放字节流。发送方和接收方各有一个窗口，接收方通过 TCP 报文段中的窗口字段告诉发送方自己的窗口大小，发送方根据这个值和其它信息设置自己的窗口大小。
-
-发送窗口内的字节都允许被发送，接收窗口内的字节都允许被接收。如果发送窗口左部的字节已经发送并且收到了确认，那么就将发送窗口向右滑动一定距离，直到左部第一个字节不是已发送并且已确认的状态；接收窗口的滑动类似，接收窗口左部字节已经发送确认并交付主机，就向右滑动接收窗口。
-
-接收窗口只会对窗口内最后一个按序到达的字节进行确认，例如接收窗口已经收到的字节为 {31, 34, 35}，其中 {31} 按序到达，而 {34, 35} 就不是，因此只对字节 31 进行确认。发送方得到一个字节的确认之后，就知道这个字节之前的所有字节都已经被接收。
+- 等待一段时间是为了让本连接持续时间内所产生的所有报文都从网络中消失，使得下一个新的连接不会出现旧的连接请求报文。
 
 ## TCP 可靠传输
 
@@ -688,6 +678,16 @@ TCP 使用超时重传来实现可靠传输：如果一个已经发送的报文
 
 其中 RTT<sub>d</sub> 为偏差。
 
+## TCP 滑动窗口
+
+窗口是缓存的一部分，用来暂时存放字节流。发送方和接收方各有一个窗口，接收方通过 TCP 报文段中的窗口字段告诉发送方自己的窗口大小，发送方根据这个值和其它信息设置自己的窗口大小。
+
+发送窗口内的字节都允许被发送，接收窗口内的字节都允许被接收。如果发送窗口左部的字节已经发送并且收到了确认，那么就将发送窗口向右滑动一定距离，直到左部第一个字节不是已发送并且已确认的状态；接收窗口的滑动类似，接收窗口左部字节已经发送确认并交付主机，就向右滑动接收窗口。
+
+接收窗口只会对窗口内最后一个按序到达的字节进行确认，例如接收窗口已经收到的字节为 {31, 34, 35}，其中 {31} 按序到达，而 {34, 35} 就不是，因此只对字节 31 进行确认。发送方得到一个字节的确认之后，就知道这个字节之前的所有字节都已经被接收。
+
+<div align="center"> <img src="../pics//a3253deb-8d21-40a1-aae4-7d178e4aa319.jpg" width="800"/> </div><br>
+
 ## TCP 流量控制
 
 流量控制是为了控制发送方发送速率，保证接收方来得及接收。
@@ -725,7 +725,9 @@ TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、
 
 在发送方，如果收到三个重复确认，那么可以知道下一个报文段丢失，此时执行快重传，立即重传下一个报文段。例如收到三个 M<sub>2</sub>，则 M<sub>3</sub> 丢失，立即重传 M<sub>3</sub>。
 
-在这种情况下，只是丢失个别报文段，而不是网络拥塞。因此执行快恢复，令 ssthresh = cwnd/2 ，cwnd = ssthresh，注意到此时直接进入拥塞避免。慢开始和快恢复的快慢指的是 cwnd 的设定值，而不是 cwnd 的增长速率。慢开始 cwnd 设定为 1，而快恢复 cwnd 设定为 ssthresh。
+在这种情况下，只是丢失个别报文段，而不是网络拥塞。因此执行快恢复，令 ssthresh = cwnd/2 ，cwnd = ssthresh，注意到此时直接进入拥塞避免。
+
+慢开始和快恢复的快慢指的是 cwnd 的设定值，而不是 cwnd 的增长速率。慢开始 cwnd 设定为 1，而快恢复 cwnd 设定为 ssthresh。
 
 <div align="center"> <img src="../pics//f61b5419-c94a-4df1-8d4d-aed9ae8cc6d5.png" width="600"/> </div><br>
 
@@ -733,23 +735,23 @@ TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、
 
 ## 域名系统
 
-DNS 是一个分布式数据库，提供了主机名和 IP 地址之间的转换。这里的分布式数据库是指，每个站点只保留它自己的那部分数据。
+DNS 是一个分布式数据库，提供了主机名和 IP 地址之间相互转换的服务。这里的分布式数据库是指，每个站点只保留它自己的那部分数据。
 
-域名具有层次结构，从上到下依次为：根域名、顶级域名、第二级域名。
+域名具有层次结构，从上到下依次为：根域名、顶级域名、二级域名。
 
 <div align="center"> <img src="../pics//b54eeb16-0b0e-484c-be62-306f57c40d77.jpg"/> </div><br>
 
 DNS 可以使用 UDP 或者 TCP 进行传输，使用的端口号都为 53。大多数情况下 DNS 使用 UDP 进行传输，这就要求域名解析器和域名服务器都必须自己处理超时和重传来保证可靠性。在两种情况下会使用 TCP 进行传输：
 
-- 因为 UDP 最大只支持 512 字节的数据，如果返回的响应超过的 512 字节就改用 TCP 进行传输。
-- 区域传送是主域名服务器向辅助域名服务器传送变化的那部分数据，区域传送需要使用 TCP 进行传输。
+- 如果返回的响应超过的 512 字节就改用 TCP 进行传输（UDP 最大只支持 512 字节的数据）。
+- 区域传送需要使用 TCP 进行传输（区域传送是主域名服务器向辅助域名服务器传送变化的那部分数据）。
 
 ## 文件传送协议
 
 FTP 使用 TCP 进行连接，它需要两个连接来传送一个文件：
 
-- 控制连接：服务器以打开端口号 21 等待客户端的连接，客户端主动建立连接后，使用这个连接将客户端的命令传送给服务器，并传回服务器的应答。
-- 数据连接：用来传送一个文件。
+- 控制连接：服务器打开端口号 21 等待客户端的连接，客户端主动建立连接后，使用这个连接将客户端的命令传送给服务器，并传回服务器的应答。
+- 数据连接：用来传送一个文件数据。
 
 根据数据连接是否是服务器端主动建立，FTP 有主动和被动两种模式：
 
@@ -763,6 +765,21 @@ FTP 使用 TCP 进行连接，它需要两个连接来传送一个文件：
 
 主动模式要求客户端开放端口号给服务器端，需要去配置客户端的防火墙。被动模式只需要服务器端开放端口号即可，无需客户端配置防火墙。但是被动模式会导致服务器端的安全性减弱，因为开放了过多的端口号。
 
+## 动态主机配置协议
+
+DHCP (Dynamic Host Configuration Protocol) 提供了即插即用的连网方式，用户不再需要去手动配置 IP 地址等信息。
+
+DHCP 配置的内容不仅是 IP 地址，还包括子网掩码、网关 IP 地址。
+
+DHCP 工作过程如下：
+
+1. 客户端发送 Discover 报文，该报文的目的地址为 255.255.255.255:67，源地址为 0.0.0.0:68，被放入 UDP 中，该报文被广播到同一个子网的所有主机上。如果客户端和 DHCP 服务器不在同一个子网，就需要使用中继代理。
+2. DHCP 服务器收到 Discover 报文之后，发送 Offer 报文给客户端，该报文包含了客户端所需要的信息。因为客户端可能收到多个 DHCP 服务器提供的信息，因此客户端需要进行选择。
+3. 如果客户端选择了某个 DHCP 服务器提供的信息，那么就发送 Request 报文给该 DHCP 服务器。
+4. DHCP 服务器发送 Ack 报文，表示客户端此时可以使用提供给它的信息。
+
+<div align="center"> <img src="../pics//bf16c541-0717-473b-b75d-4115864f4fbf.jpg"/> </div><br>
+
 ## 远程登录协议
 
 TELNET 用于登录到远程主机上，并且远程主机上的输出也会返回。
@@ -777,34 +794,19 @@ TELNET 可以适应许多计算机和操作系统的差异，例如不同操作
 
 <div align="center"> <img src="../pics//7b3efa99-d306-4982-8cfb-e7153c33aab4.png" width="700"/> </div><br>
 
-### 1. POP3
-
-POP3 的特点是只要用户从服务器上读取了邮件，就把该邮件删除。
-
-### 2. IMAP
-
-IMAP 协议中客户端和服务器上的邮件保持同步，如果不去手动删除邮件，那么服务器上的邮件也不会被删除。IMAP 这种做法可以让用户随时随地去访问服务器上的邮件。
-
-### 3. SMTP
+### 1. SMTP
 
 SMTP 只能发送 ASCII 码，而互联网邮件扩充 MIME 可以发送二进制文件。MIME 并没有改动或者取代 SMTP，而是增加邮件主体的结构，定义了非 ASCII 码的编码规则。
 
 <div align="center"> <img src="../pics//ed5522bb-3a60-481c-8654-43e7195a48fe.png" width=""/> </div><br>
 
-## 动态主机配置协议
+### 2. POP3
 
-DHCP (Dynamic Host Configuration Protocol) 提供了即插即用的连网方式，用户不再需要去手动配置 IP 地址等信息。DHCP 配置的内容不仅是 IP 地址，还包括子网掩码、网关 IP 地址。
+POP3 的特点是只要用户从服务器上读取了邮件，就把该邮件删除。
 
-DHCP 工作过程如下：
+### 3. IMAP
 
-1. 客户端发送 Discover 报文，该报文的目的地址为 255.255.255.255:67，源地址为 0.0.0.0:68，被放入 UDP 中，该报文被广播到同一个子网的所有主机上。
-2. DHCP 服务器收到 Discover 报文之后，发送 Offer 报文给客户端，该报文包含了客户端所需要的信息。因为客户端可能收到多个 DHCP 服务器提供的信息，因此客户端需要进行选择。
-3. 如果客户端选择了某个 DHCP 服务器提供的信息，那么就发送 Request 报文给该 DHCP 服务器。
-4. DHCP 服务器发送 Ack 报文，表示客户端此时可以使用提供给它的信息。
-
-<div align="center"> <img src="../pics//bf16c541-0717-473b-b75d-4115864f4fbf.jpg"/> </div><br>
-
-如果客户端和 DHCP 服务器不在同一个子网，就需要使用中继代理。
+IMAP 协议中客户端和服务器上的邮件保持同步，如果不去手动删除邮件，那么服务器上的邮件也不会被删除。IMAP 这种做法可以让用户随时随地去访问服务器上的邮件。
 
 ## 常用端口
 

From a17a00fb22e38c47f53c339f8a980dfbf9852d63 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 7 Aug 2018 12:37:46 +0800
Subject: [PATCH 072/172] auto commit

---
 notes/Java IO.md        |  549 ---------------
 notes/Java 基础.md      |   34 +-
 notes/Java 容器.md      | 1052 -----------------------------
 notes/Redis.md          |  602 -----------------
 notes/SQL.md            |  763 ---------------------
 notes/分布式.md         |  355 ----------
 notes/数据库系统原理.md |  576 ----------------
 notes/正则表达式.md     |  385 -----------
 notes/消息队列.md       |   76 ---
 notes/算法.md           |    2 +-
 notes/系统设计基础.md   |  100 ---
 notes/缓存.md           |   13 +-
 notes/计算机网络.md     |  886 ------------------------
 notes/重构.md           | 1416 ---------------------------------------
 notes/集群.md           |  190 ------
 15 files changed, 27 insertions(+), 6972 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index 08dc80f4..649f563a 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -1,553 +1,4 @@
 <!-- GFM-TOC -->
-* [一、概览](#一概览)
-* [二、磁盘操作](#二磁盘操作)
-* [三、字节操作](#三字节操作)
-* [四、字符操作](#四字符操作)
-* [五、对象操作](#五对象操作)
-* [六、网络操作](#六网络操作)
-    * [InetAddress](#inetaddress)
-    * [URL](#url)
-    * [Sockets](#sockets)
-    * [Datagram](#datagram)
-* [七、NIO](#七nio)
-    * [流与块](#流与块)
-    * [通道与缓冲区](#通道与缓冲区)
-    * [缓冲区状态变量](#缓冲区状态变量)
-    * [文件 NIO 实例](#文件-nio-实例)
-    * [选择器](#选择器)
-    * [套接字 NIO 实例](#套接字-nio-实例)
-    * [内存映射文件](#内存映射文件)
-    * [对比](#对比)
-* [八、参考资料](#八参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、概览
-
-Java 的 I/O 大概可以分成以下几类：
-
-- 磁盘操作：File
-- 字节操作：InputStream 和 OutputStream
-- 字符操作：Reader 和 Writer
-- 对象操作：Serializable
-- 网络操作：Socket
-- 新的输入/输出：NIO
-
-# 二、磁盘操作
-
-File 类可以用于表示文件和目录的信息，但是它不表示文件的内容。
-
-递归地输出一个目录下所有文件：
-
-```java
-public static void listAllFiles(File dir)
-{
-    if (dir == null || !dir.exists()) {
-        return;
-    }
-    if (dir.isFile()) {
-        System.out.println(dir.getName());
-        return;
-    }
-    for (File file : dir.listFiles()) {
-        listAllFiles(file);
-    }
-}
-```
-
-# 三、字节操作
-
-使用字节流操作进行文件复制：
-
-```java
-public static void copyFile(String src, String dist) throws IOException
-{
-    FileInputStream in = new FileInputStream(src);
-    FileOutputStream out = new FileOutputStream(dist);
-    byte[] buffer = new byte[20 * 1024];
-    // read() 最多读取 buffer.length 个字节
-    // 返回的是实际读取的个数
-    // 返回 -1 的时候表示读到 eof，即文件尾
-    while (in.read(buffer, 0, buffer.length) != -1) {
-        out.write(buffer);
-    }
-    in.close();
-    out.close();
-}
-```
-
-<div align="center"> <img src="../pics//DP-Decorator-java.io.png" width="500"/> </div><br>
-
-Java I/O 使用了装饰者模式来实现。以 InputStream 为例，InputStream 是抽象组件，FileInputStream 是 InputStream 的子类，属于具体组件，提供了字节流的输入操作。FilterInputStream 属于抽象装饰者，装饰者用于装饰组件，为组件提供额外的功能，例如 BufferedInputStream 为 FileInputStream 提供缓存的功能。
-
-实例化一个具有缓存功能的字节流对象时，只需要在 FileInputStream 对象上再套一层 BufferedInputStream 对象即可。
-
-```java
-FileInputStream fileInputStream = new FileInputStream(filePath);
-BufferedInputStream bufferedInputStream = new BufferedInputStream(fileInputStream);
-```
-
-DataInputStream 装饰者提供了对更多数据类型进行输入的操作，比如 int、double 等基本类型。
-
-# 四、字符操作
-
-不管是磁盘还是网络传输，最小的存储单元都是字节，而不是字符。但是在程序中操作的通常是字符形式的数据，因此需要提供对字符进行操作的方法。
-
-- InputStreamReader 实现从字节流解码成字符流；
-- OutputStreamWriter 实现字符流编码成为字节流。
-
-逐行输出文本文件的内容：
-
-```java
-public static void readFileContent(String filePath) throws IOException
-{
-    FileReader fileReader = new FileReader(filePath);
-    BufferedReader bufferedReader = new BufferedReader(fileReader);
-    String line;
-    while ((line = bufferedReader.readLine()) != null) {
-        System.out.println(line);
-    }
-    // 装饰者模式使得 BufferedReader 组合了一个 Reader 对象
-    // 在调用 BufferedReader 的 close() 方法时会去调用 fileReader 的 close() 方法
-    // 因此只要一个 close() 调用即可
-    bufferedReader.close();
-}
-```
-
-编码就是把字符转换为字节，而解码是把字节重新组合成字符。
-
-如果编码和解码过程使用不同的编码方式那么就出现了乱码。
-
-- GBK 编码中，中文字符占 2 个字节，英文字符占 1 个字节；
-- UTF-8 编码中，中文字符占 3 个字节，英文字符占 1 个字节；
-- UTF-16be 编码中，中文字符和英文字符都占 2 个字节。
-
-UTF-16be 中的 be 指的是 Big Endian，也就是大端。相应地也有 UTF-16le，le 指的是 Little Endian，也就是小端。
-
-Java 使用双字节编码 UTF-16be，这不是指 Java 只支持这一种编码方式，而是说 char 这种类型使用 UTF-16be 进行编码。char 类型占 16 位，也就是两个字节，Java 使用这种双字节编码是为了让一个中文或者一个英文都能使用一个 char 来存储。
-
-String 可以看成一个字符序列，可以指定一个编码方式将它转换为字节序列，也可以指定一个编码方式将一个字节序列转换为 String。
-
-```java
-String str1 = "中文";
-byte[] bytes = str1.getBytes("UTF-8");
-String str2 = new String(bytes, "UTF-8");
-System.out.println(str2);
-```
-
-在调用无参数 getBytes() 方法时，默认的编码方式不是 UTF-16be。双字节编码的好处是可以使用一个 char 存储中文和英文，而将 String 转为 bytes[] 字节数组就不再需要这个好处，因此也就不再需要双字节编码。getBytes() 的默认编码方式与平台有关，一般为 UTF-8。
-
-```java
-byte[] bytes = str1.getBytes();
-```
-
-# 五、对象操作
-
-序列化就是将一个对象转换成字节序列，方便存储和传输。
-
-- 序列化：ObjectOutputStream.writeObject()
-- 反序列化：ObjectInputStream.readObject()
-
-序列化的类需要实现 Serializable 接口，它只是一个标准，没有任何方法需要实现，但是如果不去实现它的话而进行序列化，会抛出异常。
-
-```java
-public static void main(String[] args) throws IOException, ClassNotFoundException
-{
-    A a1 = new A(123, "abc");
-    String objectFile = "file/a1";
-    ObjectOutputStream objectOutputStream = new ObjectOutputStream(new FileOutputStream(objectFile));
-    objectOutputStream.writeObject(a1);
-    objectOutputStream.close();
-
-    ObjectInputStream objectInputStream = new ObjectInputStream(new FileInputStream(objectFile));
-    A a2 = (A) objectInputStream.readObject();
-    objectInputStream.close();
-    System.out.println(a2);
-}
-
-private static class A implements Serializable
-{
-    private int x;
-    private String y;
-
-    A(int x, String y)
-    {
-        this.x = x;
-        this.y = y;
-    }
-
-    @Override
-    public String toString()
-    {
-        return "x = " + x + "  " + "y = " + y;
-    }
-}
-```
-
-不会对静态变量进行序列化，因为序列化只是保存对象的状态，静态变量属于类的状态。
-
-transient 关键字可以使一些属性不会被序列化。
-
-**ArrayList 序列化和反序列化的实现** ：ArrayList 中存储数据的数组是用 transient 修饰的，因为这个数组是动态扩展的，并不是所有的空间都被使用，因此就不需要所有的内容都被序列化。通过重写序列化和反序列化方法，使得可以只序列化数组中有内容的那部分数据。
-
-```java
-private transient Object[] elementData;
-```
-
-# 六、网络操作
-
-Java 中的网络支持：
-
-- InetAddress：用于表示网络上的硬件资源，即 IP 地址；
-- URL：统一资源定位符；
-- Sockets：使用 TCP 协议实现网络通信；
-- Datagram：使用 UDP 协议实现网络通信。
-
-## InetAddress
-
-没有公有构造函数，只能通过静态方法来创建实例。
-
-```java
-InetAddress.getByName(String host);
-InetAddress.getByAddress(byte[] address);
-```
-
-## URL
-
-可以直接从 URL 中读取字节流数据。
-
-```java
-public static void main(String[] args) throws IOException
-{
-    URL url = new URL("http://www.baidu.com");
-    // 字节流
-    InputStream is = url.openStream();
-    // 字符流
-    InputStreamReader isr = new InputStreamReader(is, "utf-8");
-    BufferedReader br = new BufferedReader(isr);
-    String line = br.readLine();
-    while (line != null) {
-        System.out.println(line);
-        line = br.readLine();
-    }
-    br.close();
-}
-```
-
-## Sockets
-
-- ServerSocket：服务器端类
-- Socket：客户端类
-- 服务器和客户端通过 InputStream 和 OutputStream 进行输入输出。
-
-<div align="center"> <img src="../pics//ClienteServidorSockets1521731145260.jpg"/> </div><br>
-
-## Datagram
-
-- DatagramPacket：数据包类
-- DatagramSocket：通信类
-
-# 七、NIO
-
-- [Java NIO Tutorial](http://tutorials.jenkov.com/java-nio/index.html)
-- [Java NIO 浅析](https://tech.meituan.com/nio.html)
-- [IBM: NIO 入门](https://www.ibm.com/developerworks/cn/education/java/j-nio/j-nio.html)
-
-新的输入/输出 (NIO) 库是在 JDK 1.4 中引入的。NIO 弥补了原来的 I/O 的不足，提供了高速的、面向块的 I/O。
-
-## 流与块
-
-I/O 与 NIO 最重要的区别是数据打包和传输的方式，I/O 以流的方式处理数据，而 NIO 以块的方式处理数据。
-
-面向流的 I/O 一次处理一个字节数据：一个输入流产生一个字节数据，一个输出流消费一个字节数据。为流式数据创建过滤器非常容易，链接几个过滤器，以便每个过滤器只负责复杂处理机制的一部分。不利的一面是，面向流的 I/O 通常相当慢。
-
-面向块的 I/O 一次处理一个数据块，按块处理数据比按流处理数据要快得多。但是面向块的 I/O 缺少一些面向流的 I/O 所具有的优雅性和简单性。
-
-I/O 包和 NIO 已经很好地集成了，java.io.\* 已经以 NIO 为基础重新实现了，所以现在它可以利用 NIO 的一些特性。例如，java.io.\* 包中的一些类包含以块的形式读写数据的方法，这使得即使在面向流的系统中，处理速度也会更快。
-
-## 通道与缓冲区
-
-### 1. 通道
-
-通道 Channel 是对原 I/O 包中的流的模拟，可以通过它读取和写入数据。
-
-通道与流的不同之处在于，流只能在一个方向上移动，(一个流必须是 InputStream 或者 OutputStream 的子类)，而通道是双向的，可以用于读、写或者同时用于读写。
-
-通道包括以下类型：
-
-- FileChannel：从文件中读写数据；
-- DatagramChannel：通过 UDP 读写网络中数据；
-- SocketChannel：通过 TCP 读写网络中数据；
-- ServerSocketChannel：可以监听新进来的 TCP 连接，对每一个新进来的连接都会创建一个 SocketChannel。
-
-### 2. 缓冲区
-
-发送给一个通道的所有数据都必须首先放到缓冲区中，同样地，从通道中读取的任何数据都要先读到缓冲区中。也就是说，不会直接对通道进行读写数据，而是要先经过缓冲区。
-
-缓冲区实质上是一个数组，但它不仅仅是一个数组。缓冲区提供了对数据的结构化访问，而且还可以跟踪系统的读/写进程。
-
-缓冲区包括以下类型：
-
-- ByteBuffer
-- CharBuffer
-- ShortBuffer
-- IntBuffer
-- LongBuffer
-- FloatBuffer
-- DoubleBuffer
-
-## 缓冲区状态变量
-
-- capacity：最大容量；
-- position：当前已经读写的字节数；
-- limit：还可以读写的字节数。
-
-状态变量的改变过程举例：
-
-① 新建一个大小为 8 个字节的缓冲区，此时 position 为 0，而 limit = capacity = 8。capacity 变量不会改变，下面的讨论会忽略它。
-
-<div align="center"> <img src="../pics//1bea398f-17a7-4f67-a90b-9e2d243eaa9a.png"/> </div><br>
-
-② 从输入通道中读取 5 个字节数据写入缓冲区中，此时 position 移动设置为 5，limit 保持不变。
-
-<div align="center"> <img src="../pics//80804f52-8815-4096-b506-48eef3eed5c6.png"/> </div><br>
-
-③ 在将缓冲区的数据写到输出通道之前，需要先调用 flip() 方法，这个方法将 limit 设置为当前 position，并将 position 设置为 0。
-
-<div align="center"> <img src="../pics//952e06bd-5a65-4cab-82e4-dd1536462f38.png"/> </div><br>
-
-④ 从缓冲区中取 4 个字节到输出缓冲中，此时 position 设为 4。
-
-<div align="center"> <img src="../pics//b5bdcbe2-b958-4aef-9151-6ad963cb28b4.png"/> </div><br>
-
-⑤ 最后需要调用 clear() 方法来清空缓冲区，此时 position 和 limit 都被设置为最初位置。
-
-<div align="center"> <img src="../pics//67bf5487-c45d-49b6-b9c0-a058d8c68902.png"/> </div><br>
-
-## 文件 NIO 实例
-
-以下展示了使用 NIO 快速复制文件的实例：
-
-```java
-public static void fastCopy(String src, String dist) throws IOException
-{
-    FileInputStream fin = new FileInputStream(src);      /* 获取源文件的输入字节流 */
-    FileChannel fcin = fin.getChannel();                 /* 获取输入字节流的文件通道 */
-    FileOutputStream fout = new FileOutputStream(dist);  /* 获取目标文件的输出字节流 */
-    FileChannel fcout = fout.getChannel();               /* 获取输出字节流的通道 */
-    ByteBuffer buffer = ByteBuffer.allocateDirect(1024); /* 为缓冲区分配 1024 个字节 */
-    while (true) {
-        int r = fcin.read(buffer);                       /* 从输入通道中读取数据到缓冲区中 */
-        if (r == -1) {                                   /* read() 返回 -1 表示 EOF */
-            break;
-        }
-        buffer.flip();                                   /* 切换读写 */
-        fcout.write(buffer);                             /* 把缓冲区的内容写入输出文件中 */
-        buffer.clear();                                  /* 清空缓冲区 */
-    }
-}
-```
-
-## 选择器
-
-NIO 常常被叫做非阻塞 IO，主要是因为 NIO 在网络通信中的非阻塞特性被广泛使用。
-
-NIO 实现了 IO 多路复用中的 Reactor 模型，一个线程 Thread 使用一个选择器 Selector 通过轮询的方式去监听多个通道 Channel 上的事件，从而让一个线程就可以处理多个事件。
-
-通过配置监听的通道 Channel 为非阻塞，那么当 Channel 上的 IO 事件还未到达时，就不会进入阻塞状态一直等待，而是继续轮询其它 Channel，找到 IO 事件已经到达的 Channel 执行。
-
-因为创建和切换线程的开销很大，因此使用一个线程来处理多个事件而不是一个线程处理一个事件具有更好的性能。
-
-应该注意的是，只有套接字 Channel 才能配置为非阻塞，而 FileChannel 不能，为 FileChannel 配置非阻塞也没有意义。
-
-<div align="center"> <img src="../pics//4d930e22-f493-49ae-8dff-ea21cd6895dc.png"/> </div><br>
-
-### 1. 创建选择器
-
-```java
-Selector selector = Selector.open();
-```
-
-### 2. 将通道注册到选择器上
-
-```java
-ServerSocketChannel ssChannel = ServerSocketChannel.open();
-ssChannel.configureBlocking(false);
-ssChannel.register(selector, SelectionKey.OP_ACCEPT);
-```
-
-通道必须配置为非阻塞模式，否则使用选择器就没有任何意义了，因为如果通道在某个事件上被阻塞，那么服务器就不能响应其它事件，必须等待这个事件处理完毕才能去处理其它事件，显然这和选择器的作用背道而驰。
-
-在将通道注册到选择器上时，还需要指定要注册的具体事件，主要有以下几类：
-
-- SelectionKey.OP_CONNECT
-- SelectionKey.OP_ACCEPT
-- SelectionKey.OP_READ
-- SelectionKey.OP_WRITE
-
-它们在 SelectionKey 的定义如下：
-
-```java
-public static final int OP_READ = 1 << 0;
-public static final int OP_WRITE = 1 << 2;
-public static final int OP_CONNECT = 1 << 3;
-public static final int OP_ACCEPT = 1 << 4;
-```
-
-可以看出每个事件可以被当成一个位域，从而组成事件集整数。例如：
-
-```java
-int interestSet = SelectionKey.OP_READ | SelectionKey.OP_WRITE;
-```
-
-### 3. 监听事件
-
-```java
-int num = selector.select();
-```
-
-使用 select() 来监听到达的事件，它会一直阻塞直到有至少一个事件到达。
-
-### 4. 获取到达的事件
-
-```java
-Set<SelectionKey> keys = selector.selectedKeys();
-Iterator<SelectionKey> keyIterator = keys.iterator();
-while (keyIterator.hasNext()) {
-    SelectionKey key = keyIterator.next();
-    if (key.isAcceptable()) {
-        // ...
-    } else if (key.isReadable()) {
-        // ...
-    }
-    keyIterator.remove();
-}
-```
-
-### 5. 事件循环
-
-因为一次 select() 调用不能处理完所有的事件，并且服务器端有可能需要一直监听事件，因此服务器端处理事件的代码一般会放在一个死循环内。
-
-```java
-while (true) {
-    int num = selector.select();
-    Set<SelectionKey> keys = selector.selectedKeys();
-    Iterator<SelectionKey> keyIterator = keys.iterator();
-    while (keyIterator.hasNext()) {
-        SelectionKey key = keyIterator.next();
-        if (key.isAcceptable()) {
-            // ...
-        } else if (key.isReadable()) {
-            // ...
-        }
-        keyIterator.remove();
-    }
-}
-```
-
-## 套接字 NIO 实例
-
-```java
-public class NIOServer
-{
-    public static void main(String[] args) throws IOException
-    {
-        Selector selector = Selector.open();
-
-        ServerSocketChannel ssChannel = ServerSocketChannel.open();
-        ssChannel.configureBlocking(false);
-        ssChannel.register(selector, SelectionKey.OP_ACCEPT);
-
-        ServerSocket serverSocket = ssChannel.socket();
-        InetSocketAddress address = new InetSocketAddress("127.0.0.1", 8888);
-        serverSocket.bind(address);
-
-        while (true) {
-            selector.select();
-            Set<SelectionKey> keys = selector.selectedKeys();
-            Iterator<SelectionKey> keyIterator = keys.iterator();
-            while (keyIterator.hasNext()) {
-                SelectionKey key = keyIterator.next();
-                if (key.isAcceptable()) {
-                    ServerSocketChannel ssChannel1 = (ServerSocketChannel) key.channel();
-                    // 服务器会为每个新连接创建一个 SocketChannel
-                    SocketChannel sChannel = ssChannel1.accept();
-                    sChannel.configureBlocking(false);
-                    // 这个新连接主要用于从客户端读取数据
-                    sChannel.register(selector, SelectionKey.OP_READ);
-                } else if (key.isReadable()) {
-                    SocketChannel sChannel = (SocketChannel) key.channel();
-                    System.out.println(readDataFromSocketChannel(sChannel));
-                    sChannel.close();
-                }
-                keyIterator.remove();
-            }
-        }
-    }
-
-    private static String readDataFromSocketChannel(SocketChannel sChannel) throws IOException {
-        ByteBuffer buffer = ByteBuffer.allocate(1024);
-        StringBuilder data = new StringBuilder();
-        while (true) {
-            buffer.clear();
-            int n = sChannel.read(buffer);
-            if (n == -1) {
-                break;
-            }
-            buffer.flip();
-            int limit = buffer.limit();
-            char[] dst = new char[limit];
-            for (int i = 0; i < limit; i++) {
-                dst[i] = (char) buffer.get(i);
-            }
-            data.append(dst);
-            buffer.clear();
-        }
-        return data.toString();
-    }
-}
-```
-
-```java
-public class NIOClient
-{
-    public static void main(String[] args) throws IOException 
-    {
-        Socket socket = new Socket("127.0.0.1", 8888);
-        OutputStream out = socket.getOutputStream();
-        String s = "hello world";
-        out.write(s.getBytes());
-        out.close();
-    }
-}
-```
-
-## 内存映射文件
-
-内存映射文件 I/O 是一种读和写文件数据的方法，它可以比常规的基于流或者基于通道的 I/O 快得多。
-
-向内存映射文件写入可能是危险的，仅只是改变数组的单个元素这样的简单操作，就可能会直接修改磁盘上的文件。修改数据与将数据保存到磁盘是没有分开的。
-
-下面代码行将文件的前 1024 个字节映射到内存中，map() 方法返回一个 MappedByteBuffer，它是 ByteBuffer 的子类。因此，您可以像使用其他任何 ByteBuffer 一样使用新映射的缓冲区，操作系统会在需要时负责执行映射。
-
-```java
-MappedByteBuffer mbb = fc.map(FileChannel.MapMode.READ_WRITE, 0, 1024);
-```
-
-## 对比
-
-NIO 与普通 I/O 的区别主要有以下两点：
-
-- NIO 是非阻塞的
-- NIO 面向块，I/O 面向流
-
-# 八、参考资料
-
-- Eckel B, 埃克尔, 昊鹏, 等. Java 编程思想 [M]. 机械工业出版社, 2002.
-- [IBM: NIO 入门](https://www.ibm.com/developerworks/cn/education/java/j-nio/j-nio.html)
-- [IBM: 深入分析 Java I/O 的工作机制](https://www.ibm.com/developerworks/cn/java/j-lo-javaio/index.html)
-- [IBM: 深入分析 Java 中的中文编码问题](https://www.ibm.com/developerworks/cn/java/j-lo-chinesecoding/index.htm)
-- [IBM: Java 序列化的高级认识](https://www.ibm.com/developerworks/cn/java/j-lo-serial/index.html)
-- [NIO 与传统 IO 的区别](http://blog.csdn.net/shimiso/article/details/24990499)
-- [Decorator Design Pattern](http://stg-tud.github.io/sedc/Lecture/ws13-14/5.3-Decorator.html#mode=document)
-- [Socket Multicast](http://labojava.blogspot.com/2012/12/socket-multicast.html)
diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index f5133a6f..bcb94acc 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -4,7 +4,7 @@
     * [缓存池](#缓存池)
 * [二、String](#二string)
     * [概览](#概览)
-    * [String 不可变的好处](#string-不可变的好处)
+    * [不可变的好处](#不可变的好处)
     * [String, StringBuffer and StringBuilder](#string,-stringbuffer-and-stringbuilder)
     * [String.intern()](#stringintern)
 * [三、运算](#三运算)
@@ -153,7 +153,7 @@ public final class String
     private final char value[];
 ```
 
-## String 不可变的好处
+## 不可变的好处
 
 **1. 可以缓存 hash 值** 
 
@@ -212,7 +212,7 @@ String s5 = "bbb";
 System.out.println(s4 == s5);  // true
 ```
 
-在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被放在堆中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
+在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到 Native Method 中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
 
 - [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning) 
 - [深入解析 String#intern](https://tech.meituan.com/in_depth_understanding_string_intern.html)
@@ -583,19 +583,7 @@ protected void finalize() throws Throwable {}
 
 ## equals()
 
-**1. equals() 与 == 的区别** 
-
-- 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
-- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用的对象的 equals() 方法的具体实现来进行比较。
-
-```java
-Integer x = new Integer(1);
-Integer y = new Integer(1);
-System.out.println(x.equals(y)); // true
-System.out.println(x == y);      // false
-```
-
-**2. 等价关系** 
+**1. 等价关系** 
 
 （一）自反性
 
@@ -632,6 +620,18 @@ x.equals(y) == x.equals(y); // true
 x.equals(null); // false;
 ```
 
+**2. equals() 与 ==** 
+
+- 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
+- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用对象 equals() 方法的具体实现来进行比较。
+
+```java
+Integer x = new Integer(1);
+Integer y = new Integer(1);
+System.out.println(x.equals(y)); // true
+System.out.println(x == y);      // false
+```
+
 **3. 实现** 
 
 - 检查是否为同一个对象的引用，如果是直接返回 true；
@@ -1005,7 +1005,7 @@ public class A {
 
 **4. 静态内部类** 
 
-非静态内部类依赖于需要外部类的实例，而静态内部类不需要。
+非静态内部类依赖于外部类的实例，而静态内部类不需要。
 
 ```java
 public class OuterClass {
diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index c6fc88f4..649f563a 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -1,1056 +1,4 @@
 <!-- GFM-TOC -->
-* [一、概览](#一概览)
-    * [Collection](#collection)
-    * [Map](#map)
-* [二、容器中的设计模式](#二容器中的设计模式)
-    * [迭代器模式](#迭代器模式)
-    * [适配器模式](#适配器模式)
-* [三、源码分析](#三源码分析)
-    * [ArrayList](#arraylist)
-    * [Vector](#vector)
-    * [CopyOnWriteArrayList](#copyonwritearraylist)
-    * [LinkedList](#linkedlist)
-    * [HashMap](#hashmap)
-    * [ConcurrentHashMap](#concurrenthashmap)
-    * [LinkedHashMap](#linkedhashmap)
-    * [WeekHashMap](#weekhashmap)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、概览
-
-容器主要包括 Collection 和 Map 两种，Collection 又包含了 List、Set 以及 Queue。
-
-## Collection
-
-<div align="center"> <img src="../pics//java-collections.png"/> </div><br>
-
-### 1. Set
-
-- HashSet：基于哈希实现，支持快速查找，但不支持有序性操作，例如根据一个范围查找元素的操作。并且失去了元素的插入顺序信息，也就是说使用 Iterator 遍历 HashSet 得到的结果是不确定的。
-
-- TreeSet：基于红黑树实现，支持有序性操作，但是查找效率不如 HashSet，HashSet 查找时间复杂度为 O(1)，TreeSet 则为 O(logN)。
-
-- LinkedHashSet：具有 HashSet 的查找效率，且内部使用链表维护元素的插入顺序。
-
-### 2. List
-
-- ArrayList：基于动态数组实现，支持随机访问。
-
-- Vector：和 ArrayList 类似，但它是线程安全的。
-
-- LinkedList：基于双向链表实现，只能顺序访问，但是可以快速地在链表中间插入和删除元素。不仅如此，LinkedList 还可以用作栈、队列和双向队列。
-
-### 3. Queue
-
-- LinkedList：可以用它来实现双向队列。
-
-- PriorityQueue：基于堆结构实现，可以用它来实现优先队列。
-
-## Map
-
-<div align="center"> <img src="../pics//java-collections1.png"/> </div><br>
-
-- HashMap：基于哈希实现；
-
-- HashTable：和 HashMap 类似，但它是线程安全的，这意味着同一时刻多个线程可以同时写入 HashTable 并且不会导致数据不一致。它是遗留类，不应该去使用它。现在可以使用 ConcurrentHashMap 来支持线程安全，并且 ConcurrentHashMap 的效率会更高，因为 ConcurrentHashMap 引入了分段锁。
-
-- LinkedHashMap：使用链表来维护元素的顺序，顺序为插入顺序或者最近最少使用（LRU）顺序。
-
-- TreeMap：基于红黑树实现。
-
-# 二、容器中的设计模式
-
-## 迭代器模式
-
-<div align="center"> <img src="../pics//Iterator-1.jpg"/> </div><br>
-
-Collection 实现了 Iterable 接口，其中的 iterator() 方法能够产生一个 Iterator 对象，通过这个对象就可以迭代遍历 Collection 中的元素。
-
-从 JDK 1.5 之后可以使用 foreach 方法来遍历实现了 Iterable 接口的聚合对象。
-
-```java
-List<String> list = new ArrayList<>();
-list.add("a");
-list.add("b");
-for (String item : list) {
-    System.out.println(item);
-}
-```
-
-## 适配器模式
-
-java.util.Arrays#asList() 可以把数组类型转换为 List 类型。
-
-```java
-@SafeVarargs
-public static <T> List<T> asList(T... a)
-```
-
-应该注意的是 asList() 的参数为泛型的变长参数，不能使用基本类型数组作为参数，只能使用相应的包装类型数组。
-
-```java
-Integer[] arr = {1, 2, 3};
-List list = Arrays.asList(arr);
-```
-
-也可以使用以下方式使用 asList()：
-
-```java
-List list = Arrays.asList(1,2,3);
-```
-
-# 三、源码分析
-
-如果没有特别说明，以下源码分析基于 JDK 1.8。
-
-在 IDEA 中 double shift 调出 Search EveryWhere，查找源码文件，找到之后就可以阅读源码。
-
-## ArrayList
-
-### 1. 概览
-
-实现了 RandomAccess 接口，因此支持随机访问，这是理所当然的，因为 ArrayList 是基于数组实现的。
-
-```java
-public class ArrayList<E> extends AbstractList<E>
-        implements List<E>, RandomAccess, Cloneable, java.io.Serializable
-```
-
-数组的默认大小为 10。
-
-```java
-private static final int DEFAULT_CAPACITY = 10;
-```
-
-### 2. 序列化
-
-基于数组实现，保存元素的数组使用 transient 修饰，该关键字声明数组默认不会被序列化。ArrayList 具有动态扩容特性，因此保存元素的数组不一定都会被使用，那么就没必要全部进行序列化。ArrayList 重写了 writeObject() 和 readObject() 来控制只序列化数组中有元素填充那部分内容。
-
-```java
-transient Object[] elementData; // non-private to simplify nested class access
-```
-
-### 3. 扩容
-
-添加元素时使用 ensureCapacityInternal() 方法来保证容量足够，如果不够时，需要使用 grow() 方法进行扩容，新容量的大小为 `oldCapacity + (oldCapacity >> 1)`，也就是旧容量的 1.5 倍。
-
-扩容操作需要调用 `Arrays.copyOf()` 把原数组整个复制到新数组中，这个操作代价很高，因此最好在创建 ArrayList 对象时就指定大概的容量大小，减少扩容操作的次数。
-
-```java
-public boolean add(E e) {
-    ensureCapacityInternal(size + 1);  // Increments modCount!!
-    elementData[size++] = e;
-    return true;
-}
-
-private void ensureCapacityInternal(int minCapacity) {
-    if (elementData == DEFAULTCAPACITY_EMPTY_ELEMENTDATA) {
-        minCapacity = Math.max(DEFAULT_CAPACITY, minCapacity);
-    }
-    ensureExplicitCapacity(minCapacity);
-}
-
-private void ensureExplicitCapacity(int minCapacity) {
-    modCount++;
-    // overflow-conscious code
-    if (minCapacity - elementData.length > 0)
-        grow(minCapacity);
-}
-
-private void grow(int minCapacity) {
-    // overflow-conscious code
-    int oldCapacity = elementData.length;
-    int newCapacity = oldCapacity + (oldCapacity >> 1);
-    if (newCapacity - minCapacity < 0)
-        newCapacity = minCapacity;
-    if (newCapacity - MAX_ARRAY_SIZE > 0)
-        newCapacity = hugeCapacity(minCapacity);
-    // minCapacity is usually close to size, so this is a win:
-    elementData = Arrays.copyOf(elementData, newCapacity);
-}
-```
-
-### 4. 删除元素
-
-需要调用 System.arraycopy() 将 index+1 后面的元素都复制到 index 位置上。
-
-```java
-public E remove(int index) {
-    rangeCheck(index);
-    modCount++;
-    E oldValue = elementData(index);
-    int numMoved = size - index - 1;
-    if (numMoved > 0)
-        System.arraycopy(elementData, index+1, elementData, index, numMoved);
-    elementData[--size] = null; // clear to let GC do its work
-    return oldValue;
-}
-```
-
-### 5. Fail-Fast
-
-modCount 用来记录 ArrayList 结构发生变化的次数。结构发生变化是指添加或者删除至少一个元素的所有操作，或者是调整内部数组的大小，仅仅只是设置元素的值不算结构发生变化。
-
-在进行序列化或者迭代等操作时，需要比较操作前后 modCount 是否改变，如果改变了需要抛出 ConcurrentModificationException。
-
-```java
-private void writeObject(java.io.ObjectOutputStream s)
-    throws java.io.IOException{
-    // Write out element count, and any hidden stuff
-    int expectedModCount = modCount;
-    s.defaultWriteObject();
-
-    // Write out size as capacity for behavioural compatibility with clone()
-    s.writeInt(size);
-
-    // Write out all elements in the proper order.
-    for (int i=0; i<size; i++) {
-        s.writeObject(elementData[i]);
-    }
-
-    if (modCount != expectedModCount) {
-        throw new ConcurrentModificationException();
-    }
-}
-```
-
-## Vector
-
-### 1. 同步
-
-它的实现与 ArrayList 类似，但是使用了 synchronized 进行同步。
-
-```java
-public synchronized boolean add(E e) {
-    modCount++;
-    ensureCapacityHelper(elementCount + 1);
-    elementData[elementCount++] = e;
-    return true;
-}
-
-public synchronized E get(int index) {
-    if (index >= elementCount)
-        throw new ArrayIndexOutOfBoundsException(index);
-
-    return elementData(index);
-}
-```
-
-### 2. 与 ArrayList 的区别
-
-- Vector 是同步的，因此开销就比 ArrayList 要大，访问速度更慢。最好使用 ArrayList 而不是 Vector，因为同步操作完全可以由程序员自己来控制；
-- Vector 每次扩容请求其大小的 2 倍空间，而 ArrayList 是 1.5 倍。
-
-### 3. 替代方案
-
-为了获得线程安全的 ArrayList，可以使用 `Collections.synchronizedList();` 得到一个线程安全的 ArrayList。
-
-```java
-List<String> list = new ArrayList<>();
-List<String> synList = Collections.synchronizedList(list);
-```
-
-也可以使用 concurrent 并发包下的 CopyOnWriteArrayList 类。
-
-```java
-List<String> list = new CopyOnWriteArrayList<>();
-```
-
-## CopyOnWriteArrayList
-
-### 读写分离
-
-写操作在一个复制的数组上进行，读操作还是在原始数组中进行，读写分离，互不影响。
-
-写操作需要加锁，防止同时并发写入时导致的写入数据丢失。
-
-写操作结束之后需要把原始数组指向新的复制数组。
-
-```java
-public boolean add(E e) {
-    final ReentrantLock lock = this.lock;
-    lock.lock();
-    try {
-        Object[] elements = getArray();
-        int len = elements.length;
-        Object[] newElements = Arrays.copyOf(elements, len + 1);
-        newElements[len] = e;
-        setArray(newElements);
-        return true;
-    } finally {
-        lock.unlock();
-    }
-}
-
-final void setArray(Object[] a) {
-    array = a;
-}
-```
-
-```java
-@SuppressWarnings("unchecked")
-private E get(Object[] a, int index) {
-    return (E) a[index];
-}
-```
-
-### 适用场景
-
-CopyOnWriteArrayList 在写操作的同时允许读操作，大大提高了读操作的性能，因此很适合读多写少的应用场景。
-
-但是 CopyOnWriteArrayList 有其缺陷：
-
-- 内存占用：在写操作时需要复制一个新的数组，使得内存占用为原来的两倍左右；
-- 数据不一致：读操作不能读取实时性的数据，因为部分写操作的数据还未同步到读数组中。
-
-所以 CopyOnWriteArrayList 不适合内存敏感以及对实时性要求很高的场景。
-
-## LinkedList
-
-### 1. 概览
-
-基于双向链表实现，内部使用 Node 来存储链表节点信息。
-
-```java
-private static class Node<E> {
-    E item;
-    Node<E> next;
-    Node<E> prev;
-}
-```
-
-每个链表存储了 first 和 last 指针：
-
-```java
-transient Node<E> first;
-transient Node<E> last;
-```
-
-<div align="center"> <img src="../pics//5158bc2f-83a6-4351-817e-c9b07f955d76.png"/> </div><br>
-
-### 2. ArrayList 与 LinkedList
-
-- ArrayList 基于动态数组实现，LinkedList 基于双向链表实现；
-- ArrayList 支持随机访问，LinkedList 不支持；
-- LinkedList 在任意位置添加删除元素更快。
-
-## HashMap
-
-为了便于理解，以下源码分析以 JDK 1.7 为主。
-
-### 1. 存储结构
-
-内部包含了一个 Entry 类型的数组 table。
-
-```java
-transient Entry[] table;
-```
-
-其中，Entry 就是存储数据的键值对，它包含了四个字段。从 next 字段我们可以看出 Entry 是一个链表，即数组中的每个位置被当成一个桶，一个桶存放一个链表，链表中存放哈希值相同的 Entry。也就是说，HashMap 使用拉链法来解决冲突。
-
-<div align="center"> <img src="../pics//8fe838e3-ef77-4f63-bf45-417b6bc5c6bb.png" width="600"/> </div><br>
-
-```java
-static class Entry<K,V> implements Map.Entry<K,V> {
-    final K key;
-    V value;
-    Entry<K,V> next;
-    int hash;
-
-    Entry(int h, K k, V v, Entry<K,V> n) {
-        value = v;
-        next = n;
-        key = k;
-        hash = h;
-    }
-
-    public final K getKey() {
-        return key;
-    }
-
-    public final V getValue() {
-        return value;
-    }
-
-    public final V setValue(V newValue) {
-        V oldValue = value;
-        value = newValue;
-        return oldValue;
-    }
-
-    public final boolean equals(Object o) {
-        if (!(o instanceof Map.Entry))
-            return false;
-        Map.Entry e = (Map.Entry)o;
-        Object k1 = getKey();
-        Object k2 = e.getKey();
-        if (k1 == k2 || (k1 != null && k1.equals(k2))) {
-            Object v1 = getValue();
-            Object v2 = e.getValue();
-            if (v1 == v2 || (v1 != null && v1.equals(v2)))
-                return true;
-        }
-        return false;
-    }
-
-    public final int hashCode() {
-        return Objects.hashCode(getKey()) ^ Objects.hashCode(getValue());
-    }
-
-    public final String toString() {
-        return getKey() + "=" + getValue();
-    }
-
-    /**
-     * This method is invoked whenever the value in an entry is
-     * overwritten by an invocation of put(k,v) for a key k that's already
-     * in the HashMap.
-     */
-    void recordAccess(HashMap<K,V> m) {
-    }
-
-    /**
-     * This method is invoked whenever the entry is
-     * removed from the table.
-     */
-    void recordRemoval(HashMap<K,V> m) {
-    }
-}
-```
-
-### 2. 拉链法的工作原理
-
-```java
-HashMap<String, String> map = new HashMap<>();
-map.put("K1", "V1");
-map.put("K2", "V2");
-map.put("K3", "V3");
-```
-
-- 新建一个 HashMap，默认大小为 16；
-- 插入 &lt;K1,V1> 键值对，先计算 K1 的 hashCode 为 115，使用除留余数法得到所在的桶下标 115%16=3。
-- 插入 &lt;K2,V2> 键值对，先计算 K2 的 hashCode 为 118，使用除留余数法得到所在的桶下标 118%16=6。
-- 插入 &lt;K3,V3> 键值对，先计算 K3 的 hashCode 为 118，使用除留余数法得到所在的桶下标 118%16=6，插在 &lt;K2,V2> 前面。
-
-应该注意到链表的插入是以头插法方式进行的，例如上面的 &lt;K3,V3> 不是插在 &lt;K2,V2> 后面，而是插入在链表头部。
-
-查找需要分成两步进行：
-
-- 计算键值对所在的桶；
-- 在链表上顺序查找，时间复杂度显然和链表的长度成正比。
-
-<div align="center"> <img src="../pics//49d6de7b-0d0d-425c-9e49-a1559dc23b10.png" width="600"/> </div><br>
-
-### 3. put 操作
-
-```java
-public V put(K key, V value) {
-    if (table == EMPTY_TABLE) {
-        inflateTable(threshold);
-    }
-    // 键为 null 单独处理
-    if (key == null)
-        return putForNullKey(value);
-    int hash = hash(key);
-    // 确定桶下标
-    int i = indexFor(hash, table.length);
-    // 先找出是否已经存在键为 key 的键值对，如果存在的话就更新这个键值对的值为 value
-    for (Entry<K,V> e = table[i]; e != null; e = e.next) {
-        Object k;
-        if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
-            V oldValue = e.value;
-            e.value = value;
-            e.recordAccess(this);
-            return oldValue;
-        }
-    }
-
-    modCount++;
-    // 插入新键值对
-    addEntry(hash, key, value, i);
-    return null;
-}
-```
-
-HashMap 允许插入键为 null 的键值对。因为无法调用 null 的 hashCode()，也就无法确定该键值对的桶下标，只能通过强制指定一个桶下标来存放。HashMap 使用第 0 个桶存放键为 null 的键值对。
-
-```java
-private V putForNullKey(V value) {
-    for (Entry<K,V> e = table[0]; e != null; e = e.next) {
-        if (e.key == null) {
-            V oldValue = e.value;
-            e.value = value;
-            e.recordAccess(this);
-            return oldValue;
-        }
-    }
-    modCount++;
-    addEntry(0, null, value, 0);
-    return null;
-}
-```
-
-使用链表的头插法，也就是新的键值对插在链表的头部，而不是链表的尾部。
-
-```java
-void addEntry(int hash, K key, V value, int bucketIndex) {
-    if ((size >= threshold) && (null != table[bucketIndex])) {
-        resize(2 * table.length);
-        hash = (null != key) ? hash(key) : 0;
-        bucketIndex = indexFor(hash, table.length);
-    }
-
-    createEntry(hash, key, value, bucketIndex);
-}
-
-void createEntry(int hash, K key, V value, int bucketIndex) {
-    Entry<K,V> e = table[bucketIndex];
-    // 头插法，链表头部指向新的键值对
-    table[bucketIndex] = new Entry<>(hash, key, value, e);
-    size++;
-}
-```
-
-```java
-Entry(int h, K k, V v, Entry<K,V> n) {
-    value = v;
-    next = n;
-    key = k;
-    hash = h;
-}
-```
-
-### 4. 确定桶下标
-
-很多操作都需要先确定一个键值对所在的桶下标。
-
-```java
-int hash = hash(key);
-int i = indexFor(hash, table.length);
-```
-
-（一）计算 hash 值
-
-```java
-final int hash(Object k) {
-    int h = hashSeed;
-    if (0 != h && k instanceof String) {
-        return sun.misc.Hashing.stringHash32((String) k);
-    }
-
-    h ^= k.hashCode();
-
-    // This function ensures that hashCodes that differ only by
-    // constant multiples at each bit position have a bounded
-    // number of collisions (approximately 8 at default load factor).
-    h ^= (h >>> 20) ^ (h >>> 12);
-    return h ^ (h >>> 7) ^ (h >>> 4);
-}
-```
-
-```java
-public final int hashCode() {
-    return Objects.hashCode(key) ^ Objects.hashCode(value);
-}
-```
-
-（二）取模
-
-令 x = 1<<4，即 x 为 2 的 4 次方，它具有以下性质：
-
-```
-x   : 00010000
-x-1 : 00001111
-```
-
-令一个数 y 与 x-1 做与运算，可以去除 y 位级表示的第 4 位以上数：
-
-```
-y       : 10110010
-x-1     : 00001111
-y&(x-1) : 00000010
-```
-
-这个性质和 y 对 x 取模效果是一样的：
-
-```
-x   : 00010000
-y   : 10110010
-y%x : 00000010
-```
-
-我们知道，位运算的代价比求模运算小的多，因此在进行这种计算时用位运算的话能带来更高的性能。
-
-确定桶下标的最后一步是将 key 的 hash 值对桶个数取模：hash%capacity，如果能保证 capacity 为 2 的 n 次方，那么就可以将这个操作转换为位运算。
-
-```java
-static int indexFor(int h, int length) {
-    return h & (length-1);
-}
-```
-
-### 5. 扩容-基本原理
-
-设 HashMap 的 table 长度为 M，需要存储的键值对数量为 N，如果哈希函数满足均匀性的要求，那么每条链表的长度大约为 N/M，因此平均查找次数的复杂度为 O(N/M)。
-
-为了让查找的成本降低，应该尽可能使得 N/M 尽可能小，因此需要保证 M 尽可能大，也就是说 table 要尽可能大。HashMap 采用动态扩容来根据当前的 N 值来调整 M 值，使得空间效率和时间效率都能得到保证。
-
-和扩容相关的参数主要有：capacity、size、threshold 和 load_factor。
-
-| 参数 | 含义 |
-| :--: | :-- |
-| capacity | table 的容量大小，默认为 16，需要注意的是 capacity 必须保证为 2 的 n 次方。|
-| size | table 的实际使用量。 |
-| threshold | size 的临界值，size 必须小于 threshold，如果大于等于，就必须进行扩容操作。 |
-| load_factor | 装载因子，table 能够使用的比例，threshold = capacity * load_factor。|
-
-```java
-static final int DEFAULT_INITIAL_CAPACITY = 16;
-
-static final int MAXIMUM_CAPACITY = 1 << 30;
-
-static final float DEFAULT_LOAD_FACTOR = 0.75f;
-
-transient Entry[] table;
-
-transient int size;
-
-int threshold;
-
-final float loadFactor;
-
-transient int modCount;
-```
-
-从下面的添加元素代码中可以看出，当需要扩容时，令 capacity 为原来的两倍。
-
-```java
-void addEntry(int hash, K key, V value, int bucketIndex) {
-    Entry<K,V> e = table[bucketIndex];
-    table[bucketIndex] = new Entry<>(hash, key, value, e);
-    if (size++ >= threshold)
-        resize(2 * table.length);
-}
-```
-
-扩容使用 resize() 实现，需要注意的是，扩容操作同样需要把旧 table 的所有键值对重新插入新的 table 中，因此这一步是很费时的。
-
-```java
-void resize(int newCapacity) {
-    Entry[] oldTable = table;
-    int oldCapacity = oldTable.length;
-    if (oldCapacity == MAXIMUM_CAPACITY) {
-        threshold = Integer.MAX_VALUE;
-        return;
-    }
-    Entry[] newTable = new Entry[newCapacity];
-    transfer(newTable);
-    table = newTable;
-    threshold = (int)(newCapacity * loadFactor);
-}
-
-void transfer(Entry[] newTable) {
-    Entry[] src = table;
-    int newCapacity = newTable.length;
-    for (int j = 0; j < src.length; j++) {
-        Entry<K,V> e = src[j];
-        if (e != null) {
-            src[j] = null;
-            do {
-                Entry<K,V> next = e.next;
-                int i = indexFor(e.hash, newCapacity);
-                e.next = newTable[i];
-                newTable[i] = e;
-                e = next;
-            } while (e != null);
-        }
-    }
-}
-```
-
-### 6. 扩容-重新计算桶下标
-
-在进行扩容时，需要把键值对重新放到对应的桶上。HashMap 使用了一个特殊的机制，可以降低重新计算桶下标的操作。
-
-假设原数组长度 capacity 为 8，扩容之后 new capacity 为 16：
-
-```html
-capacity     : 00010000
-new capacity : 00100000
-```
-
-对于一个 Key，它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；如果为 1，那么得到的结果为原来的结果 + 8。
-
-### 7. 扩容-计算数组容量
-
-HashMap 构造函数允许用户传入的容量不是 2 的 n 次方，因为它可以自动地将传入的容量转换为 2 的 n 次方。
-
-先考虑如何求一个数的掩码，对于 10010000，它的掩码为 11111111，可以使用以下方法得到：
-
-```
-mask |= mask >> 1    11011000
-mask |= mask >> 2    11111100
-mask |= mask >> 4    11111111
-```
-
-mask+1 是大于原始数字的最小的 2 的 n 次方。
-
-```
-num     10010000
-mask+1 100000000
-```
-
-以下是 HashMap 中计算数组容量的代码：
-
-```java
-static final int tableSizeFor(int cap) {
-    int n = cap - 1;
-    n |= n >>> 1;
-    n |= n >>> 2;
-    n |= n >>> 4;
-    n |= n >>> 8;
-    n |= n >>> 16;
-    return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
-}
-```
-
-### 8. 链表转红黑树
-
-从 JDK 1.8 开始，一个桶存储的链表长度大于 8 时会将链表转换为红黑树。
-
-### 9. HashMap 与 HashTable
-
-- HashTable 使用 synchronized 来进行同步。
-- HashMap 可以插入键为 null 的 Entry。
-- HashMap 的迭代器是 fail-fast 迭代器。
-- HashMap 不能保证随着时间的推移 Map 中的元素次序是不变的。
-
-## ConcurrentHashMap
-
-### 1. 存储结构
-
-```java
-static final class HashEntry<K,V> {
-    final int hash;
-    final K key;
-    volatile V value;
-    volatile HashEntry<K,V> next;
-}
-```
-
-ConcurrentHashMap 和 HashMap 实现上类似，最主要的差别是 ConcurrentHashMap 采用了分段锁（Segment），每个分段锁维护着几个桶（HashEntry），多个线程可以同时访问不同分段锁上的桶，从而使其并发度更高（并发度就是 Segment 的个数）。
-
-Segment 继承自 ReentrantLock。
-
-```java
-static final class Segment<K,V> extends ReentrantLock implements Serializable {
-
-    private static final long serialVersionUID = 2249069246763182397L;
-
-    static final int MAX_SCAN_RETRIES =
-        Runtime.getRuntime().availableProcessors() > 1 ? 64 : 1;
-
-    transient volatile HashEntry<K,V>[] table;
-
-    transient int count;
-
-    transient int modCount;
-
-    transient int threshold;
-
-    final float loadFactor;
-}
-```
-
-```java
-final Segment<K,V>[] segments;
-```
-
-默认的并发级别为 16，也就是说默认创建 16 个 Segment。
-
-```java
-static final int DEFAULT_CONCURRENCY_LEVEL = 16;
-```
-
-<div align="center"> <img src="../pics//3fdfc89d-719e-4d93-b518-29fa612b3b18.png"/> </div><br>
-
-### 2. size 操作
-
-每个 Segment 维护了一个 count 变量来统计该 Segment 中的键值对个数。
-
-```java
-/**
- * The number of elements. Accessed only either within locks
- * or among other volatile reads that maintain visibility.
- */
-transient int count;
-```
-
-在执行 size 操作时，需要遍历所有 Segment 然后把 count 累计起来。
-
-ConcurrentHashMap 在执行 size 操作时先尝试不加锁，如果连续两次不加锁操作得到的结果一致，那么可以认为这个结果是正确的。
-
-尝试次数使用 RETRIES_BEFORE_LOCK 定义，该值为 2，retries 初始值为 -1，因此尝试次数为 3。
-
-如果尝试的次数超过 3 次，就需要对每个 Segment 加锁。
-
-```java
-
-/**
- * Number of unsynchronized retries in size and containsValue
- * methods before resorting to locking. This is used to avoid
- * unbounded retries if tables undergo continuous modification
- * which would make it impossible to obtain an accurate result.
- */
-static final int RETRIES_BEFORE_LOCK = 2;
-
-public int size() {
-    // Try a few times to get accurate count. On failure due to
-    // continuous async changes in table, resort to locking.
-    final Segment<K,V>[] segments = this.segments;
-    int size;
-    boolean overflow; // true if size overflows 32 bits
-    long sum;         // sum of modCounts
-    long last = 0L;   // previous sum
-    int retries = -1; // first iteration isn't retry
-    try {
-        for (;;) {
-            // 超过尝试次数，则对每个 Segment 加锁
-            if (retries++ == RETRIES_BEFORE_LOCK) {
-                for (int j = 0; j < segments.length; ++j)
-                    ensureSegment(j).lock(); // force creation
-            }
-            sum = 0L;
-            size = 0;
-            overflow = false;
-            for (int j = 0; j < segments.length; ++j) {
-                Segment<K,V> seg = segmentAt(segments, j);
-                if (seg != null) {
-                    sum += seg.modCount;
-                    int c = seg.count;
-                    if (c < 0 || (size += c) < 0)
-                        overflow = true;
-                }
-            }
-            // 连续两次得到的结果一致，则认为这个结果是正确的
-            if (sum == last)
-                break;
-            last = sum;
-        }
-    } finally {
-        if (retries > RETRIES_BEFORE_LOCK) {
-            for (int j = 0; j < segments.length; ++j)
-                segmentAt(segments, j).unlock();
-        }
-    }
-    return overflow ? Integer.MAX_VALUE : size;
-}
-```
-
-### 3. JDK 1.8 的改动
-
-JDK 1.7 使用分段锁机制来实现并发更新操作，核心类为 Segment，它继承自重入锁 ReentrantLock，并发程度与 Segment 数量相等。
-
-JDK 1.8 使用了 CAS 操作来支持更高的并发度，在 CAS 操作失败时使用内置锁 synchronized。
-
-并且 JDK 1.8 的实现也在链表过长时会转换为红黑树。
-
-## LinkedHashMap
-
-### 存储结构
-
-继承自 HashMap，因此具有和 HashMap 一样的快速查找特性。
-
-```java
-public class LinkedHashMap<K,V> extends HashMap<K,V> implements Map<K,V>
-```
-
-内存维护了一个双向循环链表，用来维护插入顺序或者 LRU 顺序。
-
-```java
-/**
- * The head (eldest) of the doubly linked list.
- */
-transient LinkedHashMap.Entry<K,V> head;
-
-/**
- * The tail (youngest) of the doubly linked list.
- */
-transient LinkedHashMap.Entry<K,V> tail;
-```
-
-顺序使用 accessOrder 来决定，默认为 false，此时使用的是插入顺序。
-
-```java
-final boolean accessOrder;
-```
-
-LinkedHashMap 最重要的是以下用于记录顺序的函数，它们会在 put、get 等方法中调用。
-
-```java
-void afterNodeAccess(Node<K,V> p) { }
-void afterNodeInsertion(boolean evict) { }
-```
-
-### afterNodeAccess()
-
-当一个 Node 被访问时，如果 accessOrder 为 true，会将它移到链表尾部。也就是说指定为 LRU 顺序之后，在每次访问一个节点时，会将这个节点移到链表尾部，保证链表尾部是最近访问的节点，那么链表首部就是最近最久未使用的节点。
-
-```java
-void afterNodeAccess(Node<K,V> e) { // move node to last
-    LinkedHashMap.Entry<K,V> last;
-    if (accessOrder && (last = tail) != e) {
-        LinkedHashMap.Entry<K,V> p =
-            (LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
-        p.after = null;
-        if (b == null)
-            head = a;
-        else
-            b.after = a;
-        if (a != null)
-            a.before = b;
-        else
-            last = b;
-        if (last == null)
-            head = p;
-        else {
-            p.before = last;
-            last.after = p;
-        }
-        tail = p;
-        ++modCount;
-    }
-}
-```
-
-### afterNodeInsertion()
-
-在 put 等操作之后执行，当 removeEldestEntry() 方法返回 ture 时会移除最晚的节点，也就是链表首部节点 first。
-
-evict 只有在构建 Map 的时候才为 false，在这里为 true。
-
-```java
-void afterNodeInsertion(boolean evict) { // possibly remove eldest
-    LinkedHashMap.Entry<K,V> first;
-    if (evict && (first = head) != null && removeEldestEntry(first)) {
-        K key = first.key;
-        removeNode(hash(key), key, null, false, true);
-    }
-}
-```
-
-removeEldestEntry() 默认为 false，如果需要让它为 true，需要继承 LinkedHashMap 并且覆盖这个方法的实现，这在实现 LRU 的缓存中特别有用，通过移除最近最久未使用的节点，从而保证缓存空间足够，并且缓存的数据都是热点数据。
-
-```java
-protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
-        return false;
-    }
-```
-
-### LRU 缓存
-
-以下是使用 LinkedHashMap 实现的一个 LRU 缓存，设定最大缓存空间 MAX_ENTRIES  为 3。使用 LinkedHashMap 的构造函数将 accessOrder 设置为 true，开启 LUR 顺序。覆盖 removeEldestEntry() 方法实现，在节点多于 MAX_ENTRIES 就会将最近最久未使用的数据移除。
-
-```java
-class LRUCache<K, V> extends LinkedHashMap<K, V> {
-    private static final int MAX_ENTRIES = 3;
-
-    protected boolean removeEldestEntry(Map.Entry eldest) {
-        return size() > MAX_ENTRIES;
-    }
-
-    LRUCache() {
-        super(MAX_ENTRIES, 0.75f, true);
-    }
-}
-```
-
-```java
-public static void main(String[] args) {
-    LRUCache<Integer, String> cache = new LRUCache<>();
-    cache.put(1, "a");
-    cache.put(2, "b");
-    cache.put(3, "c");
-    cache.get(1);
-    cache.put(4, "d");
-    System.out.println(cache.keySet());
-}
-```
-
-```html
-[3, 1, 4]
-```
-
-## WeekHashMap
-
-### 存储结构
-
-WeakHashMap 的 Entry 继承自 WeakReference，被 WeakReference 关联的对象在下一次垃圾回收时会被回收。
-
-WeakHashMap 主要用来实现缓存，通过使用 WeakHashMap 来引用缓存对象，由 JVM 对这部分缓存进行回收。
-
-```java
-private static class Entry<K,V> extends WeakReference<Object> implements Map.Entry<K,V>
-```
-
-### ConcurrentCache
-
-Tomcat 中的 ConcurrentCache 就使用了 WeakHashMap 来实现缓存功能。
-
-ConcurrentCache 采取的是分代缓存：
-
-- 经常使用的对象放入 eden 中，eden 使用 ConcurrentHashMap 实现，不用担心会被回收（伊甸园）；
-- 不常用的对象放入 longterm，longterm 使用 WeakHashMap 实现，用来存放比较老的对象，这些老对象会被垃圾收集器回收。
-
-```java
-public final class ConcurrentCache<K, V> {
-
-    private final int size;
-
-    private final Map<K, V> eden;
-
-    private final Map<K, V> longterm;
-
-    public ConcurrentCache(int size) {
-        this.size = size;
-        this.eden = new ConcurrentHashMap<>(size);
-        this.longterm = new WeakHashMap<>(size);
-    }
-
-    public V get(K k) {
-        V v = this.eden.get(k);
-        if (v == null) {
-            v = this.longterm.get(k);
-            if (v != null)
-                this.eden.put(k, v);
-        }
-        return v;
-    }
-
-    public void put(K k, V v) {
-        if (this.eden.size() >= size) {
-            this.longterm.putAll(this.eden);
-            this.eden.clear();
-        }
-        this.eden.put(k, v);
-    }
-}
-```
-
-
-# 参考资料
-
-- Eckel B. Java 编程思想 [M]. 机械工业出版社, 2002.
-- [Java Collection Framework](https://www.w3resource.com/java-tutorial/java-collections.php)
-- [Iterator 模式](https://openhome.cc/Gossip/DesignPattern/IteratorPattern.htm)
-- [Java 8 系列之重新认识 HashMap](https://tech.meituan.com/java-hashmap.html)
-- [What is difference between HashMap and Hashtable in Java?](http://javarevisited.blogspot.hk/2010/10/difference-between-hashmap-and.html)
-- [Java 集合之 HashMap](http://www.zhangchangle.com/2018/02/07/Java%E9%9B%86%E5%90%88%E4%B9%8BHashMap/)
-- [The principle of ConcurrentHashMap analysis](http://www.programering.com/a/MDO3QDNwATM.html)
-- [探索 ConcurrentHashMap 高并发性的实现机制](https://www.ibm.com/developerworks/cn/java/java-lo-concurrenthashmap/)
-- [HashMap 相关面试题及其解答](https://www.jianshu.com/p/75adf47958a7)
-- [Java 集合细节（二）：asList 的缺陷](http://wiki.jikexueyuan.com/project/java-enhancement/java-thirtysix.html)
-- [Java Collection Framework – The LinkedList Class](http://javaconceptoftheday.com/java-collection-framework-linkedlist-class/)
-
diff --git a/notes/Redis.md b/notes/Redis.md
index c54287c8..649f563a 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -1,606 +1,4 @@
 <!-- GFM-TOC -->
-* [一、概述](#一概述)
-* [二、数据类型](#二数据类型)
-    * [STRING](#string)
-    * [LIST](#list)
-    * [SET](#set)
-    * [HASH](#hash)
-    * [ZSET](#zset)
-* [三、数据结构](#三数据结构)
-    * [字典](#字典)
-    * [跳跃表](#跳跃表)
-* [四、使用场景](#四使用场景)
-    * [计数器](#计数器)
-    * [缓存](#缓存)
-    * [查找表](#查找表)
-    * [消息队列](#消息队列)
-    * [会话缓存](#会话缓存)
-    * [分布式锁实现](#分布式锁实现)
-    * [其它](#其它)
-* [五、Redis 与 Memcached](#五redis-与-memcached)
-    * [数据类型](#数据类型)
-    * [数据持久化](#数据持久化)
-    * [分布式](#分布式)
-    * [内存管理机制](#内存管理机制)
-* [六、键的过期时间](#六键的过期时间)
-* [七、数据淘汰策略](#七数据淘汰策略)
-* [八、持久化](#八持久化)
-    * [RDB 持久化](#rdb-持久化)
-    * [AOF 持久化](#aof-持久化)
-* [九、事务](#九事务)
-* [十、事件](#十事件)
-    * [文件事件](#文件事件)
-    * [时间事件](#时间事件)
-    * [事件的调度与执行](#事件的调度与执行)
-* [十一、复制](#十一复制)
-    * [连接过程](#连接过程)
-    * [主从链](#主从链)
-* [十二、Sentinel](#十二sentinel)
-* [十三、分片](#十三分片)
-* [十四、一个简单的论坛系统分析](#十四一个简单的论坛系统分析)
-    * [文章信息](#文章信息)
-    * [点赞功能](#点赞功能)
-    * [对文章进行排序](#对文章进行排序)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、概述
-
-Redis 是速度非常快的非关系型（NoSQL）内存键值数据库，可以存储键和五种不同类型的值之间的映射。
-
-键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、有序集合、散列表。
-
-Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，使用复制来扩展读性能，使用分片来扩展写性能。
-
-# 二、数据类型
-
-| 数据类型 | 可以存储的值 | 操作 |
-| :--: | :--: | :--: |
-| STRING | 字符串、整数或者浮点数 | 对整个字符串或者字符串的其中一部分执行操作</br> 对整数和浮点数执行自增或者自减操作 |
-| LIST | 列表 | 从两端压入或者弹出元素</br> 读取单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
-| SET | 无序集合 | 添加、获取、移除单个元素</br> 检查一个元素是否存在于集合中</br> 计算交集、并集、差集</br> 从集合里面随机获取元素 |
-| HASH | 包含键值对的无序散列表 | 添加、获取、移除单个键值对</br> 获取所有键值对</br> 检查某个键是否存在|
-| ZSET | 有序集合 | 添加、获取、删除元素</br> 根据分值范围或者成员来获取元素</br> 计算一个键的排名 |
-
-> [What Redis data structures look like](https://redislabs.com/ebook/part-1-getting-started/chapter-1-getting-to-know-redis/1-2-what-redis-data-structures-look-like/)
-
-## STRING
-
-<div align="center"> <img src="../pics//6019b2db-bc3e-4408-b6d8-96025f4481d6.png" width="400"/> </div><br>
-
-```html
-> set hello world
-OK
-> get hello
-"world"
-> del hello
-(integer) 1
-> get hello
-(nil)
-```
-
-## LIST
-
-<div align="center"> <img src="../pics//fb327611-7e2b-4f2f-9f5b-38592d408f07.png" width="400"/> </div><br>
-
-```html
-> rpush list-key item
-(integer) 1
-> rpush list-key item2
-(integer) 2
-> rpush list-key item
-(integer) 3
-
-> lrange list-key 0 -1
-1) "item"
-2) "item2"
-3) "item"
-
-> lindex list-key 1
-"item2"
-
-> lpop list-key
-"item"
-
-> lrange list-key 0 -1
-1) "item2"
-2) "item"
-```
-
-## SET
-
-<div align="center"> <img src="../pics//cd5fbcff-3f35-43a6-8ffa-082a93ce0f0e.png" width="400"/> </div><br>
-
-```html
-> sadd set-key item
-(integer) 1
-> sadd set-key item2
-(integer) 1
-> sadd set-key item3
-(integer) 1
-> sadd set-key item
-(integer) 0
-
-> smembers set-key
-1) "item"
-2) "item2"
-3) "item3"
-
-> sismember set-key item4
-(integer) 0
-> sismember set-key item
-(integer) 1
-
-> srem set-key item2
-(integer) 1
-> srem set-key item2
-(integer) 0
-
-> smembers set-key
-1) "item"
-2) "item3"
-```
-
-## HASH
-
-<div align="center"> <img src="../pics//7bd202a7-93d4-4f3a-a878-af68ae25539a.png" width="400"/> </div><br>
-
-```html
-> hset hash-key sub-key1 value1
-(integer) 1
-> hset hash-key sub-key2 value2
-(integer) 1
-> hset hash-key sub-key1 value1
-(integer) 0
-
-> hgetall hash-key
-1) "sub-key1"
-2) "value1"
-3) "sub-key2"
-4) "value2"
-
-> hdel hash-key sub-key2
-(integer) 1
-> hdel hash-key sub-key2
-(integer) 0
-
-> hget hash-key sub-key1
-"value1"
-
-> hgetall hash-key
-1) "sub-key1"
-2) "value1"
-```
-
-## ZSET
-
-<div align="center"> <img src="../pics//1202b2d6-9469-4251-bd47-ca6034fb6116.png" width="400"/> </div><br>
-
-```html
-> zadd zset-key 728 member1
-(integer) 1
-> zadd zset-key 982 member0
-(integer) 1
-> zadd zset-key 982 member0
-(integer) 0
-
-> zrange zset-key 0 -1 withscores
-1) "member1"
-2) "728"
-3) "member0"
-4) "982"
-
-> zrangebyscore zset-key 0 800 withscores
-1) "member1"
-2) "728"
-
-> zrem zset-key member1
-(integer) 1
-> zrem zset-key member1
-(integer) 0
-
-> zrange zset-key 0 -1 withscores
-1) "member0"
-2) "982"
-```
-
-# 三、数据结构
-
-## 字典
-
-dictht 是一个散列表结构，使用拉链法保存哈希冲突的 dictEntry。
-
-```c
-/* This is our hash table structure. Every dictionary has two of this as we
- * implement incremental rehashing, for the old to the new table. */
-typedef struct dictht {
-    dictEntry **table;
-    unsigned long size;
-    unsigned long sizemask;
-    unsigned long used;
-} dictht;
-```
-
-```c
-typedef struct dictEntry {
-    void *key;
-    union {
-        void *val;
-        uint64_t u64;
-        int64_t s64;
-        double d;
-    } v;
-    struct dictEntry *next;
-} dictEntry;
-```
-
-Redis 的字典 dict 中包含两个哈希表 dictht，这是为了方便进行 rehash 操作。在扩容时，将其中一个 dictht 上的键值对 rehash 到另一个 dictht 上面，完成之后释放空间并交换两个 dictht 的角色。
-
-```c
-typedef struct dict {
-    dictType *type;
-    void *privdata;
-    dictht ht[2];
-    long rehashidx; /* rehashing not in progress if rehashidx == -1 */
-    unsigned long iterators; /* number of iterators currently running */
-} dict;
-```
-
-rehash 操作不是一次性完成，而是采用渐进方式，这是为了避免一次性执行过多的 rehash 操作给服务器带来过大的负担。
-
-渐进式 rehash 通过记录 dict 的 rehashidx 完成，它从 0 开始，然后每执行一次 rehash 都会递增。例如在一次 rehash 中，要把 dict[0] rehash 到 dict[1]，这一次会把 dict[0] 上 table[rehashidx] 的键值对 rehash 到 dict[1] 上，dict[0] 的 table[rehashidx] 指向 null，并令 rehashidx++。
-
-在 rehash 期间，每次对字典执行添加、删除、查找或者更新操作时，都会执行一次渐进式 rehash。
-
-采用渐进式 rehash 会导致字典中的数据分散在两个 dictht 上，因此对字典的操作也需要到对应的 dictht 去执行。
-
-```c
-/* Performs N steps of incremental rehashing. Returns 1 if there are still
- * keys to move from the old to the new hash table, otherwise 0 is returned.
- *
- * Note that a rehashing step consists in moving a bucket (that may have more
- * than one key as we use chaining) from the old to the new hash table, however
- * since part of the hash table may be composed of empty spaces, it is not
- * guaranteed that this function will rehash even a single bucket, since it
- * will visit at max N*10 empty buckets in total, otherwise the amount of
- * work it does would be unbound and the function may block for a long time. */
-int dictRehash(dict *d, int n) {
-    int empty_visits = n * 10; /* Max number of empty buckets to visit. */
-    if (!dictIsRehashing(d)) return 0;
-
-    while (n-- && d->ht[0].used != 0) {
-        dictEntry *de, *nextde;
-
-        /* Note that rehashidx can't overflow as we are sure there are more
-         * elements because ht[0].used != 0 */
-        assert(d->ht[0].size > (unsigned long) d->rehashidx);
-        while (d->ht[0].table[d->rehashidx] == NULL) {
-            d->rehashidx++;
-            if (--empty_visits == 0) return 1;
-        }
-        de = d->ht[0].table[d->rehashidx];
-        /* Move all the keys in this bucket from the old to the new hash HT */
-        while (de) {
-            uint64_t h;
-
-            nextde = de->next;
-            /* Get the index in the new hash table */
-            h = dictHashKey(d, de->key) & d->ht[1].sizemask;
-            de->next = d->ht[1].table[h];
-            d->ht[1].table[h] = de;
-            d->ht[0].used--;
-            d->ht[1].used++;
-            de = nextde;
-        }
-        d->ht[0].table[d->rehashidx] = NULL;
-        d->rehashidx++;
-    }
-
-    /* Check if we already rehashed the whole table... */
-    if (d->ht[0].used == 0) {
-        zfree(d->ht[0].table);
-        d->ht[0] = d->ht[1];
-        _dictReset(&d->ht[1]);
-        d->rehashidx = -1;
-        return 0;
-    }
-
-    /* More to rehash... */
-    return 1;
-}
-```
-
-## 跳跃表
-
-是有序集合的底层实现之一。
-
-跳跃表是基于多指针有序链表实现的，可以看成多个有序链表。
-
-<div align="center"> <img src="../pics//beba612e-dc5b-4fc2-869d-0b23408ac90a.png"/> </div><br>
-
-在查找时，从上层指针开始查找，找到对应的区间之后再到下一层去查找。下图演示了查找 22 的过程。
-
-<div align="center"> <img src="../pics//0ea37ee2-c224-4c79-b895-e131c6805c40.png"/> </div><br>
-
-与红黑树等平衡树相比，跳跃表具有以下优点：
-
-- 插入速度非常快速，因为不需要进行旋转等操作来维护平衡性；
-- 更容易实现；
-- 支持无锁操作。
-
-# 四、使用场景
-
-## 计数器
-
-可以对 String 进行自增自减运算，从而实现计数器功能。
-
-Redis 这种内存型数据库的读写性能非常高，很适合存储频繁读写的计数量。
-
-## 缓存
-
-将热点数据放到内存中，设置内存的最大使用量以及淘汰策略来保证缓存的命中率。
-
-## 查找表
-
-例如 DNS 记录就很适合使用 Redis 进行存储。
-
-查找表和缓存类似，也是利用了 Redis 快速的查找特性。但是查找表的内容不能失效，而缓存的内容可以失效，因为缓存不作为可靠的数据来源。
-
-## 消息队列
-
-List 是一个双向链表，可以通过 lpop 和 lpush 写入和读取消息。
-
-不过最好使用 Kafka、RabbitMQ 等消息中间件。
-
-## 会话缓存
-
-在分布式场景下具有多个应用服务器，可以使用 Redis 来统一存储这些应用服务器的会话信息。
-
-当应用服务器不再存储用户的会话信息，也就不再具有状态，一个用户可以请求任意一个应用服务器。
-
-## 分布式锁实现
-
-在分布式场景下，无法使用单机环境下的锁来对多个节点上的进程进行同步。
-
-可以使用 Reids 自带的 SETNX 命令实现分布式锁，除此之外，还可以使用官方提供的 RedLock 分布式锁实现。
-
-## 其它
-
-Set 可以实现交集、并集等操作，从而实现共同好友等功能。
-
-ZSet 可以实现有序性操作，从而实现排行榜等功能。
-
-# 五、Redis 与 Memcached
-
-两者都是非关系型内存键值数据库，主要有以下不同：
-
-## 数据类型
-
-Memcached 仅支持字符串类型，而 Redis 支持五种不同的数据类型，可以更灵活地解决问题。
-
-## 数据持久化
-
-Redis 支持两种持久化策略：RDB 快照和 AOF 日志，而 Memcached 不支持持久化。
-
-## 分布式
-
-Memcached 不支持分布式，只能通过在客户端使用一致性哈希来实现分布式存储，这种方式在存储和查询时都需要先在客户端计算一次数据所在的节点。
-
-Redis Cluster 实现了分布式的支持。
-
-## 内存管理机制
-
-- 在 Redis 中，并不是所有数据都一直存储在内存中，可以将一些很久没用的 value 交换到磁盘，而 Memcached 的数据则会一直在内存中。
-
-- Memcached 将内存分割成特定长度的块来存储数据，以完全解决内存碎片的问题，但是这种方式会使得内存的利用率不高，例如块的大小为 128 bytes，只存储 100 bytes 的数据，那么剩下的 28 bytes 就浪费掉了。
-
-# 六、键的过期时间
-
-Redis 可以为每个键设置过期时间，当键过期时，会自动删除该键。
-
-对于散列表这种容器，只能为整个键设置过期时间（整个散列表），而不能为键里面的单个元素设置过期时间。
-
-# 七、数据淘汰策略
-
-可以设置内存最大使用量，当内存使用量超出时，会施行数据淘汰策略。
-
-Reids 具体有 6 种淘汰策略：
-
-| 策略 | 描述 |
-| :--: | :--: |
-| volatile-lru | 从已设置过期时间的数据集中挑选最近最少使用的数据淘汰 |
-| volatile-ttl | 从已设置过期时间的数据集中挑选将要过期的数据淘汰 |
-|volatile-random | 从已设置过期时间的数据集中任意选择数据淘汰 |
-| allkeys-lru | 从所有数据集中挑选最近最少使用的数据淘汰 |
-| allkeys-random | 从所有数据集中任意选择数据进行淘汰 |
-| noeviction | 禁止驱逐数据 |
-
-作为内存数据库，出于对性能和内存消耗的考虑，Redis 的淘汰算法实际实现上并非针对所有 key，而是抽样一小部分 key 从中选出被淘汰 key。
-
-使用 Redis 缓存数据时，为了提高缓存命中率，需要保证缓存数据都是热点数据。可以将内存最大使用量设置为热点数据占用的内存量，然后启用 allkeys-lru 淘汰策略，将最近最少使用的数据淘汰。
-
-Redis 4.0 引入了 volatile-lfu 和 allkeys-lfu 淘汰策略，LFU 策略通过统计访问频率，将访问频率最少的键值对淘汰。
-
-# 八、持久化
-
-Redis 是内存型数据库，为了保证数据在断电后不会丢失，需要将内存中的数据持久化到硬盘上。
-
-## RDB 持久化
-
-将某个时间点的所有数据都存放到硬盘上。
-
-可以将快照复制到其它服务器从而创建具有相同数据的服务器副本。
-
-如果系统发生故障，将会丢失最后一次创建快照之后的数据。
-
-如果数据量很大，保存快照的时间会很长。
-
-## AOF 持久化
-
-将写命令添加到 AOF 文件（Append Only File）的末尾。
-
-使用 AOF 持久化需要设置同步选项，从而确保写命令什么时候会同步到磁盘文件上。这是因为对硬盘的文件进行写入并不会马上将内容同步到磁盘文件上，而是先存储到缓冲区，然后由操作系统决定什么时候同步到硬盘。有以下同步选项：
-
-| 选项 | 同步频率 |
-| :--: | :--: |
-| always | 每个写命令都同步 |
-| everysec | 每秒同步一次 |
-| no | 让操作系统来决定何时同步 |
-
-- always 选项会严重减低服务器的性能；
-- everysec 选项比较合适，可以保证系统奔溃时只会丢失一秒左右的数据，并且 Redis 每秒执行一次同步对服务器性能几乎没有任何影响；
-- no 选项并不能给服务器性能带来多大的提升，而且也会增加系统奔溃时数据丢失的数量。
-
-随着服务器写请求的增多，AOF 文件会越来越大。Redis 提供了一种将 AOF 重写的特性，能够去除 AOF 文件中的冗余写命令。
-
-# 九、事务
-
-一个事务包含了多个命令，服务器在执行事务期间，不会改去执行其它客户端的命令请求。
-
-事务中的多个命令被一次性发送给服务器，而不是一条一条发送，这种方式被称为流水线，它可以减少客户端与服务器之间的网络通信次数从而提升性能。
-
-Redis 最简单的事务实现方式是使用 MULTI 和 EXEC 命令将事务操作包围起来。
-
-# 十、事件
-
-Redis 服务器是一个事件驱动程序。
-
-## 文件事件
-
-服务器通过套接字与客户端或者其它服务器进行通信，文件事件就是对套接字操作的抽象。
-
-Redis 基于 Reactor 模式开发了自己的网络事件处理器，使用 I/O 多路复用程序来同时监听多个套接字，并将到达的事件传送给文件事件分派器，分派器会根据套接字产生的事件类型调用相应的事件处理器。
-
-<div align="center"> <img src="../pics//9ea86eb5-000a-4281-b948-7b567bd6f1d8.png"/> </div><br>
-
-## 时间事件
-
-服务器有一些操作需要在给定的时间点执行，时间事件是对这类定时操作的抽象。
-
-时间事件又分为：
-
-- 定时事件：是让一段程序在指定的时间之内执行一次；
-- 周期性事件：是让一段程序每隔指定时间就执行一次。
-
-Redis 将所有时间事件都放在一个无序链表中，通过遍历整个链表查找出已到达的时间事件，并调用相应的事件处理器。
-
-## 事件的调度与执行
-
-服务器需要不断监听文件事件的套接字才能得到待处理的文件事件，但是不能一直监听，否则时间事件无法在规定的时间内执行，因此监听时间应该根据距离现在最近的时间事件来决定。
-
-事件调度与执行由 aeProcessEvents 函数负责，伪代码如下：
-
-```python
-def aeProcessEvents():
-    # 获取到达时间离当前时间最接近的时间事件
-    time_event = aeSearchNearestTimer()
-    # 计算最接近的时间事件距离到达还有多少毫秒
-    remaind_ms = time_event.when - unix_ts_now()
-    # 如果事件已到达，那么 remaind_ms 的值可能为负数，将它设为 0
-    if remaind_ms < 0:
-        remaind_ms = 0
-    # 根据 remaind_ms 的值，创建 timeval
-    timeval = create_timeval_with_ms(remaind_ms)
-    # 阻塞并等待文件事件产生，最大阻塞时间由传入的 timeval 决定
-    aeApiPoll(timeval)
-    # 处理所有已产生的文件事件
-    procesFileEvents()
-    # 处理所有已到达的时间事件
-    processTimeEvents()
-```
-
-将 aeProcessEvents 函数置于一个循环里面，加上初始化和清理函数，就构成了 Redis 服务器的主函数，伪代码如下：
-
-```python
-def main():
-    # 初始化服务器
-    init_server()
-    # 一直处理事件，直到服务器关闭为止
-    while server_is_not_shutdown():
-        aeProcessEvents()
-    # 服务器关闭，执行清理操作
-    clean_server()
-```
-
-从事件处理的角度来看，服务器运行流程如下：
-
-<div align="center"> <img src="../pics//c0a9fa91-da2e-4892-8c9f-80206a6f7047.png" width="400"/> </div><br>
-
-# 十一、复制
-
-通过使用 slaveof host port 命令来让一个服务器成为另一个服务器的从服务器。
-
-一个从服务器只能有一个主服务器，并且不支持主主复制。
-
-## 连接过程
-
-1. 主服务器创建快照文件，发送给从服务器，并在发送期间使用缓冲区记录执行的写命令。快照文件发送完毕之后，开始向从服务器发送存储在缓冲区中的写命令；
-
-2. 从服务器丢弃所有旧数据，载入主服务器发来的快照文件，之后从服务器开始接受主服务器发来的写命令；
-
-3. 主服务器每执行一次写命令，就向从服务器发送相同的写命令。
-
-## 主从链
-
-随着负载不断上升，主服务器可能无法很快地更新所有从服务器，或者重新连接和重新同步从服务器将导致系统超载。为了解决这个问题，可以创建一个中间层来分担主服务器的复制工作。中间层的服务器是最上层服务器的从服务器，又是最下层服务器的主服务器。
-
-<div align="center"> <img src="../pics//395a9e83-b1a1-4a1d-b170-d081e7bb5bab.png" width="600"/> </div><br>
-
-# 十二、Sentinel
-
-Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状态时，自动从从服务器中选举出新的主服务器。
-
-# 十三、分片
-
-分片是将数据划分为多个部分的方法，可以将数据存储到多台机器里面，也可以从多台机器里面获取数据，这种方法在解决某些问题时可以获得线性级别的性能提升。
-
-假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... 等等，有不同的方式来选择一个指定的键存储在哪个实例中。最简单的方式是范围分片，例如用户 id 从 0\~1000 的存储到实例 R0 中，用户 id 从 1001\~2000 的存储到实例 R1 中，等等。但是这样需要维护一张映射范围表，维护操作代价很高。还有一种方式是哈希分片，使用 CRC32 哈希函数将键转换为一个数字，再对实例数量求模就能知道应该存储的实例。
-
-根据执行分片的位置，可以分为三种分片方式：
-
-- 客户端分片：客户端使用一致性哈希等算法决定键应当分布到哪个节点。
-- 代理分片：将客户端请求发送到代理上，由代理转发请求到正确的节点上。
-- 服务器分片：Redis Cluster。
-
-# 十四、一个简单的论坛系统分析
-
-该论坛系统功能如下：
-
-- 可以发布文章；
-- 可以对文章进行点赞；
-- 在首页可以按文章的发布时间或者文章的点赞数进行排序显示。
-
-## 文章信息
-
-文章包括标题、作者、赞数等信息，在关系型数据库中很容易构建一张表来存储这些信息，在 Redis 中可以使用 HASH 来存储每种信息以及其对应的值的映射。
-
-Redis 没有关系型数据库中的表这一概念来将同种类型的数据存放在一起，而是使用命名空间的方式来实现这一功能。键名的前面部分存储命名空间，后面部分的内容存储 ID，通常使用 : 来进行分隔。例如下面的 HASH 的键名为 article:92617，其中 article 为命名空间，ID 为 92617。
-
-<div align="center"> <img src="../pics//7c54de21-e2ff-402e-bc42-4037de1c1592.png" width="400"/> </div><br>
-
-## 点赞功能
-
-当有用户为一篇文章点赞时，除了要对该文章的 votes 字段进行加 1 操作，还必须记录该用户已经对该文章进行了点赞，防止用户点赞次数超过 1。可以建立文章的已投票用户集合来进行记录。
-
-为了节约内存，规定一篇文章发布满一周之后，就不能再对它进行投票，而文章的已投票集合也会被删除，可以为文章的已投票集合设置一个一周的过期时间就能实现这个规定。
-
-<div align="center"> <img src="../pics//485fdf34-ccf8-4185-97c6-17374ee719a0.png" width="400"/> </div><br>
-
-## 对文章进行排序
-
-为了按发布时间和点赞数进行排序，可以建立一个文章发布时间的有序集合和一个文章点赞数的有序集合。（下图中的 score 就是这里所说的点赞数；下面所示的有序集合分值并不直接是时间和点赞数，而是根据时间和点赞数间接计算出来的）
-
-<div align="center"> <img src="../pics//f7d170a3-e446-4a64-ac2d-cb95028f81a8.png" width="800"/> </div><br>
-
-# 参考资料
-
-- Carlson J L. Redis in Action[J]. Media.johnwiley.com.au, 2013.
-- [黄健宏. Redis 设计与实现 [M]. 机械工业出版社, 2014.](http://redisbook.com/index.html)
-- [REDIS IN ACTION](https://redislabs.com/ebook/foreword/)
-- [Skip Lists: Done Right](http://ticki.github.io/blog/skip-lists-done-right/)
-- [论述 Redis 和 Memcached 的差异](http://www.cnblogs.com/loveincode/p/7411911.html)
-- [Redis 3.0 中文版- 分片](http://wiki.jikexueyuan.com/project/redis-guide)
-- [Redis 应用场景](http://www.scienjus.com/redis-use-case/)
-- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
-- [Using Redis as an LRU cache](https://redis.io/topics/lru-cache)
diff --git a/notes/SQL.md b/notes/SQL.md
index ca9b7b29..649f563a 100644
--- a/notes/SQL.md
+++ b/notes/SQL.md
@@ -1,767 +1,4 @@
 <!-- GFM-TOC -->
-* [一、基础](#一基础)
-* [二、创建表](#二创建表)
-* [三、修改表](#三修改表)
-* [四、插入](#四插入)
-* [五、更新](#五更新)
-* [六、删除](#六删除)
-* [七、查询](#七查询)
-* [八、排序](#八排序)
-* [九、过滤](#九过滤)
-* [十、通配符](#十通配符)
-* [十一、计算字段](#十一计算字段)
-* [十二、函数](#十二函数)
-* [十三、分组](#十三分组)
-* [十四、子查询](#十四子查询)
-* [十五、连接](#十五连接)
-* [十六、组合查询](#十六组合查询)
-* [十七、视图](#十七视图)
-* [十八、存储过程](#十八存储过程)
-* [十九、游标](#十九游标)
-* [二十、触发器](#二十触发器)
-* [二十一、事务处理](#二十一事务处理)
-* [二十二、字符集](#二十二字符集)
-* [二十三、权限管理](#二十三权限管理)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、基础
-
-模式定义了数据如何存储、存储什么样的数据以及数据如何分解等信息，数据库和表都有模式。
-
-主键的值不允许修改，也不允许复用（不能使用已经删除的主键值赋给新数据行的主键）。
-
-SQL（Structured Query Language)，标准 SQL 由 ANSI 标准委员会管理，从而称为 ANSI SQL。各个 DBMS 都有自己的实现，如 PL/SQL、Transact-SQL 等。
-
-SQL 语句不区分大小写，但是数据库表名、列名和值是否区分依赖于具体的 DBMS 以及配置。
-
-SQL 支持以下三种注释：
-
-```sql
-# 注释
-SELECT *
-FROM mytable; -- 注释
-/* 注释1
-   注释2 */
-```
-
-数据库创建与使用：
-
-```sql
-CREATE DATABASE test;
-USE test;
-```
-
-# 二、创建表
-
-```sql
-CREATE TABLE mytable (
-  id INT NOT NULL AUTO_INCREMENT,
-  col1 INT NOT NULL DEFAULT 1,
-  col2 VARCHAR(45) NULL,
-  col3 DATE NULL,
-  PRIMARY KEY (`id`));
-```
-
-# 三、修改表
-
-添加列
-
-```sql
-ALTER TABLE mytable
-ADD col CHAR(20);
-```
-
-删除列
-
-```sql
-ALTER TABLE mytable
-DROP COLUMN col;
-```
-
-删除表
-
-```sql
-DROP TABLE mytable;
-```
-
-# 四、插入
-
-普通插入
-
-```sql
-INSERT INTO mytable(col1, col2)
-VALUES(val1, val2);
-```
-
-插入检索出来的数据
-
-```sql
-INSERT INTO mytable1(col1, col2)
-SELECT col1, col2
-FROM mytable2;
-```
-
-将一个表的内容插入到一个新表
-
-```sql
-CREATE TABLE newtable AS
-SELECT * FROM mytable;
-```
-
-# 五、更新
-
-```sql
-UPDATE mytable
-SET col = val
-WHERE id = 1;
-```
-
-# 六、删除
-
-```sql
-DELETE FROM mytable
-WHERE id = 1;
-```
-
-**TRUNCATE TABLE**  可以清空表，也就是删除所有行。
-
-```sql
-TRUNCATE TABLE mytable;
-```
-
-使用更新和删除操作时一定要用 WHERE 子句，不然会把整张表的数据都破坏。可以先用 SELECT 语句进行测试，防止错误删除。
-
-# 七、查询
-
-## DISTINCT
-
-相同值只会出现一次。它作用于所有列，也就是说所有列的值都相同才算相同。
-
-```sql
-SELECT DISTINCT col1, col2
-FROM mytable;
-```
-
-## LIMIT
-
-限制返回的行数。可以有两个参数，第一个参数为起始行，从 0 开始；第二个参数为返回的总行数。
-
-返回前 5 行：
-
-```sql
-SELECT *
-FROM mytable
-LIMIT 5;
-```
-
-```sql
-SELECT *
-FROM mytable
-LIMIT 0, 5;
-```
-
-返回第 3 \~ 5 行：
-
-```sql
-SELECT *
-FROM mytable
-LIMIT 2, 3;
-```
-
-# 八、排序
-
--  **ASC** ：升序（默认）
--  **DESC** ：降序
-
-可以按多个列进行排序，并且为每个列指定不同的排序方式：
-
-```sql
-SELECT *
-FROM mytable
-ORDER BY col1 DESC, col2 ASC;
-```
-
-# 九、过滤
-
-不进行过滤的数据非常大，导致通过网络传输了多余的数据，从而浪费了网络带宽。因此尽量使用 SQL 语句来过滤不必要的数据，而不是传输所有的数据到客户端中然后由客户端进行过滤。
-
-```sql
-SELECT *
-FROM mytable
-WHERE col IS NULL;
-```
-
-下表显示了 WHERE 子句可用的操作符
-
-|  操作符 | 说明  |
-| :---: | :---: |
-| = | 等于 |
-| &lt; | 小于 |
-| &gt; | 大于 |
-| &lt;&gt; != | 不等于 |
-| &lt;= !&gt; | 小于等于 |
-| &gt;= !&lt; | 大于等于 |
-| BETWEEN | 在两个值之间 |
-| IS NULL | 为 NULL 值 |
-
-应该注意到，NULL 与 0、空字符串都不同。
-
-**AND 和 OR**  用于连接多个过滤条件。优先处理 AND，当一个过滤表达式涉及到多个 AND 和 OR 时，可以使用 () 来决定优先级，使得优先级关系更清晰。
-
-**IN**  操作符用于匹配一组值，其后也可以接一个 SELECT 子句，从而匹配子查询得到的一组值。
-
-**NOT**  操作符用于否定一个条件。
-
-# 十、通配符
-
-通配符也是用在过滤语句中，但它只能用于文本字段。
-
--  **%**  匹配 >=0 个任意字符；
-
--  **\_**  匹配 ==1 个任意字符；
-
--  **[ ]**  可以匹配集合内的字符，例如 [ab] 将匹配字符 a 或者 b。用脱字符 ^ 可以对其进行否定，也就是不匹配集合内的字符。
-
-使用 Like 来进行通配符匹配。
-
-```sql
-SELECT *
-FROM mytable
-WHERE col LIKE '[^AB]%'; -- 不以 A 和 B 开头的任意文本
-```
-
-不要滥用通配符，通配符位于开头处匹配会非常慢。
-
-# 十一、计算字段
-
-在数据库服务器上完成数据的转换和格式化的工作往往比客户端上快得多，并且转换和格式化后的数据量更少的话可以减少网络通信量。
-
-计算字段通常需要使用  **AS**  来取别名，否则输出的时候字段名为计算表达式。
-
-```sql
-SELECT col1 * col2 AS alias
-FROM mytable;
-```
-
-**CONCAT()**  用于连接两个字段。许多数据库会使用空格把一个值填充为列宽，因此连接的结果会出现一些不必要的空格，使用 **TRIM()** 可以去除首尾空格。
-
-```sql
-SELECT CONCAT(TRIM(col1), '(', TRIM(col2), ')') AS concat_col
-FROM mytable;
-```
-
-# 十二、函数
-
-各个 DBMS 的函数都是不相同的，因此不可移植，以下主要是 MySQL 的函数。
-
-## 汇总
-
-|函 数 |说 明|
-| :---: | :---: |
-| AVG() | 返回某列的平均值 |
-| COUNT() | 返回某列的行数 |
-| MAX() | 返回某列的最大值 |
-| MIN() | 返回某列的最小值 |
-| SUM() |返回某列值之和 |
-
-AVG() 会忽略 NULL 行。
-
-使用 DISTINCT 可以让汇总函数值汇总不同的值。
-
-```sql
-SELECT AVG(DISTINCT col1) AS avg_col
-FROM mytable;
-```
-
-## 文本处理
-
-| 函数  | 说明  |
-| :---: | :---: |
-|  LEFT() |  左边的字符 |
-| RIGHT() | 右边的字符 |
-| LOWER() | 转换为小写字符 |
-| UPPER() | 转换为大写字符 |
-| LTRIM() | 去除左边的空格 |
-| RTRIM() | 去除右边的空格 |
-| LENGTH() | 长度 |
-| SOUNDEX() | 转换为语音值 |
-
-其中， **SOUNDEX()**  可以将一个字符串转换为描述其语音表示的字母数字模式。
-
-```sql
-SELECT *
-FROM mytable
-WHERE SOUNDEX(col1) = SOUNDEX('apple')
-```
-
-## 日期和时间处理
-
-- 日期格式：YYYY-MM-DD
-- 时间格式：HH:MM:SS
-
-|函 数 | 说 明|
-| :---: | :---: |
-| AddDate() | 增加一个日期（天、周等）|
-| AddTime() | 增加一个时间（时、分等）|
-| CurDate() | 返回当前日期 |
-| CurTime() | 返回当前时间 |
-| Date() |返回日期时间的日期部分|
-| DateDiff() |计算两个日期之差|
-| Date_Add() |高度灵活的日期运算函数|
-| Date_Format() |返回一个格式化的日期或时间串|
-| Day()| 返回一个日期的天数部分|
-| DayOfWeek() |对于一个日期，返回对应的星期几|
-| Hour() |返回一个时间的小时部分|
-| Minute() |返回一个时间的分钟部分|
-| Month() |返回一个日期的月份部分|
-| Now() |返回当前日期和时间|
-| Second() |返回一个时间的秒部分|
-| Time() |返回一个日期时间的时间部分|
-| Year() |返回一个日期的年份部分|
-
-```sql
-mysql> SELECT NOW();
-```
-
-```
-2018-4-14 20:25:11
-```
-
-## 数值处理
-
-| 函数 | 说明 |
-| :---: | :---: |
-| SIN() | 正弦 |
-| COS() | 余弦 |
-| TAN() | 正切 |
-| ABS() | 绝对值 |
-| SQRT() | 平方根 |
-| MOD() | 余数 |
-| EXP() | 指数 |
-| PI() | 圆周率 |
-| RAND() | 随机数 |
-
-# 十三、分组
-
-分组就是把具有相同的数据值的行放在同一组中。
-
-可以对同一分组数据使用汇总函数进行处理，例如求分组数据的平均值等。
-
-指定的分组字段除了能按该字段进行分组，也会自动按该字段进行排序。
-
-```sql
-SELECT col, COUNT(*) AS num
-FROM mytable
-GROUP BY col;
-```
-
-GROUP BY 自动按分组字段进行排序，ORDER BY 也可以按汇总字段来进行排序。
-
-```sql
-SELECT col, COUNT(*) AS num
-FROM mytable
-GROUP BY col
-ORDER BY num;
-```
-
-WHERE 过滤行，HAVING 过滤分组，行过滤应当先于分组过滤。
-
-```sql
-SELECT col, COUNT(*) AS num
-FROM mytable
-WHERE col > 2
-GROUP BY col
-HAVING num >= 2;
-```
-
-分组规定：
-
-- GROUP BY 子句出现在 WHERE 子句之后，ORDER BY 子句之前；
-- 除了汇总字段外，SELECT 语句中的每一字段都必须在 GROUP BY 子句中给出；
-- NULL 的行会单独分为一组；
-- 大多数 SQL 实现不支持 GROUP BY 列具有可变长度的数据类型。
-
-# 十四、子查询
-
-子查询中只能返回一个字段的数据。
-
-可以将子查询的结果作为 WHRER 语句的过滤条件：
-
-```sql
-SELECT *
-FROM mytable1
-WHERE col1 IN (SELECT col2
-               FROM mytable2);
-```
-
-下面的语句可以检索出客户的订单数量，子查询语句会对第一个查询检索出的每个客户执行一次：
-
-```sql
-SELECT cust_name, (SELECT COUNT(*)
-                   FROM Orders
-                   WHERE Orders.cust_id = Customers.cust_id)
-                   AS orders_num
-FROM Customers
-ORDER BY cust_name;
-```
-
-# 十五、连接
-
-连接用于连接多个表，使用 JOIN 关键字，并且条件语句使用 ON 而不是 WHERE。
-
-连接可以替换子查询，并且比子查询的效率一般会更快。
-
-可以用 AS 给列名、计算字段和表名取别名，给表名取别名是为了简化 SQL 语句以及连接相同表。
-
-## 内连接
-
-内连接又称等值连接，使用 INNER JOIN 关键字。
-
-```sql
-SELECT A.value, B.value
-FROM tablea AS A INNER JOIN tableb AS B
-ON A.key = B.key;
-```
-
-可以不明确使用 INNER JOIN，而使用普通查询并在 WHERE 中将两个表中要连接的列用等值方法连接起来。
-
-```sql
-SELECT A.value, B.value
-FROM tablea AS A, tableb AS B
-WHERE A.key = B.key;
-```
-
-在没有条件语句的情况下返回笛卡尔积。
-
-## 自连接
-
-自连接可以看成内连接的一种，只是连接的表是自身而已。
-
-一张员工表，包含员工姓名和员工所属部门，要找出与 Jim 处在同一部门的所有员工姓名。
-
-子查询版本
-
-```sql
-SELECT name
-FROM employee
-WHERE department = (
-      SELECT department
-      FROM employee
-      WHERE name = "Jim");
-```
-
-自连接版本
-
-```sql
-SELECT e1.name
-FROM employee AS e1 INNER JOIN employee AS e2
-ON e1.department = e2.department
-      AND e2.name = "Jim";
-```
-
-## 自然连接
-
-自然连接是把同名列通过等值测试连接起来的，同名列可以有多个。
-
-内连接和自然连接的区别：内连接提供连接的列，而自然连接自动连接所有同名列。
-
-```sql
-SELECT A.value, B.value
-FROM tablea AS A NATURAL JOIN tableb AS B;
-```
-
-## 外连接
-
-外连接保留了没有关联的那些行。分为左外连接，右外连接以及全外连接，左外连接就是保留左表没有关联的行。
-
-检索所有顾客的订单信息，包括还没有订单信息的顾客。
-
-```sql
-SELECT Customers.cust_id, Orders.order_num
-FROM Customers LEFT OUTER JOIN Orders
-ON Customers.cust_id = Orders.cust_id;
-```
-
-customers 表：
-
-| cust_id | cust_name |
-| :---: | :---: |
-| 1 | a |
-| 2 | b |
-| 3 | c |
-
-orders 表：
-
-| order_id | cust_id |
-| :---: | :---: |
-|1    | 1 |
-|2    | 1 |
-|3    | 3 |
-|4    | 3 |
-
-结果：
-
-| cust_id | cust_name | order_id |
-| :---: | :---: | :---: |
-| 1 | a | 1 |
-| 1 | a | 2 |
-| 3 | c | 3 |
-| 3 | c | 4 |
-| 2 | b | Null |
-
-# 十六、组合查询
-
-使用  **UNION**  来组合两个查询，如果第一个查询返回 M 行，第二个查询返回 N 行，那么组合查询的结果一般为 M+N 行。
-
-每个查询必须包含相同的列、表达式和聚集函数。
-
-默认会去除相同行，如果需要保留相同行，使用 UNION ALL。
-
-只能包含一个 ORDER BY 子句，并且必须位于语句的最后。
-
-```sql
-SELECT col
-FROM mytable
-WHERE col = 1
-UNION
-SELECT col
-FROM mytable
-WHERE col =2;
-```
-
-# 十七、视图
-
-视图是虚拟的表，本身不包含数据，也就不能对其进行索引操作。
-
-对视图的操作和对普通表的操作一样。
-
-视图具有如下好处：
-
-- 简化复杂的 SQL 操作，比如复杂的连接；
-- 只使用实际表的一部分数据；
-- 通过只给用户访问视图的权限，保证数据的安全性；
-- 更改数据格式和表示。
-
-```sql
-CREATE VIEW myview AS
-SELECT Concat(col1, col2) AS concat_col, col3*col4 AS compute_col
-FROM mytable
-WHERE col5 = val;
-```
-
-# 十八、存储过程
-
-存储过程可以看成是对一系列 SQL 操作的批处理；
-
-使用存储过程的好处：
-
-- 代码封装，保证了一定的安全性；
-- 代码复用；
-- 由于是预先编译，因此具有很高的性能。
-
-命令行中创建存储过程需要自定义分隔符，因为命令行是以 ; 为结束符，而存储过程中也包含了分号，因此会错误把这部分分号当成是结束符，造成语法错误。
-
-包含 in、out 和 inout 三种参数。
-
-给变量赋值都需要用 select into 语句。
-
-每次只能给一个变量赋值，不支持集合的操作。
-
-```sql
-delimiter //
-
-create procedure myprocedure( out ret int )
-    begin
-        declare y int;
-        select sum(col1)
-        from mytable
-        into y;
-        select y*y into ret;
-    end //
-
-delimiter ;
-```
-
-```sql
-call myprocedure(@ret);
-select @ret;
-```
-
-# 十九、游标
-
-在存储过程中使用游标可以对一个结果集进行移动遍历。
-
-游标主要用于交互式应用，其中用户需要对数据集中的任意行进行浏览和修改。
-
-使用游标的四个步骤：
-
-1. 声明游标，这个过程没有实际检索出数据；
-2. 打开游标；
-3. 取出数据；
-4. 关闭游标；
-
-```sql
-delimiter //
-create procedure myprocedure(out ret int)
-    begin
-        declare done boolean default 0;
-
-        declare mycursor cursor for
-        select col1 from mytable;
-        # 定义了一个 continue handler，当 sqlstate '02000' 这个条件出现时，会执行 set done = 1
-        declare continue handler for sqlstate '02000' set done = 1;
-
-        open mycursor;
-
-        repeat
-            fetch mycursor into ret;
-            select ret;
-        until done end repeat;
-
-        close mycursor;
-    end //
- delimiter ;
-```
-
-# 二十、触发器
-
-触发器会在某个表执行以下语句时而自动执行：DELETE、INSERT、UPDATE。
-
-触发器必须指定在语句执行之前还是之后自动执行，之前执行使用 BEFORE 关键字，之后执行使用 AFTER 关键字。BEFORE 用于数据验证和净化，AFTER 用于审计跟踪，将修改记录到另外一张表中。
-
-INSERT 触发器包含一个名为 NEW 的虚拟表。
-
-```sql
-CREATE TRIGGER mytrigger AFTER INSERT ON mytable
-FOR EACH ROW SELECT NEW.col into @result;
-
-SELECT @result; -- 获取结果
-```
-
-DELETE 触发器包含一个名为 OLD 的虚拟表，并且是只读的。
-
-UPDATE 触发器包含一个名为 NEW 和一个名为 OLD 的虚拟表，其中 NEW 是可以被修改地，而 OLD 是只读的。
-
-MySQL 不允许在触发器中使用 CALL 语句，也就是不能调用存储过程。
-
-# 二十一、事务处理
-
-基本术语：
-
-- 事务（transaction）指一组 SQL 语句；
-- 回退（rollback）指撤销指定 SQL 语句的过程；
-- 提交（commit）指将未存储的 SQL 语句结果写入数据库表；
-- 保留点（savepoint）指事务处理中设置的临时占位符（placeholder），你可以对它发布回退（与回退整个事务处理不同）。
-
-不能回退 SELECT 语句，回退 SELECT 语句也没意义；也不能回退 CREATE 和 DROP 语句。
-
-MySQL 的事务提交默认是隐式提交，每执行一条语句就把这条语句当成一个事务然后进行提交。当出现 START TRANSACTION 语句时，会关闭隐式提交；当 COMMIT 或 ROLLBACK 语句执行后，事务会自动关闭，重新恢复隐式提交。
-
-通过设置 autocommit 为 0 可以取消自动提交；autocommit 标记是针对每个连接而不是针对服务器的。
-
-如果没有设置保留点，ROLLBACK 会回退到 START TRANSACTION 语句处；如果设置了保留点，并且在 ROLLBACK 中指定该保留点，则会回退到该保留点。
-
-```sql
-START TRANSACTION
-// ...
-SAVEPOINT delete1
-// ...
-ROLLBACK TO delete1
-// ...
-COMMIT
-```
-
-# 二十二、字符集
-
-基本术语：
-
-- 字符集为字母和符号的集合；
-- 编码为某个字符集成员的内部表示；
-- 校对字符指定如何比较，主要用于排序和分组。
-
-除了给表指定字符集和校对外，也可以给列指定：
-
-```sql
-CREATE TABLE mytable
-(col VARCHAR(10) CHARACTER SET latin COLLATE latin1_general_ci )
-DEFAULT CHARACTER SET hebrew COLLATE hebrew_general_ci;
-```
-
-可以在排序、分组时指定校对：
-
-```sql
-SELECT *
-FROM mytable
-ORDER BY col COLLATE latin1_general_ci;
-```
-
-# 二十三、权限管理
-
-MySQL 的账户信息保存在 mysql 这个数据库中。
-
-```sql
-USE mysql;
-SELECT user FROM user;
-```
-
-**创建账户** 
-
-```sql
-CREATE USER myuser IDENTIFIED BY 'mypassword';
-```
-
-新创建的账户没有任何权限。
-
-**修改账户名** 
-
-```sql
-RENAME myuser TO newuser;
-```
-
-**删除账户** 
-
-```sql
-DROP USER myuser;
-```
-
-**查看权限** 
-
-```sql
-SHOW GRANTS FOR myuser;
-```
-
-**授予权限** 
-
-```sql
-GRANT SELECT, INSERT ON mydatabase.* TO myuser;
-```
-
-账户用 username@host 的形式定义，username@% 使用的是默认主机名。
-
-**删除权限** 
-
-```sql
-REVOKE SELECT, INSERT ON mydatabase.* FROM myuser;
-```
-
-GRANT 和 REVOKE 可在几个层次上控制访问权限：
-
-- 整个服务器，使用 GRANT ALL 和 REVOKE ALL；
-- 整个数据库，使用 ON database.\*；
-- 特定的表，使用 ON database.table；
-- 特定的列；
-- 特定的存储过程。
-
-**更改密码** 
-
-必须使用 Password() 函数
-
-```sql
-SET PASSWROD FOR myuser = Password('new_password');
-```
-
-# 参考资料
-
-- BenForta. SQL 必知必会 [M]. 人民邮电出版社, 2013.
diff --git a/notes/分布式.md b/notes/分布式.md
index 0f136855..649f563a 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -1,359 +1,4 @@
 <!-- GFM-TOC -->
-* [一、分布式锁](#一分布式锁)
-    * [数据库的唯一索引](#数据库的唯一索引)
-    * [Redis 的 SETNX 指令](#redis-的-setnx-指令)
-    * [Redis 的 RedLock 算法](#redis-的-redlock-算法)
-    * [Zookeeper 的有序节点](#zookeeper-的有序节点)
-* [二、分布式事务](#二分布式事务)
-    * [本地消息表](#本地消息表)
-    * [2PC](#2pc)
-* [三、CAP](#三cap)
-    * [一致性](#一致性)
-    * [可用性](#可用性)
-    * [分区容忍性](#分区容忍性)
-    * [权衡](#权衡)
-* [四、BASE](#四base)
-    * [基本可用](#基本可用)
-    * [软状态](#软状态)
-    * [最终一致性](#最终一致性)
-* [五、Paxos](#五paxos)
-    * [执行过程](#执行过程)
-    * [约束条件](#约束条件)
-* [五、Raft](#五raft)
-    * [单个 Candidate 的竞选](#单个-candidate-的竞选)
-    * [多个 Candidate 竞选](#多个-candidate-竞选)
-    * [日志复制](#日志复制)
 <!-- GFM-TOC -->
 
 
-# 一、分布式锁
-
-在单机场景下，可以使用 Java 提供的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
-
-阻塞锁通常使用互斥量来实现：
-
-- 互斥量为 1 表示有其它进程在使用锁，此时处于锁定状态；
-- 互斥量为 0 表示未锁定状态。
-
-1 和 0 可以用一个整型值表示，也可以用某个数据存在或者不存在表示，存在表示互斥量为 1。
-
-## 数据库的唯一索引
-
-当想要获得锁时，就向表中插入一条记录，释放锁时就删除这条记录。唯一索引可以保证该记录只被插入一次，那么就可以用这个记录是否存在来判断是否存于锁定状态。
-
-存在以下几个问题：
-
-- 锁没有失效时间，解锁失败的话其它进程无法再获得锁。
-- 只能是非阻塞锁，插入失败直接就报错了，无法重试。
-- 不可重入，已经获得锁的进程也必须重新获取锁。
-
-## Redis 的 SETNX 指令
-
-使用 SETNX（set if not exist）指令插入一个键值对，如果 Key 已经存在，那么会返回 False，否则插入成功并返回 True。
-
-SETNX 指令和数据库的唯一索引类似，保证了只存在一个 Key 的键值对，那么可以用一个 Key 的键值对是否存在来判断是否存于锁定状态。
-
-EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了数据库唯一索引实现方式中释放锁失败的问题。
-
-## Redis 的 RedLock 算法
-
-使用了多个 Redis 实例来实现分布式锁，这是为了保证在发生单点故障时仍然可用。
-
-- 尝试从 N 个相互独立 Redis 实例获取锁，如果一个实例不可用，应该尽快尝试下一个；
-- 计算获取锁消耗的时间，只有当这个时间小于锁的过期时间，并且从大多数（N / 2 + 1）实例上获取了锁，那么就认为锁获取成功了；
-- 如果锁获取失败，会到每个实例上释放锁。
-
-## Zookeeper 的有序节点
-
-### 1. Zookeeper 抽象模型
-
-Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示它的父节点为 /app1。
-
-<div align="center"> <img src="../pics//31d99967-1171-448e-8531-bccf5c14cffe.jpg" width="400"/> </div><br>
-
-### 2. 节点类型
-
-- 永久节点：不会因为会话结束或者超时而消失；
-- 临时节点：如果会话结束或者超时就会消失；
-- 有序节点：会在节点名的后面加一个数字后缀，并且是有序的，例如生成的有序节点为 /lock/node-0000000000，它的下一个有序节点则为 /lock/node-0000000001，以此类推。
-
-### 3. 监听器
-
-为一个节点注册监听器，在节点状态发生改变时，会给客户端发送消息。
-
-### 4. 分布式锁实现
-
-- 创建一个锁目录 /lock；
-- 当一个客户端需要获取锁时，在 /lock 下创建临时的且有序的子节点；
--  客户端获取 /lock 下的子节点列表，判断自己创建的子节点是否为当前子节点列表中序号最小的子节点，如果是则认为获得锁；否则监听自己的前一个子节点，获得子节点的变更通知后重复此步骤直至获得锁；
-- 执行业务代码，完成后，删除对应的子节点。
-
-### 5. 会话超时
-
-如果一个已经获得锁的会话超时了，因为创建的是临时节点，所以该会话对应的临时节点会被删除，其它会话就可以获得锁了。可以看到，Zookeeper 分布式锁不会出现数据库的唯一索引实现分布式锁的释放锁失败问题。
-
-### 6. 羊群效应
-
-一个节点未获得锁，需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
-
-参考：
-
-- [Distributed locks with Redis](https://redis.io/topics/distlock)
-- [浅谈分布式锁](http://www.linkedkeeper.com/detail/blog.action?bid=1023)
-- [基于 Zookeeper 的分布式锁](http://www.dengshenyu.com/java/%E5%88%86%E5%B8%83%E5%BC%8F%E7%B3%BB%E7%BB%9F/2017/10/23/zookeeper-distributed-lock.html)
-
-# 二、分布式事务
-
-指事务的操作位于不同的节点上，需要保证事务的 AICD 特性。例如在下单场景下，库存和订单如果不在同一个节点上，就涉及分布式事务。
-
-## 本地消息表
-
-### 1. 原理
-
-本地消息表与业务数据表处于同一个数据库中，这样就能利用本地事务来保证在对这两个表的操作满足事务特性。
-
-1. 在分布式事务操作的一方完成写业务数据的操作之后向本地消息表发送一个消息，本地事务能保证这个消息一定会被写入本地消息表中。
-2. 之后将本地消息表中的消息转发到 Kafka 等消息队列中，如果转发成功则将消息从本地消息表中删除，否则继续重新转发。
-3. 在分布式事务操作的另一方从消息队列中读取一个消息，并执行消息中的操作。
-
-<div align="center"> <img src="../pics//e3bf5de4-ab1e-4a9b-896d-4b0ad7e9220a.jpg"/> </div><br>
-
-### 2. 分析
-
-本地消息表利用了本地事务来实现分布式事务，并且使用了消息队列来保证最终一致性。
-
-## 2PC
-
-两阶段提交（Two-phase Commit，2PC）
-
-通过引入协调者（Coordinator）来协调参与者的行为，并最终决定这些参与者是否要真正执行事务。
-
-### 1. 运行过程
-
-（一）准备阶段
-
-协调者询问参与者事务是否执行成功，参与者发回事务执行结果。
-
-<div align="center"> <img src="../pics//04f41228-375d-4b7d-bfef-738c5a7c8f07.jpg"/> </div><br>
-
-（二）提交阶段
-
-如果事务在每个参与者上都执行成功，事务协调者发送通知让参与者提交事务；否则，协调者发送通知让参与者回滚事务。
-
-需要注意的是，在准备阶段，参与者执行了事务，但是还未提交。只有在提交阶段接收到协调者发来的通知后，才进行提交或者回滚。
-
-<div align="center"> <img src="../pics//2991c772-fb1c-4051-a9c7-932b68e76bd7.jpg"/> </div><br>
-
-### 2. 存在的问题
-
-（一）同步阻塞
-
-所有事务参与者在等待其它参与者响应的时候都处于同步阻塞状态，无法进行其它操作。
-
-（二）单点问题
-
-协调者在 2PC 中起到非常大的作用，发生故障将会造成很大影响，特别是在阶段二发生故障，所有参与者会一直等待状态，无法完成其它操作。
-
-（三）数据不一致
-
-在阶段二，如果协调者只发送了部分 Commit 消息，此时网络发生异常，那么只有部分参与者接收到 Commit 消息，也就是说只有部分参与者提交了事务，使得系统数据不一致。
-
-（四）太过保守
-
-任意一个节点失败就会导致整个事务失败，没有完善的容错机制。
-
-参考：
-
-- [聊聊分布式事务，再说说解决方案](https://www.cnblogs.com/savorboard/p/distributed-system-transaction-consistency.html)
-- [分布式系统的事务处理](https://coolshell.cn/articles/10910.html)
-- [深入理解分布式事务](https://juejin.im/entry/577c6f220a2b5800573492be)
-
-# 三、CAP
-
-分布式系统不可能同时满足一致性（C：Consistency）、可用性（A：Availability）和分区容忍性（P：Partition Tolerance），最多只能同时满足其中两项。
-
-<div align="center"> <img src="../pics//f1109d04-3c67-48a3-9963-2c475f94e175.jpg"/> </div><br>
-
-## 一致性
-
-一致性指的是多个数据副本是否能保持一致的特性。
-
-在一致性的条件下，系统在执行数据更新操作之后能够从一致性状态转移到另一个一致性状态。
-
-对系统的一个数据更新成功之后，如果所有用户都能够读取到最新的值，该系统就被认为具有强一致性。
-
-## 可用性
-
-可用性指分布式系统在面对各种异常时可以提供正常服务的能力，可以用系统可用时间占总时间的比值来衡量，4 个 9 的可用性表示系统 99.99% 的时间是可用的。
-
-在可用性条件下，系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
-
-## 分区容忍性
-
-网络分区指分布式系统中的节点被划分为多个区域，每个区域内部可以通信，但是区域之间无法通信。
-
-在分区容忍性条件下，分布式系统在遇到任何网络分区故障的时候，仍然需要能对外提供一致性和可用性的服务，除非是整个网络环境都发生了故障。
-
-## 权衡
-
-在分布式系统中，分区容忍性必不可少，因为需要总是假设网络是不可靠的。因此，CAP 理论实际在是要在可用性和一致性之间做权衡。
-
-可用性和一致性往往是冲突的，很难都使它们同时满足。在多个节点之间进行数据同步时，
-
-*   为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
-*   为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
-
-<div align="center"> <img src="../pics//0b587744-c0a8-46f2-8d72-e8f070d67b4b.jpg"/> </div><br>
-
-参考：
-
-- 倪超. 从 Paxos 到 ZooKeeper : 分布式一致性原理与实践 [M]. 电子工业出版社, 2015.
-- [What is CAP theorem in distributed database system?](http://www.colooshiki.com/index.php/2017/04/20/what-is-cap-theorem-in-distributed-database-system/)
-
-# 四、BASE
-
-BASE 是基本可用（Basically Available）、软状态（Soft State）和最终一致性（Eventually Consistent）三个短语的缩写。
-
-BASE 理论是对 CAP 中一致性和可用性权衡的结果，它的理论的核心思想是：即使无法做到强一致性，但每个应用都可以根据自身业务特点，采用适当的方式来使系统达到最终一致性。
-
-<div align="center"> <img src="../pics//bc603930-d74d-4499-a3e7-2d740fc07f33.png"/> </div><br>
-
-## 基本可用
-
-指分布式系统在出现故障的时候，保证核心可用，允许损失部分可用性。
-
-例如，电商在做促销时，为了保证购物系统的稳定性，部分消费者可能会被引导到一个降级的页面。
-
-## 软状态
-
-指允许系统中的数据存在中间状态，并认为该中间状态不会影响系统整体可用性，即允许系统不同节点的数据副本之间进行同步的过程存在延时。
-
-## 最终一致性
-
-最终一致性强调的是系统中所有的数据副本，在经过一段时间的同步后，最终能达到一致的状态。
-
-ACID 要求强一致性，通常运用在传统的数据库系统上。而 BASE 要求最终一致性，通过牺牲强一致性来达到可用性，通常运用在大型分布式系统中。
-
-在实际的分布式场景中，不同业务单元和组件对一致性的要求是不同的，因此 ACID 和 BASE 往往会结合在一起使用。
-
-# 五、Paxos
-
-用于达成共识性问题，即对多个节点产生的值，该算法能保证只选出唯一一个值。
-
-主要有三类节点：
-
-- 提议者（Proposer）：提议一个值；
-- 接受者（Acceptor）：对每个提议进行投票；
-- 告知者（Learner）：被告知投票的结果，不参与投票过程。
-
-<div align="center"> <img src="../pics//b988877c-0f0a-4593-916d-de2081320628.jpg"/> </div><br>
-
-## 执行过程
-
-规定一个提议包含两个字段：[n, v]，其中 n 为序号（具有唯一性），v 为提议值。
-
-下图演示了两个 Proposer 和三个 Acceptor 的系统中运行该算法的初始过程，每个 Proposer 都会向所有 Acceptor 发送提议请求。
-
-<div align="center"> <img src="../pics//1a9977e4-2f5c-49a6-aec9-f3027c9f46a7.png"/> </div><br>
-
-当 Acceptor 接收到一个提议请求，包含的提议为 [n1, v1]，并且之前还未接收过提议请求，那么发送一个提议响应，设置当前接收到的提议为 [n1, v1]，并且保证以后不会再接受序号小于 n1 的提议。
-
-如下图，Acceptor X 在收到 [n=2, v=8] 的提议请求时，由于之前没有接收过提议，因此就发送一个 [no previous] 的提议响应，设置当前接收到的提议为 [n=2, v=8]，并且保证以后不会再接受序号小于 2 的提议。其它的 Acceptor 类似。
-
-<div align="center"> <img src="../pics//fb44307f-8e98-4ff7-a918-31dacfa564b4.jpg"/> </div><br>
-
-如果 Acceptor 接收到一个提议请求，包含的提议为 [n2, v2]，并且之前已经接收过提议 [n1, v1]。如果 n1 > n2，那么就丢弃该提议请求；否则，发送提议响应，该提议响应包含之前已经接收过的提议 [n1, v1]，设置当前接收到的提议为 [n2, v2]，并且保证以后不会再接受序号小于 n2 的提议。
-
-如下图，Acceptor Z 收到 Proposer A 发来的 [n=2, v=8] 的提议请求，由于之前已经接收过 [n=4, v=5] 的提议，并且 n > 2，因此就抛弃该提议请求；Acceptor X 收到 Proposer B 发来的 [n=4, v=5] 的提议请求，因为之前接收到的提议为 [n=2, v=8]，并且 2 <= 4，因此就发送 [n=2, v=8] 的提议响应，设置当前接收到的提议为 [n=4, v=5]，并且保证以后不会再接受序号小于 4 的提议。Acceptor Y 类似。
-
-<div align="center"> <img src="../pics//2bcc58ad-bf7f-485c-89b5-e7cafc211ce2.jpg"/> </div><br>
-
-当一个 Proposer 接收到超过一半 Acceptor 的提议响应时，就可以发送接受请求。
-
-Proposer A 接收到两个提议响应之后，就发送 [n=2, v=8] 接受请求。该接受请求会被所有 Acceptor 丢弃，因为此时所有 Acceptor 都保证不接受序号小于 4 的提议。
-
-Proposer B 过后也收到了两个提议响应，因此也开始发送接受请求。需要注意的是，接受请求的 v 需要取它收到的最大 v 值，也就是 8。因此它发送 [n=4, v=8] 的接受请求。
-
-<div align="center"> <img src="../pics//9b838aee-0996-44a5-9b0f-3d1e3e2f5100.png"/> </div><br>
-
-Acceptor 接收到接受请求时，如果序号大于等于该 Acceptor 承诺的最小序号，那么就发送通知给所有的 Learner。当 Learner 发现有大多数的 Acceptor 接收了某个提议，那么该提议的提议值就被 Paxos 选择出来。
-
-<div align="center"> <img src="../pics//bf667594-bb4b-4634-bf9b-0596a45415ba.jpg"/> </div><br>
-
-## 约束条件
-
-### 1\. 正确性
-
-指只有一个提议值会生效。
-
-因为 Paxos 协议要求每个生效的提议被多数 Acceptor 接收，并且 Acceptor 不会接受两个不同的提议，因此可以保证正确性。
-
-### 2\. 可终止性
-
-指最后总会有一个提议生效。
-
-Paxos 协议能够让 Proposer 发送的提议朝着能被大多数 Acceptor 接受的那个提议靠拢，因此能够保证可终止性。
-
-参考：
-
-- [NEAT ALGORITHMS - PAXOS](http://harry.me/blog/2014/12/27/neat-algorithms-paxos/)
-- [Paxos By Example](https://angus.nyc/2012/paxos-by-example/)
-
-# 五、Raft
-
-Raft 和 Paxos 类似，但是更容易理解，也更容易实现。
-
-Raft 主要是用来竞选主节点。
-
-## 单个 Candidate 的竞选
-
-有三种节点：Follower、Candidate 和 Leader。Leader 会周期性的发送心跳包给 Follower。每个 Follower 都设置了一个随机的竞选超时时间，一般为 150ms\~300ms，如果在这个时间内没有收到 Leader 的心跳包，就会变成 Candidate，进入竞选阶段。
-
-*   下图表示一个分布式系统的最初阶段，此时只有 Follower，没有 Leader。Follower A 等待一个随机的竞选超时时间之后，没收到 Leader 发来的心跳包，因此进入竞选阶段。
-
-<div align="center"> <img src="../pics//111521118015898.gif"/> </div><br>
-
-*   此时 A 发送投票请求给其它所有节点。
-
-<div align="center"> <img src="../pics//111521118445538.gif"/> </div><br>
-
-*   其它节点会对请求进行回复，如果超过一半的节点回复了，那么该 Candidate 就会变成 Leader。
-
-<div align="center"> <img src="../pics//111521118483039.gif"/> </div><br>
-
-*   之后 Leader 会周期性地发送心跳包给 Follower，Follower 接收到心跳包，会重新开始计时。
-
-<div align="center"> <img src="../pics//111521118640738.gif"/> </div><br>
-
-## 多个 Candidate 竞选
-
-*   如果有多个 Follower 成为 Candidate，并且所获得票数相同，那么就需要重新开始投票，例如下图中 Candidate B 和 Candidate D 都获得两票，因此需要重新开始投票。
-
-<div align="center"> <img src="../pics//111521119203347.gif"/> </div><br>
-
-*   当重新开始投票时，由于每个节点设置的随机竞选超时时间不同，因此能下一次再次出现多个 Candidate 并获得同样票数的概率很低。
-
-<div align="center"> <img src="../pics//111521119368714.gif"/> </div><br>
-
-## 日志复制
-
-*   来自客户端的修改都会被传入 Leader。注意该修改还未被提交，只是写入日志中。
-
-<div align="center"> <img src="../pics//7.gif"/> </div><br>
-
-*   Leader 会把修改复制到所有 Follower。
-
-<div align="center"> <img src="../pics//9.gif"/> </div><br>
-
-*   Leader 会等待大多数的 Follower 也进行了修改，然后才将修改提交。
-
-<div align="center"> <img src="../pics//10.gif"/> </div><br>
-
-*   此时 Leader 会通知的所有 Follower 让它们也提交修改，此时所有节点的值达成一致。
-
-<div align="center"> <img src="../pics//11.gif"/> </div><br>
-
-参考：
-
-- [Raft: Understandable Distributed Consensus](http://thesecretlivesofdata.com/raft)
-
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index cabc2f97..649f563a 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -1,580 +1,4 @@
 <!-- GFM-TOC -->
-* [一、事务](#一事务)
-    * [概念](#概念)
-    * [ACID](#acid)
-    * [AUTOCOMMIT](#autocommit)
-* [二、并发一致性问题](#二并发一致性问题)
-    * [丢失修改](#丢失修改)
-    * [读脏数据](#读脏数据)
-    * [不可重复读](#不可重复读)
-    * [幻影读](#幻影读)
-* [三、封锁](#三封锁)
-    * [封锁粒度](#封锁粒度)
-    * [封锁类型](#封锁类型)
-    * [封锁协议](#封锁协议)
-    * [MySQL 隐式与显示锁定](#mysql-隐式与显示锁定)
-* [四、隔离级别](#四隔离级别)
-    * [未提交读（READ UNCOMMITTED）](#未提交读read-uncommitted)
-    * [提交读（READ COMMITTED）](#提交读read-committed)
-    * [可重复读（REPEATABLE READ）](#可重复读repeatable-read)
-    * [可串行化（SERIALIZABLE）](#可串行化serializable)
-* [五、多版本并发控制](#五多版本并发控制)
-    * [版本号](#版本号)
-    * [Undo 日志](#undo-日志)
-    * [实现过程](#实现过程)
-    * [快照读与当前读](#快照读与当前读)
-* [六、Next-Key Locks](#六next-key-locks)
-    * [Record Locks](#record-locks)
-    * [Gap Locks](#gap-locks)
-    * [Next-Key Locks](#next-key-locks)
-* [七、关系数据库设计理论](#七关系数据库设计理论)
-    * [函数依赖](#函数依赖)
-    * [异常](#异常)
-    * [范式](#范式)
-* [八、ER 图](#八er-图)
-    * [实体的三种联系](#实体的三种联系)
-    * [表示出现多次的关系](#表示出现多次的关系)
-    * [联系的多向性](#联系的多向性)
-    * [表示子类](#表示子类)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、事务
-
-## 概念
-
-<div align="center"> <img src="../pics//185b9c49-4c13-4241-a848-fbff85c03a64.png" width="400"/> </div><br>
-
-事务指的是满足 ACID 特性的一组操作，可以通过 Commit 提交一个事务，也可以使用 Rollback 进行回滚。
-
-## ACID
-
-### 1. 原子性（Atomicity）
-
-事务被视为不可分割的最小单元，事务的所有操作要么全部提交成功，要么全部失败回滚。
-
-回滚可以用日志来实现，日志记录着事务所执行的修改操作，在回滚时反向执行这些修改操作即可。
-
-### 2. 一致性（Consistency）
-
-数据库在事务执行前后都保持一致性状态。
-
-在一致性状态下，所有事务对一个数据的读取结果都是相同的。
-
-### 3. 隔离性（Isolation）
-
-一个事务所做的修改在最终提交以前，对其它事务是不可见的。
-
-### 4. 持久性（Durability）
-
-一旦事务提交，则其所做的修改将会永远保存到数据库中。即使系统发生崩溃，事务执行的结果也不能丢失。
-
-可以通过数据库备份和恢复来实现，在系统发生崩溃时，使用备份的数据库进行数据恢复。
-
-----
-
-事务的 ACID 特性概念简单，但不是很好理解，主要是因为这几个特性不是一种平级关系：
-
-- 只有满足一致性，事务的执行结果才是正确的。
-- 在无并发的情况下，事务串行执行，隔离性一定能够满足。此时要只要能满足原子性，就一定能满足一致性。
-- 在并发的情况下，多个事务并发执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
-- 事务满足持久化是为了能应对数据库崩溃的情况。
-
-<div align="center"> <img src="../pics//a58e294a-615d-4ea0-9fbf-064a6daec4b2.png" width="400"/> </div><br>
-
-## AUTOCOMMIT
-
-MySQL 默认采用自动提交模式。也就是说，如果不显式使用`START TRANSACTION`语句来开始一个事务，那么每个查询都会被当做一个事务自动提交。
-
-# 二、并发一致性问题
-
-在并发环境下，事务的隔离性很难保证，因此会出现很多并发一致性问题。
-
-## 丢失修改
-
-T<sub>1</sub> 和 T<sub>2</sub> 两个事务都对一个数据进行修改，T<sub>1</sub> 先修改，T<sub>2</sub> 随后修改，T<sub>2</sub> 的修改覆盖了 T<sub>1</sub> 的修改。
-
-<div align="center"> <img src="../pics//88ff46b3-028a-4dbb-a572-1f062b8b96d3.png" width="300"/> </div><br>
-
-## 读脏数据
-
-T<sub>1</sub> 修改一个数据，T<sub>2</sub> 随后读取这个数据。如果 T<sub>1</sub> 撤销了这次修改，那么 T<sub>2</sub> 读取的数据是脏数据。
-
-<div align="center"> <img src="../pics//dd782132-d830-4c55-9884-cfac0a541b8e.png" width="300"/> </div><br>
-
-## 不可重复读
-
-T<sub>2</sub> 读取一个数据，T<sub>1</sub> 对该数据做了修改。如果 T<sub>2</sub> 再次读取这个数据，此时读取的结果和第一次读取的结果不同。
-
-<div align="center"> <img src="../pics//c8d18ca9-0b09-441a-9a0c-fb063630d708.png" width="300"/> </div><br>
-
-## 幻影读
-
-T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插入新的数据，T<sub>1</sub> 再次读取这个范围的数据，此时读取的结果和和第一次读取的结果不同。
-
-<div align="center"> <img src="../pics//72fe492e-f1cb-4cfc-92f8-412fb3ae6fec.png" width="300"/> </div><br>
-
-----
-
-产生并发不一致性问题主要原因是破坏了事务的隔离性，解决方法是通过并发控制来保证隔离性。并发控制可以通过封锁来实现，但是封锁操作需要用户自己控制，相当复杂。数据库管理系统提供了事务的隔离级别，让用户以一种更轻松的方式处理并发一致性问题。
-
-# 三、封锁
-
-## 封锁粒度
-
-<div align="center"> <img src="../pics//1a851e90-0d5c-4d4f-ac54-34c20ecfb903.jpg" width="300"/> </div><br>
-
-MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
-
-应该尽量只锁定需要修改的那部分数据，而不是所有的资源。锁定的数据量越少，发生锁争用的可能就越小，系统的并发程度就越高。
-
-但是加锁需要消耗资源，锁的各种操作（包括获取锁、释放锁、以及检查锁状态）都会增加系统开销。因此封锁粒度越小，系统开销就越大。
-
-在选择封锁粒度时，需要在锁开销和并发程度之间做一个权衡。
-
-## 封锁类型
-
-### 1. 读写锁
-
-- 排它锁（Exclusive），简写为 X 锁，又称写锁。
-- 共享锁（Shared），简写为 S 锁，又称读锁。
-
-有以下两个规定：
-
-- 一个事务对数据对象 A 加了 X 锁，就可以对 A 进行读取和更新。加锁期间其它事务不能对 A 加任何锁。
-- 一个事务对数据对象 A 加了 S 锁，可以对 A 进行读取操作，但是不能进行更新操作。加锁期间其它事务能对 A 加 S 锁，但是不能加 X 锁。
-
-锁的兼容关系如下：
-
-| - | X | S |
-| :--: | :--: | :--: |
-|X|NO|NO|
-|S|NO|YES|
-
-### 2. 意向锁
-
-使用意向锁（Intention Locks）可以更容易地支持多粒度封锁。
-
-在存在行级锁和表级锁的情况下，事务 T 想要对表 A 加 X 锁，就需要先检测是否有其它事务对表 A 或者表 A 中的任意一行加了锁，那么就需要对表 A 的每一行都检测一次，这是非常耗时的。
-
-意向锁在原来的 X/S 锁之上引入了 IX/IS，IX/IS 都是表锁，用来表示一个事务想要在表中的某个数据行上加 X 锁或 S 锁。有以下两个规定：
-
-- 一个事务在获得某个数据行对象的 S 锁之前，必须先获得表的 IS 锁或者更强的锁；
-- 一个事务在获得某个数据行对象的 X 锁之前，必须先获得表的 IX 锁。
-
-通过引入意向锁，事务 T 想要对表 A 加 X 锁，只需要先检测是否有其它事务对表 A 加了 X/IX/S/IS 锁，如果加了就表示有其它事务正在使用这个表或者表中某一行的锁，因此事务 T 加 X 锁失败。
-
-各种锁的兼容关系如下：
-
-| - | X | IX | S | IS |
-| :--: | :--: | :--: | :--: | :--: |
-|X     |NO    |NO    |NO   | NO|
-|IX    |NO    |YES   |NO   | YES|
-|S     |NO    |NO    |YES  | YES|
-|IS    |NO    |YES  |YES  | YES|
-
-解释如下：
-
-- 任意 IS/IX 锁之间都是兼容的，因为它们只是表示想要对表加锁，而不是真正加锁；
-- S 锁只与 S 锁和 IS 锁兼容，也就是说事务 T 想要对数据行加 S 锁，其它事务可以已经获得对表或者表中的行的 S 锁。
-
-## 封锁协议
-
-### 1. 三级封锁协议
-
-**一级封锁协议** 
-
-事务 T 要修改数据 A 时必须加 X 锁，直到 T 结束才释放锁。
-
-可以解决丢失修改问题，因为不能同时有两个事务对同一个数据进行修改，那么事务的修改就不会被覆盖。
-
-| T<sub>1</sub> | T<sub>2</sub> |
-| :--: | :--: |
-| lock-x(A) | |
-| read A=20 | |
-| | lock-x(A) |
-|  | wait |
-| write A=19 |. |
-| commit |. |
-| unlock-x(A) |. |
-| | obtain |
-| | read A=19 |
-| | write A=21 |
-| | commit |
-| | unlock-x(A)|
-
-**二级封锁协议** 
-
-在一级的基础上，要求读取数据 A 时必须加 S 锁，读取完马上释放 S 锁。
-
-可以解决读脏数据问题，因为如果一个事务在对数据 A 进行修改，根据 1 级封锁协议，会加 X 锁，那么就不能再加 S 锁了，也就是不会读入数据。
-
-| T<sub>1</sub> | T<sub>2</sub> |
-| :--: | :--: |
-| lock-x(A) | |
-| read A=20 | |
-| write A=19 | |
-| | lock-s(A) |
-|  | wait |
-| rollback | .|
-| A=20 |. |
-| unlock-x(A) |. |
-| | obtain |
-| | read A=20 |
-| | commit |
-| | unlock-s(A)|
-
-**三级封锁协议** 
-
-在二级的基础上，要求读取数据 A 时必须加 S 锁，直到事务结束了才能释放 S 锁。
-
-可以解决不可重复读的问题，因为读 A 时，其它事务不能对 A 加 X 锁，从而避免了在读的期间数据发生改变。
-
-| T<sub>1</sub> | T<sub>2</sub> |
-| :--: | :--: |
-| lock-s(A) | |
-| read A=20 | |
-|  |lock-x(A) |
-| | wait |
-|  read A=20| . |
-| commit | .|
-| unlock-s(A) |. |
-| | obtain |
-| | read A=20 |
-| | write A=19|
-| | commit |
-| | unlock-X(A)|
-
-### 2. 两段锁协议
-
-加锁和解锁分为两个阶段进行。
-
-可串行化调度是指，通过并发控制，使得并发执行的事务结果与某个串行执行的事务结果相同。
-
-事务遵循两段锁协议是保证可串行化调度的充分条件。例如以下操作满足两段锁协议，它是可串行化调度。
-
-```html
-lock-x(A)...lock-s(B)...lock-s(C)...unlock(A)...unlock(C)...unlock(B)
-```
-
-但不是必要条件，例如以下操作不满足两段锁协议，但是它还是可串行化调度。
-
-```html
-lock-x(A)...unlock(A)...lock-s(B)...unlock(B)...lock-s(C)...unlock(C)
-```
-
-## MySQL 隐式与显示锁定
-
-MySQL 的 InnoDB 存储引擎采用两段锁协议，会根据隔离级别在需要的时候自动加锁，并且所有的锁都是在同一时刻被释放，这被称为隐式锁定。
-
-InnoDB 也可以使用特定的语句进行显示锁定：
-
-```sql
-SELECT ... LOCK In SHARE MODE;
-SELECT ... FOR UPDATE;
-```
-
-# 四、隔离级别
-
-## 未提交读（READ UNCOMMITTED）
-
-事务中的修改，即使没有提交，对其它事务也是可见的。
-
-## 提交读（READ COMMITTED）
-
-一个事务只能读取已经提交的事务所做的修改。换句话说，一个事务所做的修改在提交之前对其它事务是不可见的。
-
-## 可重复读（REPEATABLE READ）
-
-保证在同一个事务中多次读取同样数据的结果是一样的。
-
-## 可串行化（SERIALIZABLE）
-
-强制事务串行执行。
-
-----
-
-| 隔离级别 | 脏读 | 不可重复读 | 幻影读 |
-| :---: | :---: | :---:| :---: |
-| 未提交读 | YES | YES | YES |
-| 提交读 | NO | YES | YES |
-| 可重复读 | NO | NO | YES |
-| 可串行化 | NO | NO | NO |
-
-# 五、多版本并发控制
-
-多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC；可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
-
-## 版本号
-
-- 系统版本号：是一个递增的数字，每开始一个新的事务，系统版本号就会自动递增。
-- 事务版本号：事务开始时的系统版本号。
-
-InooDB 的 MVCC 在每行记录后面都保存着两个隐藏的列，用来存储两个版本号：
-
-- 创建版本号：指示创建一个数据行的快照时的系统版本号；
-- 删除版本号：如果该快照的删除版本号大于当前事务版本号表示该快照有效，否则表示该快照已经被删除了。
-
-## Undo 日志
-
-InnoDB 的 MVCC 使用到的快照存储在 Undo 日志中，该日志通过回滚指针把一个数据行（Record）的所有快照连接起来。
-
-<div align="center"> <img src="../pics//e41405a8-7c05-4f70-8092-e961e28d3112.jpg" width=""/> </div><br>
-
-## 实现过程
-
-以下实现过程针对可重复读隔离级别。
-
-### 1. SELECT
-
-当开始新一个事务时，该事务的版本号肯定会大于当前所有数据行快照的创建版本号，理解这一点很关键。
-
-多个事务必须读取到同一个数据行的快照，并且这个快照是距离现在最近的一个有效快照。但是也有例外，如果有一个事务正在修改该数据行，那么它可以读取事务本身所做的修改，而不用和其它事务的读取结果一致。
-
-把没有对一个数据行做修改的事务称为 T，T 所要读取的数据行快照的创建版本号必须小于 T 的版本号，因为如果大于或者等于 T 的版本号，那么表示该数据行快照是其它事务的最新修改，因此不能去读取它。
-
-除了上面的要求，T 所要读取的数据行快照的删除版本号必须大于 T 的版本号，因为如果小于等于 T 的版本号，那么表示该数据行快照是已经被删除的，不应该去读取它。
-
-### 2. INSERT
-
-将当前系统版本号作为数据行快照的创建版本号。
-
-### 3. DELETE
-
-将当前系统版本号作为数据行快照的删除版本号。
-
-### 4. UPDATE
-
-将当前系统版本号作为更新前的数据行快照的删除版本号，并将当前系统版本号作为更新后的数据行快照的创建版本号。可以理解为先执行 DELETE 后执行 INSERT。
-
-## 快照读与当前读
-
-### 1. 快照读
-
-使用 MVCC 读取的是快照中的数据，这样可以减少加锁所带来的开销。
-
-```sql
-select * from table ...;
-```
-
-### 2. 当前读
-
-读取的是最新的数据，需要加锁。以下第一个语句需要加 S 锁，其它都需要加 X 锁。
-
-```sql
-select * from table where ? lock in share mode;
-select * from table where ? for update;
-insert;
-update;
-delete;
-```
-
-# 六、Next-Key Locks
-
-Next-Key Locks 也是 MySQL 的 InnoDB 存储引擎的一种锁实现。MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题而存在的。在可重复读（REPEATABLE READ）隔离级别下，使用 MVCC + Next-Key Locks 可以解决幻读问题。
-
-## Record Locks
-
-锁定整个记录（行）。锁定的对象是记录的索引，而不是记录本身。如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
-
-## Gap Locks
-
-锁定一个范围内的索引，例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
-
-```sql
-SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
-```
-
-## Next-Key Locks
-
-它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录，也锁定范围内的索引。在 user 中有以下记录：
-
-```sql
-|   id | last_name   | first_name   |   age |
-|------|-------------|--------------|-------|
-|    4 | stark       | tony         |    21 |
-|    1 | tom         | hiddleston   |    30 |
-|    3 | morgan      | freeman      |    40 |
-|    5 | jeff        | dean         |    50 |
-|    2 | donald      | trump        |    80 |
-+------|-------------|--------------|-------+
-```
-
-那么就需要锁定以下范围：
-
-```sql
-(-∞, 21]
-(21, 30]
-(30, 40]
-(40, 50]
-(50, 80]
-(80, ∞)
-```
-
-# 七、关系数据库设计理论
-
-## 函数依赖
-
-记 A->B 表示 A 函数决定 B，也可以说 B 函数依赖于 A。
-
-如果 {A1，A2，... ，An} 是关系的一个或多个属性的集合，该集合函数决定了关系的其它所有属性并且是最小的，那么该集合就称为键码。
-
-对于 A->B，如果能找到 A 的真子集 A'，使得 A'-> B，那么 A->B 就是部分函数依赖，否则就是完全函数依赖；
-
-对于 A->B，B->C，则 A->C 是一个传递函数依赖。
-
-## 异常
-
-以下的学生课程关系的函数依赖为 Sno, Cname -> Sname, Sdept, Mname, Grade，键码为 {Sno, Cname}。也就是说，确定学生和课程之后，就能确定其它信息。
-
-| Sno | Sname | Sdept | Mname | Cname | Grade |
-| :---: | :---: | :---: | :---: | :---: |:---:|
-| 1 | 学生-1 | 学院-1 | 院长-1 | 课程-1 | 90 |
-| 2 | 学生-2 | 学院-2 | 院长-2 | 课程-2 | 80 |
-| 2 | 学生-2 | 学院-2 | 院长-2 | 课程-1 | 100 |
-| 3 | 学生-3 | 学院-2 | 院长-2 | 课程-2 | 95 |
-
-不符合范式的关系，会产生很多异常，主要有以下四种异常：
-
-- 冗余数据：例如 学生-2 出现了两次。
-- 修改异常：修改了一个记录中的信息，但是另一个记录中相同的信息却没有被修改。
-- 删除异常：删除一个信息，那么也会丢失其它信息。例如如果删除了 课程-1，需要删除第一行和第三行，那么 学生-1 的信息就会丢失。
-- 插入异常，例如想要插入一个学生的信息，如果这个学生还没选课，那么就无法插入。
-
-## 范式
-
-范式理论是为了解决以上提到四种异常。
-
-高级别范式的依赖于低级别的范式，1NF 是最低级别的范式。
-
-<div align="center"> <img src="../pics//c2d343f7-604c-4856-9a3c-c71d6f67fecc.png" width="300"/> </div><br>
-
-### 1. 第一范式 (1NF)
-
-属性不可分；
-
-### 2. 第二范式 (2NF)
-
-每个非主属性完全函数依赖于键码。
-
-可以通过分解来满足。
-
-<font size=4> **分解前** </font><br>
-
-| Sno | Sname | Sdept | Mname | Cname | Grade |
-| :---: | :---: | :---: | :---: | :---: |:---:|
-| 1 | 学生-1 | 学院-1 | 院长-1 | 课程-1 | 90 |
-| 2 | 学生-2 | 学院-2 | 院长-2 | 课程-2 | 80 |
-| 2 | 学生-2 | 学院-2 | 院长-2 | 课程-1 | 100 |
-| 3 | 学生-3 | 学院-2 | 院长-2 | 课程-2 | 95 |
-
-以上学生课程关系中，{Sno, Cname} 为键码，有如下函数依赖：
-
-- Sno -> Sname, Sdept
-- Sdept -> Mname
-- Sno, Cname-> Grade
-
-Grade 完全函数依赖于键码，它没有任何冗余数据，每个学生的每门课都有特定的成绩。
-
-Sname, Sdept 和 Mname 都部分依赖于键码，当一个学生选修了多门课时，这些数据就会出现多次，造成大量冗余数据。
-
-<font size=4> **分解后** </font><br>
-
-关系-1
-
-| Sno | Sname | Sdept | Mname |
-| :---: | :---: | :---: | :---: |
-| 1 | 学生-1 | 学院-1 | 院长-1 |
-| 2 | 学生-2 | 学院-2 | 院长-2 |
-| 3 | 学生-3 | 学院-2 | 院长-2 |
-
-有以下函数依赖：
-
-- Sno -> Sname, Sdept
-- Sdept -> Mname
-
-关系-2
-
-| Sno | Cname | Grade |
-| :---: | :---: |:---:|
-| 1 | 课程-1 | 90 |
-| 2 | 课程-2 | 80 |
-| 2 | 课程-1 | 100 |
-| 3 | 课程-2 | 95 |
-
-有以下函数依赖：
-
-- Sno, Cname ->  Grade
-
-### 3. 第三范式 (3NF)
-
-非主属性不传递函数依赖于键码。
-
-上面的 关系-1 中存在以下传递函数依赖：Sno -> Sdept -> Mname，可以进行以下分解：
-
-关系-11
-
-| Sno | Sname | Sdept |
-| :---: | :---: | :---: |
-| 1 | 学生-1 | 学院-1 |
-| 2 | 学生-2 | 学院-2 |
-| 3 | 学生-3 | 学院-2 |
-
-关系-12
-
-| Sdept | Mname |
-| :---: | :---: |
-| 学院-1 | 院长-1 |
-| 学院-2 | 院长-2 |
-
-# 八、ER 图
-
-Entity-Relationship，有三个组成部分：实体、属性、联系。
-
-用来进行关系型数据库系统的概念设计。
-
-## 实体的三种联系
-
-包含一对一，一对多，多对多三种。
-
-如果 A 到 B 是一对多关系，那么画个带箭头的线段指向 B；如果是一对一，画两个带箭头的线段；如果是多对多，画两个不带箭头的线段。下图的 Course 和 Student 是一对多的关系。
-
-<div align="center"> <img src="../pics//292b4a35-4507-4256-84ff-c218f108ee31.jpg" width=""/> </div><br>
-
-## 表示出现多次的关系
-
-一个实体在联系出现几次，就要用几条线连接。下图表示一个课程的先修关系，先修关系出现两个 Course 实体，第一个是先修课程，后一个是后修课程，因此需要用两条线来表示这种关系。
-
-<div align="center"> <img src="../pics//8b798007-e0fb-420c-b981-ead215692417.jpg" width=""/> </div><br>
-
-## 联系的多向性
-
-虽然老师可以开设多门课，并且可以教授多名学生，但是对于特定的学生和课程，只有一个老师教授，这就构成了一个三元联系。
-
-<div align="center"> <img src="../pics//423f2a40-bee1-488e-b460-8e76c48ee560.png" width=""/> </div><br>
-
-一般只使用二元联系，可以把多元关系转换为二元关系。
-
-<div align="center"> <img src="../pics//de9b9ea0-1327-4865-93e5-6f805c48bc9e.png" width=""/> </div><br>
-
-## 表示子类
-
-用一个三角形和两条线来连接类和子类，与子类有关的属性和联系都连到子类上，而与父类和子类都有关的连到父类上。
-
-<div align="center"> <img src="../pics//7ec9d619-fa60-4a2b-95aa-bf1a62aad408.jpg" width=""/> </div><br>
-
-# 参考资料
-
-- AbrahamSilberschatz, HenryF.Korth, S.Sudarshan, 等. 数据库系统概念 [M]. 机械工业出版社, 2006.
-- 施瓦茨. 高性能 MYSQL(第3版)[M]. 电子工业出版社, 2013.
-- 史嘉权. 数据库系统概论[M]. 清华大学出版社有限公司, 2006.
-- [The InnoDB Storage Engine](https://dev.mysql.com/doc/refman/5.7/en/innodb-storage-engine.html)
-- [Transaction isolation levels](https://www.slideshare.net/ErnestoHernandezRodriguez/transaction-isolation-levels)
-- [Concurrency Control](http://scanftree.com/dbms/2-phase-locking-protocol)
-- [The Nightmare of Locking, Blocking and Isolation Levels!](https://www.slideshare.net/brshristov/the-nightmare-of-locking-blocking-and-isolation-levels-46391666)
-- [Database Normalization and Normal Forms with an Example](https://aksakalli.github.io/2012/03/12/database-normalization-and-normal-forms-with-an-example.html)
-- [The basics of the InnoDB undo logging and history system](https://blog.jcole.us/2014/04/16/the-basics-of-the-innodb-undo-logging-and-history-system/)
-- [MySQL locking for the busy web developer](https://www.brightbox.com/blog/2013/10/31/on-mysql-locks/)
-- [浅入浅出 MySQL 和 InnoDB](https://draveness.me/mysql-innodb)
-- [Innodb 中的事务隔离级别和锁的关系](https://tech.meituan.com/innodb-lock.html)
diff --git a/notes/正则表达式.md b/notes/正则表达式.md
index c6c21f2c..649f563a 100644
--- a/notes/正则表达式.md
+++ b/notes/正则表达式.md
@@ -1,389 +1,4 @@
 <!-- GFM-TOC -->
-* [一、概述](#一概述)
-* [二、匹配单个字符](#二匹配单个字符)
-* [三、匹配一组字符](#三匹配一组字符)
-* [四、使用元字符](#四使用元字符)
-* [五、重复匹配](#五重复匹配)
-* [六、位置匹配](#六位置匹配)
-* [七、使用子表达式](#七使用子表达式)
-* [八、回溯引用](#八回溯引用)
-* [九、前后查找](#九前后查找)
-* [十、嵌入条件](#十嵌入条件)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、概述
-
-正则表达式用于文本内容的查找和替换。
-
-正则表达式内置于其它语言或者软件产品中，它本身不是一种语言或者软件。
-
-[正则表达式在线工具](https://regexr.com/)
-
-# 二、匹配单个字符
-
-正则表达式一般是区分大小写的，但是也有些实现是不区分。
-
-**.**  可以用来匹配任何的单个字符，但是在绝大多数实现里面，不能匹配换行符；
-
-**\\**  是元字符，表示它有特殊的含义，而不是字符本身的含义。如果需要匹配 . ，那么要用 \ 进行转义，即在 . 前面加上 \ 。
-
-**正则表达式** 
-
-```
-nam.
-```
-
-**匹配结果** 
-
-My  **name**  is Zheng.
-
-# 三、匹配一组字符
-
-**[ ]**  定义一个字符集合；
-
-0-9、a-z 定义了一个字符区间，区间使用 ASCII 码来确定，字符区间只能用在 [ ] 之间。
-
-**-**  元字符只有在 [ ] 之间才是元字符，在 [ ] 之外就是一个普通字符；
-
-**^**  在 [ ] 字符集合中是取非操作。
-
-**应用** 
-
-匹配以 abc 为开头，并且最后一个字母不为数字的字符串：
-
-**正则表达式** 
-
-```
-abc[^0-9]
-```
-
-**匹配结果** 
-
-1.  **abcd** 
-2. abc1
-3. abc2
-
-# 四、使用元字符
-
-## 匹配空白字符
-
-|  元字符 | 说明  |
-| :---: | :---: |
-|  [\b] | 回退（删除）一个字符   |
-|  \f |  换页符 |
-|  \n |  换行符 |
-|  \r |  回车符 |
-|  \t |  制表符 |
-|  \v |  垂直制表符 |
-
-\r\n 是 Windows 中的文本行结束标签，在 Unix/Linux 则是 \n ；\r\n\r\n 可以匹配 Windows 下的空白行，因为它将匹配两个连续的行尾标签，而这正是两条记录之间的空白行；
-
-. 是元字符，前提是没有对它们进行转义；f 和 n 也是元字符，但是前提是对它们进行了转义。
-
-## 匹配特定的字符类别
-
-### 1. 数字元字符
-
-|  元字符 | 说明  |
-| :---: | :---: |
-| \d  | 数字字符，等价于 [0-9]  |
-| \D  | 非数字字符，等价于 [^0-9]   |
-
-### 2. 字母数字元字符
-
-|  元字符 | 说明  |
-| :---: | :---: |
-| \w  |  大小写字母，下划线和数字，等价于 [a-zA-Z0-9\_] |
-|  \W |  对 \w 取非 |
-
-### 3. 空白字符元字符
-
-| 元字符  | 说明  |
-| :---: | :---: |
-|  \s | 任何一个空白字符，等价于 [\f\n\r\t\v]  |
-| \S  |  对 \s 取非  |
-
-\x 匹配十六进制字符，\0 匹配八进制，例如 \x0A 对应 ASCII 字符 10 ，等价于 \n，也就是它会匹配 \n 。
-
-# 五、重复匹配
-
-**\+**  匹配 1 个或者多个字符， **\*** 匹配 0 个或者多个，**?** 匹配 0 个或者 1 个。
-
-**应用** 
-
-匹配邮箱地址。
-
-**正则表达式** 
-
-```
-[\w.]+@\w+\.\w+
-```
-
-[\w.] 匹配的是字母数字或者 . ，在其后面加上 + ，表示匹配多次。在字符集合 [ ] 里，. 不是元字符；
-
-**匹配结果** 
-
-**abc.def<span>@</span>qq.com** 
-
-为了可读性，常常把转义的字符放到字符集合 [ ] 中，但是含义是相同的。
-
-```
-[\w.]+@\w+\.\w+
-[\w.]+@[\w]+[\.][\w]+
-```
-
-**{n}**  匹配 n 个字符，**{m, n}** 匹配 m\~n 个字符，**{m,}** 至少匹配 m 个字符；
-
-\* 和 + 都是贪婪型元字符，会匹配最多的内容，在元字符后面加 ? 可以转换为懒惰型元字符，例如 \*?、+? 和 {m, n}? 。
-
-**正则表达式** 
-
-```
-a.+c
-```
-
-由于 + 是贪婪型的，因此 .+ 会匹配更可能多的内容，所以会把整个 abcabcabc 文本都匹配，而不是只匹配前面的 abc 文本。用懒惰型可以实现匹配前面的。
-
-**匹配结果** 
-
-**abcabcabc** 
-
-# 六、位置匹配
-
-## 单词边界
-
-**\b**  可以匹配一个单词的边界，边界是指位于 \w 和 \W 之间的位置；**\B** 匹配一个不是单词边界的位置。
-
-\b 只匹配位置，不匹配字符，因此 \babc\b 匹配出来的结果为 3 个字符。
-
-## 字符串边界
-
-**^**  匹配整个字符串的开头，**$** 匹配结尾。
-
-^ 元字符在字符集合中用作求非，在字符集合外用作匹配字符串的开头。
-
-分行匹配模式（multiline）下，换行被当做字符串的边界。
-
-**应用** 
-
-匹配代码中以 // 开始的注释行
-
-**正则表达式** 
-
-```
-^\s*\/\/.*$
-```
-
-<div align="center"> <img src="../pics//600e9c75-5033-4dad-ae2b-930957db638e.png"/> </div><br>
-
-**匹配结果** 
-
-1. public void fun() {
-2. &nbsp;&nbsp;&nbsp;&nbsp;     **// 注释 1** 
-3. &nbsp;&nbsp;&nbsp;&nbsp;    int a = 1;
-4. &nbsp;&nbsp;&nbsp;&nbsp;    int b = 2;
-5. &nbsp;&nbsp;&nbsp;&nbsp;     **// 注释 2** 
-6. &nbsp;&nbsp;&nbsp;&nbsp;    int c = a + b;
-7. }
-
-# 七、使用子表达式
-
-使用  **( )**  定义一个子表达式。子表达式的内容可以当成一个独立元素，即可以将它看成一个字符，并且使用 * 等元字符。
-
-子表达式可以嵌套，但是嵌套层次过深会变得很难理解。
-
-**正则表达式** 
-
-```
-(ab){2,}
-```
-
-**匹配结果** 
-
-**ababab** 
-
-**|**  是或元字符，它把左边和右边所有的部分都看成单独的两个部分，两个部分只要有一个匹配就行。
-
-**正则表达式** 
-
-```
-(19|20)\d{2}
-```
-
-**匹配结果** 
-
-1.  **1900** 
-2.  **2010** 
-3. 1020
-
-**应用** 
-
-匹配 IP 地址。IP 地址中每部分都是 0-255 的数字，用正则表达式匹配时以下情况是合法的：
-
-- 一位数字
-- 不以 0 开头的两位数字
-- 1 开头的三位数
-- 2 开头，第 2 位是 0-4 的三位数
-- 25 开头，第 3 位是 0-5 的三位数
-
-**正则表达式** 
-
-```
-((25[0-5]|(2[0-4]\d)|(1\d{2})|([1-9]\d)|(\d))\.){3}(25[0-5]|(2[0-4]\d)|(1\d{2})|([1-9]\d)|(\d))
-```
-
-**匹配结果** 
-
-1.  **192.168.0.1** 
-2. 00.00.00.00
-3. 555.555.555.555
-
-# 八、回溯引用
-
-回溯引用使用  **\n**  来引用某个子表达式，其中 n 代表的是子表达式的序号，从 1 开始。它和子表达式匹配的内容一致，比如子表达式匹配到 abc，那么回溯引用部分也需要匹配 abc 。
-
-**应用** 
-
-匹配 HTML 中合法的标题元素。
-
-**正则表达式** 
-
-\1 将回溯引用子表达式 (h[1-6]) 匹配的内容，也就是说必须和子表达式匹配的内容一致。
-
-```
-<(h[1-6])>\w*?<\/\1>
-```
-
-**匹配结果** 
-
-1.  **&lt;h1>x&lt;/h1>** 
-2.  **&lt;h2>x&lt;/h2>** 
-3. &lt;h3>x&lt;/h1>
-
-## 替换
-
-需要用到两个正则表达式。
-
-**应用** 
-
-修改电话号码格式。
-
-**文本** 
-
-313-555-1234
-
-**查找正则表达式** 
-
-```
-(\d{3})(-)(\d{3})(-)(\d{4})
-```
-
-**替换正则表达式** 
-
-在第一个子表达式查找的结果加上 () ，然后加一个空格，在第三个和第五个字表达式查找的结果中间加上 - 进行分隔。
-
-```
-($1) $3-$5
-```
-
-**结果** 
-
-(313) 555-1234
-
-## 大小写转换
-
-|  元字符 | 说明  |
-| :---: | :---: |
-|  \l | 把下个字符转换为小写  |
-|   \u| 把下个字符转换为大写  |
-|  \L | 把\L 和\E 之间的字符全部转换为小写  |
-|  \U | 把\U 和\E 之间的字符全部转换为大写  |
-|  \E | 结束\L 或者\U  |
-
-**应用** 
-
-把文本的第二个和第三个字符转换为大写。
-
-**文本** 
-
-abcd
-
-**查找** 
-
-```
-(\w)(\w{2})(\w)
-```
-
-**替换** 
-
-```
-$1\U$2\E$3
-```
-
-**结果** 
-
-aBCd
-
-# 九、前后查找
-
-前后查找规定了匹配的内容首尾应该匹配的内容，但是又不包含首尾匹配的内容。向前查找用  **?=**  来定义，它规定了尾部匹配的内容，这个匹配的内容在 ?= 之后定义。所谓向前查找，就是规定了一个匹配的内容，然后以这个内容为尾部向前面查找需要匹配的内容。向后匹配用 ?<= 定义（注: javaScript 不支持向后匹配, java 对其支持也不完善）。
-
-**应用** 
-
-查找出邮件地址 @ 字符前面的部分。
-
-**正则表达式** 
-
-```
-\w+(?=@)
-```
-
-**结果** 
-
-**abc** @qq.com
-
-对向前和向后查找取非，只要把 = 替换成 ! 即可，比如 (?=) 替换成 (?!) 。取非操作使得匹配那些首尾不符合要求的内容。
-
-# 十、嵌入条件
-
-## 回溯引用条件
-
-条件判断为某个子表达式是否匹配，如果匹配则需要继续匹配条件表达式后面的内容。
-
-**正则表达式** 
-
-子表达式 (\\() 匹配一个左括号，其后的 ? 表示匹配 0 个或者 1 个。 ?(1) 为条件，当子表达式 1 匹配时条件成立，需要执行 \) 匹配，也就是匹配右括号。
-
-```
-(\()?abc(?(1)\))
-```
-
-**结果** 
-
-1.  **(abc)** 
-2.  **abc** 
-3. (abc
-
-## 前后查找条件
-
-条件为定义的首尾是否匹配，如果匹配，则继续执行后面的匹配。注意，首尾不包含在匹配的内容中。
-
-**正则表达式** 
-
- ?(?=-) 为前向查找条件，只有在以 - 为前向查找的结尾能匹配 \d{5} ，才继续匹配 -\d{4} 。
-
-```
-\d{5}(?(?=-)-\d{4})
-```
-
-**结果** 
-
-1.  **11111** 
-2. 22222-
-3.  **33333-4444** 
-
-# 参考资料
-
-- BenForta. 正则表达式必知必会 [M]. 人民邮电出版社, 2007.
diff --git a/notes/消息队列.md b/notes/消息队列.md
index e28fcd3f..649f563a 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -1,80 +1,4 @@
 <!-- GFM-TOC -->
-* [一、消息模型](#一消息模型)
-    * [点对点](#点对点)
-    * [发布/订阅](#发布订阅)
-* [二、使用场景](#二使用场景)
-    * [异步处理](#异步处理)
-    * [流量削锋](#流量削锋)
-    * [应用解耦](#应用解耦)
-* [三、可靠性](#三可靠性)
-    * [发送端的可靠性](#发送端的可靠性)
-    * [接收端的可靠性](#接收端的可靠性)
 <!-- GFM-TOC -->
 
 
-# 一、消息模型
-
-## 点对点
-
-消息生产者向消息队列中发送了一个消息之后，只能被一个消费者消费一次。
-
-<div align="center"> <img src="../pics//09b52bcb-88ba-4e36-8244-b375f16ad116.jpg"/> </div><br>
-
-## 发布/订阅
-
-消息生产者向频道发送一个消息之后，多个消费者可以从该频道订阅到这条消息并消费。
-
-<div align="center"> <img src="../pics//ddb5ff4c-4ada-46aa-9bf1-140bdb5e4676.jpg"/> </div><br>
-
-发布与订阅模式和观察者模式有以下不同：
-
-- 观察者模式中，观察者和主题都知道对方的存在；而在发布与订阅模式中，发布者与订阅者不知道对方的存在，它们之间通过频道进行通信。
-- 观察者模式是同步的，当事件触发时，主题会去调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息。
-
-<div align="center"> <img src="../pics//bee1ff1d-c80f-4b3c-b58c-7073a8896ab2.jpg"/> </div><br>
-
-参考：
-
-- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
-- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
-
-# 二、使用场景
-
-## 异步处理
-
-发送者将消息发送给消息队列之后，不需要同步等待消息接收者处理完毕，而是立即返回进行其它操作。消息接收者从消息队列中订阅消息之后异步处理。
-
-例如在注册流程中通常需要发送验证邮件来确保注册用户的身份合法，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
-
-只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成操作的话，就不能再使用消息队列。
-
-## 流量削锋
-
-在高并发的场景下，如果短时间有大量的请求会压垮服务器。
-
-可以将请求发送到消息队列中，服务器按照其处理能力从消息队列中订阅消息进行处理。
-
-## 应用解耦
-
-如果模块之间不直接进行调用，模块之间耦合度很低，那么修改一个模块或者新增一个模块对其它模块的影响会很小，从而实现可扩展性。
-
-通过使用消息队列，一个模块只需要向消息队列中发送消息，其它模块可以选择性地从消息队列中订阅消息从而完成调用。
-
-# 三、可靠性
-
-## 发送端的可靠性
-
-发送端完成操作后一定能将消息成功发送到消息队列中。
-
-实现方法：
-
-- 在本地数据库建一张消息表，将消息数据与业务数据保存在同一数据库实例里，这样就可以利用本地数据库的事务机制。事务提交成功后，将消息表中的消息转移到消息队列中，若转移消息成功则删除消息表中的数据，否则继续重传。
-
-## 接收端的可靠性
-
-接收端能够从消息中间件成功消费一次消息。
-
-实现方法：
-
-- 保证接收端处理消息的业务逻辑具有幂等性：只要具有幂等性，那么消费多少次消息，最后处理的结果都是一样的。
-- 保证消息具有唯一编号，并使用一张日志表来记录已经消费的消息编号。
diff --git a/notes/算法.md b/notes/算法.md
index df5a8049..fb878742 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -2050,7 +2050,7 @@ public class SparseVector {
 
 如果只有一个圆盘，那么只需要进行一次移动操作。
 
-从上面的讨论可以知道，n 圆盘需要移动 (n-1)+1+(n-1) = 2n-1 次。
+从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然 a<sub>n</sub> = 2<sup>n</sup> - 1，n 个圆盘需要移动 2<sup>n</sup> - 1 次。
 
 <div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
 
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index e132aa2c..649f563a 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -1,104 +1,4 @@
 <!-- GFM-TOC -->
-* [一、性能](#一性能)
-* [二、伸缩性](#二伸缩性)
-* [三、扩展性](#三扩展性)
-* [四、可用性](#四可用性)
-* [五、安全性](#五安全性)
 <!-- GFM-TOC -->
 
 
-# 一、性能
-
-## 性能指标
-
-### 1. 响应时间
-
-指从某个请求发出到接收到响应消耗的时间。
-
-在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
-
-### 2. 吞吐量
-
-指系统在单位时间内可以处理的请求数量，通常使用每秒的请求数来衡量。
-
-### 3. 并发用户数
-
-指系统能同时处理的并发用户请求数量。
-
-在没有并发存在的系统中，请求被顺序执行，此时响应时间为吞吐量的倒数。例如系统支持的吞吐量为 100 req/s，那么平均响应时间应该为 0.01s。
-
-目前的大型系统都支持多线程来处理并发请求，多线程能够提高吞吐量以及缩短响应时间，主要有两个原因：
-
-- 多 CPU
-- IO 等待时间
-
-使用 IO 多路复用等方式，系统在等待一个 IO 操作完成的这段时间内不需要被阻塞，可以去处理其它请求。通过将这个等待时间利用起来，使得 CPU 利用率大大提高。
-
-并发用户数不是越高越好，因为如果并发用户数太高，系统来不及处理这么多的请求，会使得过多的请求需要等待，那么响应时间就会大大提高。
-
-## 性能优化
-
-### 1. 集群
-
-将多台服务器组成集群，使用负载均衡将请求转发到集群中，避免单一服务器的负载压力过大导致性能降低。
-
-### 2. 缓存
-
-缓存能够提高性能的原因如下：
-
-- 缓存数据通常位于内存等介质中，这种介质对于读操作特别快；
-- 缓存数据可以位于靠近用户的地理位置上；
-- 可以将计算结果进行缓存，从而避免重复计算。
-
-### 3. 异步
-
-某些流程可以将操作转换为消息，将消息发送到消息队列之后立即返回，之后这个操作会被异步处理。
-
-# 二、伸缩性
-
-指不断向集群中添加服务器来缓解不断上升的用户并发访问压力和不断增长的数据存储需求。
-
-## 伸缩性与性能
-
-如果系统存在性能问题，那么单个用户的请求总是很慢的；
-
-如果系统存在伸缩性问题，那么单个用户的请求可能会很快，但是在并发数很高的情况下系统会很慢。
-
-## 实现伸缩性
-
-应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
-
-关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
-
-对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
-
-# 三、扩展性
-
-指的是添加新功能时对现有系统的其它应用无影响，这就要求不同应用具备低耦合的特点。
-
-实现可扩展主要有两种方式：
-
-- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
-- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务器框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
-
-# 四、可用性
-
-## 冗余
-
-保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
-
-应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
-
-存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
-
-## 监控
-
-对 CPU、内存、磁盘、网络等系统负载信息进行监控，当某个数据达到一定阈值时通知运维人员，从而在系统发生故障之前及时发现问题。
-
-## 服务降级
-
-服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
-
-# 五、安全性
-
-要求系统的应对各种攻击手段时能够有可靠的应对措施。
diff --git a/notes/缓存.md b/notes/缓存.md
index e60373dd..449598ff 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -163,7 +163,11 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 分布式缓存
 
-使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存，本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。
+
+相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。
+
+不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
 
 ## 数据库缓存
 
@@ -171,7 +175,7 @@ MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查
 
 # 四、CDN
 
-内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源其它数据分发给用户。
+内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
 
 CDN 主要有以下优点：
 
@@ -190,7 +194,7 @@ CDN 主要有以下优点：
 
 ## 缓存穿透
 
-指的是对某个一定不存在的数据进行请求，该请求将会穿透缓存来到数据库。
+指的是对某个一定不存在的数据进行请求，该请求将会穿透缓存到达数据库。
 
 解决方案：
 
@@ -207,6 +211,7 @@ CDN 主要有以下优点：
 
 - 为了防止缓存在同一时间大面积过期导致的缓存雪崩，可以通过观察用户行为，合理设置缓存过期时间来实现；
 - 为了防止缓存服务器宕机出现的缓存雪崩，可以使用分布式缓存，分布式缓存中每一个节点只缓存部分的数据，当某个节点宕机时可以保证其它节点的缓存仍然可用。
+- 也可以在进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
 
 ## 缓存一致性
 
@@ -258,7 +263,7 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
 
-数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点号，从而使得数据分布也更加均匀。
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点好，从而使得数据分布也更加均匀。
 
 参考资料：
 
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index f1460844..649f563a 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -1,890 +1,4 @@
 <!-- GFM-TOC -->
-* [一、概述](#一概述)
-    * [网络的网络](#网络的网络)
-    * [ISP](#isp)
-    * [主机之间的通信方式](#主机之间的通信方式)
-    * [电路交换与分组交换](#电路交换与分组交换)
-    * [时延](#时延)
-    * [计算机网络体系结构*](#计算机网络体系结构)
-* [二、物理层](#二物理层)
-    * [通信方式](#通信方式)
-    * [带通调制](#带通调制)
-* [三、数据链路层](#三数据链路层)
-    * [基本问题](#基本问题)
-    * [信道分类](#信道分类)
-    * [信道复用技术](#信道复用技术)
-    * [CSMA/CD 协议*](#csmacd-协议)
-    * [PPP 协议](#ppp-协议)
-    * [MAC 地址](#mac-地址)
-    * [局域网](#局域网)
-    * [以太网*](#以太网)
-    * [交换机*](#交换机)
-    * [虚拟局域网](#虚拟局域网)
-* [四、网络层*](#四网络层)
-    * [概述](#概述)
-    * [IP 数据报格式](#ip-数据报格式)
-    * [IP 地址编址方式](#ip-地址编址方式)
-    * [地址解析协议 ARP](#地址解析协议-arp)
-    * [网际控制报文协议 ICMP](#网际控制报文协议-icmp)
-    * [虚拟专用网 VPN](#虚拟专用网-vpn)
-    * [网络地址转换 NAT](#网络地址转换-nat)
-    * [路由器的结构](#路由器的结构)
-    * [路由器分组转发流程](#路由器分组转发流程)
-    * [路由选择协议](#路由选择协议)
-* [五、运输层*](#五运输层)
-    * [UDP 和 TCP 的特点](#udp-和-tcp-的特点)
-    * [UDP 首部格式](#udp-首部格式)
-    * [TCP 首部格式](#tcp-首部格式)
-    * [TCP 的三次握手](#tcp-的三次握手)
-    * [TCP 的四次挥手](#tcp-的四次挥手)
-    * [TCP 滑动窗口](#tcp-滑动窗口)
-    * [TCP 可靠传输](#tcp-可靠传输)
-    * [TCP 流量控制](#tcp-流量控制)
-    * [TCP 拥塞控制](#tcp-拥塞控制)
-* [六、应用层](#六应用层)
-    * [域名系统](#域名系统)
-    * [文件传送协议](#文件传送协议)
-    * [远程登录协议](#远程登录协议)
-    * [电子邮件协议](#电子邮件协议)
-    * [动态主机配置协议](#动态主机配置协议)
-    * [常用端口](#常用端口)
-    * [Web 页面请求过程](#web-页面请求过程)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、概述
-
-## 网络的网络
-
-网络把主机连接起来，而互联网是把多种不同的网络连接起来，因此互联网是网络的网络。
-
-<div align="center"> <img src="../pics//network-of-networks.gif" width="500"/> </div><br>
-
-## ISP
-
-互联网服务提供商 ISP 可以从互联网管理机构获得许多 IP 地址，同时拥有通信线路以及路由器等联网设备，个人或机构向 ISP 缴纳一定的费用就可以接入互联网。
-
-<div align="center"> <img src="../pics//46cec213-3048-4a80-aded-fdd577542801.jpg" width="500"/> </div><br>
-
-目前的互联网是一种多层次 ISP 结构，ISP 根据覆盖面积的大小分为第一层 ISP、区域 ISP 和接入 ISP。
-
-互联网交换点 IXP 允许两个 ISP 直接相连而不用经过第三个 ISP。
-
-<div align="center"> <img src="../pics//168e893c-e4a0-4ba4-b81f-9d993483abd0.jpg" width="500"/> </div><br>
-
-## 主机之间的通信方式
-
-- 客户-服务器（C/S）：客户是服务的请求方，服务器是服务的提供方。
-
-- 对等（P2P）：不区分客户和服务器。
-
-<div align="center"> <img src="../pics//2ad244f5-939c-49fa-9385-69bc688677ab.jpg" width=""/> </div><br>
-
-## 电路交换与分组交换
-
-### 1. 电路交换
-
-电路交换用于电话通信系统，两个用户要通信之前需要建立一条专用的物理链路，并且在整个通信过程中始终占用该链路。由于通信的过程中不可能一直在使用传输线路，因此电路交换对线路的利用率很低，往往不到 10%。
-
-### 2. 分组交换
-
-每个分组都有首部和尾部，包含了源地址和目的地址等控制信息，在同一个传输线路上同时传输多个分组互相不会影响，因此在同一条传输线路上允许同时传输多个分组，也就是说分组交换不需要占用传输线路。
-
-考虑在一个邮局通信系统中，邮局收到一份邮件之后，先存储下来，然后把相同目的地的邮件一起转发到下一个目的地，这个过程就是存储转发过程，分组交换也使用了存储转发过程。
-
-## 时延
-
-总时延 = 传输时延 + 传播时延 + 处理时延 + 排队时延
-
-<div align="center"> <img src="../pics//3939369b-3a4a-48a0-b9eb-3efae26dd400.png" width="800"/> </div><br>
-
-### 1. 传输时延
-
-主机或路由器传输数据帧所需要的时间。
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?delay=\frac{l(bit)}{v(bit/s)}"/></div> <br>
-
-其中 l 表示数据帧的长度，v 表示传输速率。
-
-### 2. 传播时延
-
-电磁波在信道中传播一定的距离需要花费的时间，电磁波传播速度接近光速。
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?delay=\frac{l(m)}{v(m/s)}"/></div> <br>
-
-其中 l 表示信道长度，v 表示电磁波在信道上的传播速率。
-
-### 3. 处理时延
-
-主机或路由器收到分组时进行处理所需要的时间，例如分析首部、从分组中提取数据、进行差错检验或查找适当的路由等。
-
-### 4. 排队时延
-
-分组在路由器的输入队列和输出队列中排队等待的时间，取决于网络当前的通信量。
-
-## 计算机网络体系结构*
-
-<div align="center"> <img src="../pics//426df589-6f97-4622-b74d-4a81fcb1da8e.png" width="800"/> </div><br>
-
-### 1. 五层协议
-
--  **应用层** ：为特定应用程序提供数据传输服务，例如 HTTP、DNS 等。数据单位为报文。
-
--  **运输层** ：提供的是进程间的通用数据传输服务。由于应用层协议很多，定义通用的运输层协议就可以支持不断增多的应用层协议。运输层包括两种协议：传输控制协议 TCP，提供面向连接、可靠的数据传输服务，数据单位为报文段；用户数据报协议 UDP，提供无连接、尽最大努力的数据传输服务，数据单位为用户数据报。TCP 主要提供完整性服务，UDP 主要提供及时性服务。
-
--  **网络层** ：为主机间提供数据传输服务，而运输层协议是为主机中的进程提供服务。网络层把运输层传递下来的报文段或者用户数据报封装成分组。
-
--  **数据链路层** ：网络层针对的还是主机之间的数据传输服务，而主机之间可以有很多链路，链路层协议就是为同一链路的节点提供服务。数据链路层把网络层传来的分组封装成帧。
-
--  **物理层** ：考虑的是怎样在传输媒体上传输数据比特流，而不是指具体的传输媒体。物理层的作用是尽可能屏蔽传输媒体和通信手段的差异，使数据链路层感觉不到这些差异。
-
-### 2. OSI
-
-其中表示层和会话层用途如下：
-
--  **表示层** ：数据压缩、加密以及数据描述，这使得应用程序不必担心在各台主机中数据内部格式不同的问题。
-
--  **会话层** ：建立及管理会话。
-
-五层协议没有表示层和会话层，而是将这些功能留给应用程序开发者处理。
-
-### 3. TCP/IP
-
-它只有四层，相当于五层协议中数据链路层和物理层合并为网络接口层。
-
-现在的 TCP/IP 体系结构不严格遵循 OSI 分层概念，应用层可能会直接使用 IP 层或者网络接口层。
-
-<div align="center"> <img src="../pics//45e0e0bf-386d-4280-a341-a0b9496c7674.png" width="400"/> </div><br>
-
-TCP/IP 协议族是一种沙漏形状，中间小两边大，IP 协议在其中占用举足轻重的地位。
-
-<div align="center"> <img src="../pics//d4eef1e2-5703-4ca4-82ab-8dda93d6b81f.png" width="500"/> </div><br>
-
-### 4. 数据在各层之间的传递过程
-
-在向下的过程中，需要添加下层协议所需要的首部或者尾部，而在向上的过程中不断拆开首部和尾部。
-
-路由器只有下面三层协议，因为路由器位于网络核心中，不需要为进程或者应用程序提供服务，因此也就不需要运输层和应用层。
-
-<div align="center"> <img src="../pics//ac106e7e-489a-4082-abd9-dabebe48394c.jpg" width="800"/> </div><br>
-
-# 二、物理层
-
-## 通信方式
-
-根据信息在传输线上的传送方向，分为以下三种通信方式：
-
-- 单工通信：单向传输
-- 半双工通信：双向交替传输
-- 全双工通信：双向同时传输
-
-## 带通调制
-
-模拟信号是连续的信号，数字信号是离散的信号。带通调制把数字信号转换为模拟信号。
-
-<div align="center"> <img src="../pics//f0a31c04-6e26-408c-8395-88f4e2ae928b.jpg"/> </div><br>
-
-# 三、数据链路层
-
-## 基本问题
-
-### 1. 封装成帧
-
-将网络层传下来的分组添加首部和尾部，用于标记帧的开始和结束。
-
-<div align="center"> <img src="../pics//ea5f3efe-d5e6-499b-b278-9e898af61257.jpg" width="500"/> </div><br>
-
-### 2. 透明传输
-
-透明表示一个实际存在的事物看起来好像不存在一样。
-
-帧使用首部和尾部进行定界，如果帧的数据部分含有和首部尾部相同的内容，那么帧的开始和结束位置就会被错误的判定。需要在数据部分出现首部尾部相同的内容前面插入转义字符。如果数据部分出现转义字符，那么就在转义字符前面再加个转义字符。在接收端进行处理之后可以还原出原始数据。这个过程透明传输的内容是转义字符，用户察觉不到转义字符的存在。
-
-<div align="center"> <img src="../pics//c5022dd3-be22-4250-b9f6-38ae984a04d7.jpg" width="600"/> </div><br>
-
-### 3. 差错检测
-
-目前数据链路层广泛使用了循环冗余检验（CRC）来检查比特差错。
-
-## 信道分类
-
-### 1. 广播信道
-
-一对多通信，一个节点发送的数据能够被广播信道上所有的节点接收到。
-
-所有的节点都在同一个广播信道上发送数据，因此需要有专门的控制方法进行协调，避免发生冲突（冲突也叫碰撞）。
-
-主要有两种控制方法进行协调，一个是使用信道复用技术，一是使用 CSMA/CD 协议。
-
-### 2. 点对点信道
-
-一对一通信。
-
-因为不会发生碰撞，因此也比较简单，使用 PPP 协议进行控制。
-
-## 信道复用技术
-
-### 1. 频分复用
-
-频分复用的所有主机在相同的时间占用不同的频率带宽资源。
-
-<div align="center"> <img src="../pics//c4c14368-519c-4a0e-8331-0a553715e3e7.jpg"/> </div><br>
-
-### 2. 时分复用
-
-时分复用的所有主机在不同的时间占用相同的频率带宽资源。
-
-<div align="center"> <img src="../pics//fa2273c3-1b5f-48ce-8e8b-441a4116c1c4.jpg"/> </div><br>
-
-使用频分复用和时分复用进行通信，在通信的过程中主机会一直占用一部分信道资源。但是由于计算机数据的突发性质，通信过程没必要一直占用信道资源而不让出给其它用户使用，因此这两种方式对信道的利用率都不高。
-
-### 3. 统计时分复用
-
-是对时分复用的一种改进，不固定每个用户在时分复用帧中的位置，只要有数据就集中起来组成统计时分复用帧然后发送。
-
-<div align="center"> <img src="../pics//5999e5de-7c16-4b52-b3aa-6dc7b58c7894.png" width="700"/> </div><br>
-
-### 4. 波分复用
-
-光的频分复用。由于光的频率很高，因此习惯上用波长而不是频率来表示所使用的光载波。
-
-<div align="center"> <img src="../pics//21041ec2-babb-483f-bf47-8b8148eec162.png" width="700"/> </div><br>
-
-### 5. 码分复用
-
-为每个用户分配 m bit 的码片，并且所有的码片正交，对于任意两个码片 <img src="https://latex.codecogs.com/gif.latex?\vec{S}"/> 和 <img src="https://latex.codecogs.com/gif.latex?\vec{T}"/> 有
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?\frac{1}{m}\vec{S}\cdot\vec{T}=0"/></div> <br>
-
-为了讨论方便，取 m=8，设码片 <img src="https://latex.codecogs.com/gif.latex?\vec{S}"/> 为 00011011。在拥有该码片的用户发送比特 1 时就发送该码片，发送比特 0 时就发送该码片的反码 11100100。
-
-在计算时将 00011011 记作 (-1 -1 -1 +1 +1 -1 +1 +1)，可以得到
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?\frac{1}{m}\vec{S}\cdot\vec{S}=1"/></div> <br>
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?\frac{1}{m}\vec{S}\cdot\vec{S'}=-1"/></div> <br>
-
-其中 <img src="https://latex.codecogs.com/gif.latex?\vec{S'}"/> 为 <img src="https://latex.codecogs.com/gif.latex?\vec{S}"/> 的反码。
-
-利用上面的式子我们知道，当接收端使用码片 <img src="https://latex.codecogs.com/gif.latex?\vec{S}"/> 对接收到的数据进行内积运算时，结果为 0 的是其它用户发送的数据，结果为 1 的是用户发送的比特 1，结果为 -1 的是用户发送的比特 0。
-
-码分复用需要发送的数据量为原先的 m 倍。
-
-<div align="center"> <img src="../pics//92ad9bae-7d02-43ba-8115-a9d6f530ca28.png" width="600"/> </div><br>
-
-## CSMA/CD 协议*
-
-CSMA/CD 表示载波监听多点接入 / 碰撞检测。
-
--  **多点接入** ：说明这是总线型网络，许多主机以多点的方式连接到总线上。
--  **载波监听** ：每个主机都必须不停地监听信道。在发送前，如果监听到信道正在使用，就必须等待。
--  **碰撞检测** ：在发送中，如果监听到信道已有其它主机正在发送数据，就表示发生了碰撞。虽然每个主机在发送数据之前都已经监听到信道为空闲，但是由于电磁波的传播时延的存在，还是有可能会发生碰撞。
-
-记端到端的传播时延为 τ，最先发送的站点最多经过 2τ 就可以知道是否发生了碰撞，称 2τ 为  **争用期** 。只有经过争用期之后还没有检测到碰撞，才能肯定这次发送不会发生碰撞。
-
-当发生碰撞时，站点要停止发送，等待一段时间再发送。这个时间采用  **截断二进制指数退避算法**  来确定。从离散的整数集合 {0, 1, .., (2<sup>k</sup>-1)} 中随机取出一个数，记作 r，然后取 r 倍的争用期作为重传等待时间。
-
-<div align="center"> <img src="../pics//5aa82b89-f266-44da-887d-18f31f01d8ef.png" width="600"/> </div><br>
-
-## PPP 协议
-
-互联网用户通常需要连接到某个 ISP 之后才能接入到互联网，PPP 协议是用户计算机和 ISP 进行通信时所使用的数据链路层协议。
-
-<div align="center"> <img src="../pics//ddcf2327-8d84-425d-8535-121a94bcb88d.jpg" width="600"/> </div><br>
-
-PPP 的帧格式：
-
-- F 字段为帧的定界符
-- A 和 C 字段暂时没有意义
-- FCS 字段是使用 CRC 的检验序列
-- 信息部分的长度不超过 1500
-
-<div align="center"> <img src="../pics//69f16984-a66f-4288-82e4-79b4aa43e835.jpg" width="500"/> </div><br>
-
-## MAC 地址
-
-MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标识网络适配器（网卡）。
-
-一台主机拥有多少个适配器就有多少个 MAC 地址。例如笔记本电脑普遍存在无线网络适配器和有线网络适配器，因此就有两个 MAC 地址。
-
-## 局域网
-
-局域网是一种典型的广播信道，主要特点是网络为一个单位所拥有，且地理范围和站点数目均有限。
-
-主要有以太网、令牌环网、FDDI 和 ATM 等局域网技术，目前以太网占领着有线局域网市场。
-
-可以按照网络拓结构扑对局域网进行分类：
-
-<div align="center"> <img src="../pics//a6026bb4-3daf-439f-b1ec-a5a24e19d2fb.jpg" width="600"/> </div><br>
-
-## 以太网*
-
-以太网是一种星型拓扑结构局域网。
-
-早期使用集线器进行连接。集线器是一种物理层设备，作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离。之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
-
-目前以太网使用交换机替代了集线器。交换机是一种链路层设备，它不会发生碰撞，能根据 MAC 地址进行存储转发。
-
-以太网帧格式：
-
--  **类型** ：标记上层使用的协议；
--  **数据** ：长度在 46-1500 之间，如果太小则需要填充；
--  **FCS** ：帧检验序列，使用的是 CRC 检验方法；
--  **前同步码** ：只是为了计算 FCS 临时加入的，计算结束之后会丢弃。
-
-<div align="center"> <img src="../pics//50d38e84-238f-4081-8876-14ef6d7938b5.jpg" width="600"/> </div><br>
-
-## 交换机*
-
-交换机具有自学习能力，学习的是交换表的内容，交换表中存储着 MAC 地址到接口的映射。正是由于这种自学习能力，因此交换机是一种即插即可即用设备，不需要网络管理员手动配置交换表内容。
-
-下图中，交换机有 4 个接口，主机 A 向主机 B 发送数据帧时，交换机把主机 A 到接口 1 的映射写入交换表中。为了发送数据帧到 B，先查交换表，此时没有主机 B 的表项，那么主机 A 就发送广播帧，主机 C 和主机 D 会丢弃该帧。主机 B 收下之后，查找交换表得到主机 A 映射的接口为 1，就发送数据帧到接口 1，同时交换机添加主机 B 到接口 3 的映射。
-
-<div align="center"> <img src="../pics//c9cfcd20-c901-435f-9a07-3e46830c359f.jpg" width="800"/> </div><br>
-
-## 虚拟局域网
-
-虚拟局域网可以建立与物理位置无关的逻辑组，只有在同一个虚拟局域网中的成员才会收到链路层广播信息。例如下图中 (A1, A2, A3, A4) 属于一个虚拟局域网，A1 发送的广播会被 A2、A3、A4 收到，而其它站点收不到。
-
-使用 VLAN 干线连接来建立虚拟局域网，每台交换机上的一个特殊端口被设置为干线端口，以互连 VLAN 交换机。IEEE 定义了一种扩展的以太网帧格式 802.1Q，它在标准以太网帧上加进了 4 字节首部 VLAN 标签，用于表示该帧属于哪一个虚拟局域网。
-
-<div align="center"> <img src="../pics//a74b70ac-323a-4b31-b4d5-90569b8a944b.png" width="500"/> </div><br>
-
-# 四、网络层*
-
-## 概述
-
-因为网络层是整个互联网的核心，因此应当让网络层尽可能简单。网络层向上只提供简单灵活的、无连接的、尽最大努力交互的数据报服务。
-
-使用 IP 协议，可以把异构的物理网络连接起来，使得在网络层看起来好像是一个统一的网络。
-
-<div align="center"> <img src="../pics//7b038838-c75b-4538-ae84-6299386704e5.jpg" width="500"/> </div><br>
-
-与 IP 协议配套使用的还有三个协议：
-
-- 地址解析协议 ARP（Address Resolution Protocol）
-- 网际控制报文协议 ICMP（Internet Control Message Protocol）
-- 网际组管理协议 IGMP（Internet Group Management Protocol）
-
-<div align="center"> <img src="../pics//0a9f4125-b6ab-4e94-a807-fd7070ae726a.png" width="350"/> </div><br>
-
-## IP 数据报格式
-
-<div align="center"> <img src="../pics//85c05fb1-5546-4c50-9221-21f231cdc8c5.jpg" width="700"/> </div><br>
-
--  **版本**  : 有 4（IPv4）和 6（IPv6）两个值；
-
--  **首部长度**  : 占 4 位，因此最大值为 15。值为 1 表示的是 1 个 32 位字的长度，也就是 4 字节。因为首部固定长度为 20 字节，因此该值最小为 5。如果可选字段的长度不是 4 字节的整数倍，就用尾部的填充部分来填充。
-
--  **区分服务**  : 用来获得更好的服务，一般情况下不使用。
-
--  **总长度**  : 包括首部长度和数据部分长度。
-
--  **生存时间**  ：TTL，它的存在是为了防止无法交付的数据报在互联网中不断兜圈子。以路由器跳数为单位，当 TTL 为 0 时就丢弃数据报。
-
--  **协议** ：指出携带的数据应该上交给哪个协议进行处理，例如 ICMP、TCP、UDP 等。
-
--  **首部检验和** ：因为数据报每经过一个路由器，都要重新计算检验和，因此检验和不包含数据部分可以减少计算的工作量。
-
--  **标识**  : 在数据报长度过长从而发生分片的情况下，相同数据报的不同分片具有相同的标识符。
-
--  **片偏移**  : 和标识符一起，用于发生分片的情况。片偏移的单位为 8 字节。
-
-<div align="center"> <img src="../pics//23ba890e-e11c-45e2-a20c-64d217f83430.png" width="700"/> </div><br>
-
-## IP 地址编址方式
-
-IP 地址的编址方式经历了三个历史阶段：
-
-- 分类
-- 子网划分
-- 无分类
-
-### 1. 分类
-
-由两部分组成，网络号和主机号，其中不同分类具有不同的网络号长度，并且是固定的。
-
-IP 地址 ::= {< 网络号 >, < 主机号 >}
-
-<div align="center"> <img src="../pics//cbf50eb8-22b4-4528-a2e7-d187143d57f7.png" width="500"/> </div><br>
-
-### 2. 子网划分
-
-通过在主机号字段中拿一部分作为子网号，把两级 IP 地址划分为三级 IP 地址。注意，外部网络看不到子网的存在。
-
-IP 地址 ::= {< 网络号 >, < 子网号 >, < 主机号 >}
-
-要使用子网，必须配置子网掩码。一个 B 类地址的默认子网掩码为 255.255.0.0，如果 B 类地址的子网占两个比特，那么子网掩码为 11111111 11111111 11000000 00000000，也就是 255.255.192.0。
-
-### 3. 无分类
-
-无分类编址 CIDR 消除了传统 A 类、B 类和 C 类地址以及划分子网的概念，使用网络前缀和主机号来对 IP 地址进行编码，网络前缀的长度可以根据需要变化。
-
-IP 地址 ::= {< 网络前缀号 >, < 主机号 >}
-
-CIDR 的记法上采用在 IP 地址后面加上网络前缀长度的方法，例如 128.14.35.7/20 表示前 20 位为网络前缀。
-
-CIDR 的地址掩码可以继续称为子网掩码，子网掩码首 1 长度为网络前缀的长度。
-
-一个 CIDR 地址块中有很多地址，一个 CIDR 表示的网络就可以表示原来的很多个网络，并且在路由表中只需要一个路由就可以代替原来的多个路由，减少了路由表项的数量。把这种通过使用网络前缀来减少路由表项的方式称为路由聚合，也称为  **构成超网** 。
-
-在路由表中的项目由“网络前缀”和“下一跳地址”组成，在查找时可能会得到不止一个匹配结果，应当采用最长前缀匹配来确定应该匹配哪一个。
-
-## 地址解析协议 ARP
-
-网络层实现主机之间的通信，而链路层实现具体每段链路之间的通信。因此在通信过程中，IP 数据报的源地址和目的地址始终不变，而 MAC 地址随着链路的改变而改变。
-
-<div align="center"> <img src="../pics//66192382-558b-4b05-a35d-ac4a2b1a9811.jpg" width="700"/> </div><br>
-
-ARP 实现由 IP 地址得到 MAC 地址。
-
-<div align="center"> <img src="../pics//b9d79a5a-e7af-499b-b989-f10483e71b8b.jpg" width="500"/> </div><br>
-
-每个主机都有一个 ARP 高速缓存，里面有本局域网上的各主机和路由器的 IP 地址到硬件地址的映射表。
-
-如果主机 A 知道主机 B 的 IP 地址，但是 ARP 高速缓存中没有该 IP 地址到 MAC 地址的映射，此时主机 A 通过广播的方式发送 ARP 请求分组，主机 B 收到该请求后会发送 ARP 响应分组给主机 A 告知其 MAC 地址，随后主机 A 向其高速缓存中写入主机 B 的 IP 地址到 MAC 地址的映射。
-
-<div align="center"> <img src="../pics//8006a450-6c2f-498c-a928-c927f758b1d0.png" width="700"/> </div><br>
-
-## 网际控制报文协议 ICMP
-
-ICMP 是为了更有效地转发 IP 数据报和提高交付成功的机会。它封装在 IP 数据报中，但是不属于高层协议。
-
-<div align="center"> <img src="../pics//e3124763-f75e-46c3-ba82-341e6c98d862.jpg" width="500"/> </div><br>
-
-ICMP 报文分为差错报告报文和询问报文。
-
-<div align="center"> <img src="../pics//aa29cc88-7256-4399-8c7f-3cf4a6489559.png" width="600"/> </div><br>
-
-### 1. Ping
-
-Ping 是 ICMP 的一个重要应用，主要用来测试两台主机之间的连通性。
-
-Ping 发送的 IP 数据报封装的是无法交付的 UDP 用户数据报。
-
-### 2. Traceroute
-
-Traceroute 是 ICMP 的另一个应用，用来跟踪一个分组从源点到终点的路径。
-
-- 源主机向目的主机发送一连串的 IP 数据报。第一个数据报 P1 的生存时间 TTL 设置为 1，当 P1 到达路径上的第一个路由器 R1 时，R1 收下它并把 TTL 减 1，此时 TTL 等于 0，R1 就把 P1 丢弃，并向源主机发送一个 ICMP 时间超过差错报告报文；
-- 源主机接着发送第二个数据报 P2，并把 TTL 设置为 2。P2 先到达 R1，R1 收下后把 TTL 减 1 再转发给 R2，R2 收下后也把 TTL 减 1，由于此时 TTL 等于 0，R2 就丢弃 P2，并向源主机发送一个 ICMP 时间超过差错报文。
-- 不断执行这样的步骤，直到最后一个数据报刚刚到达目的主机，主机不转发数据报，也不把 TTL 值减 1。但是因为数据报封装的是无法交付的 UDP，因此目的主机要向源主机发送 ICMP 终点不可达差错报告报文。
-- 之后源主机知道了到达目的主机所经过的路由器 IP 地址以及到达每个路由器的往返时间。
-
-## 虚拟专用网 VPN
-
-由于 IP 地址的紧缺，一个机构能申请到的 IP 地址数往往远小于本机构所拥有的主机数。并且一个机构并不需要把所有的主机接入到外部的互联网中，机构内的计算机可以使用仅在本机构有效的 IP 地址（专用地址）。
-
-有三个专用地址块：
-
-- 10.0.0.0 \~ 10.255.255.255
-- 172.16.0.0 \~ 172.31.255.255
-- 192.168.0.0 \~ 192.168.255.255
-
-VPN 使用公用的互联网作为本机构各专用网之间的通信载体。专用指机构内的主机只与本机构内的其它主机通信；虚拟指“好像是”，而实际上并不是，它有经过公用的互联网。
-
-下图中，场所 A 和 B 的通信经过互联网，如果场所 A 的主机 X 要和另一个场所 B 的主机 Y 通信，IP 数据报的源地址是 10.1.0.1，目的地址是 10.2.0.3。数据报先发送到与互联网相连的路由器 R1，R1 对内部数据进行加密，然后重新加上数据报的首部，源地址是路由器 R1 的全球地址 125.1.2.3，目的地址是路由器 R2 的全球地址 194.4.5.6。路由器 R2 收到数据报后将数据部分进行解密，恢复原来的数据报，此时目的地址为 10.2.0.3，就交付给 Y。
-
-<div align="center"> <img src="../pics//1556770b-8c01-4681-af10-46f1df69202c.jpg" width="800"/> </div><br>
-
-## 网络地址转换 NAT
-
-专用网内部的主机使用本地 IP 地址又想和互联网上的主机通信时，可以使用 NAT 来将本地 IP 转换为全球 IP。
-
-在以前，NAT 将本地 IP 和全球 IP 一一对应，这种方式下拥有 n 个全球 IP 地址的专用网内最多只可以同时有 n 台主机接入互联网。为了更有效地利用全球 IP 地址，现在常用的 NAT 转换表把运输层的端口号也用上了，使得多个专用网内部的主机共用一个全球 IP 地址。使用端口号的 NAT 也叫做网络地址与端口转换 NAPT。
-
-<div align="center"> <img src="../pics//2719067e-b299-4639-9065-bed6729dbf0b.png" width=""/> </div><br>
-
-## 路由器的结构
-
-路由器从功能上可以划分为：路由选择和分组转发。
-
-分组转发结构由三个部分组成：交换结构、一组输入端口和一组输出端口。
-
-<div align="center"> <img src="../pics//c3369072-c740-43b0-b276-202bd1d3960d.jpg" width="600"/> </div><br>
-
-## 路由器分组转发流程
-
-- 从数据报的首部提取目的主机的 IP 地址 D，得到目的网络地址 N。
-- 若 N 就是与此路由器直接相连的某个网络地址，则进行直接交付；
-- 若路由表中有目的地址为 D 的特定主机路由，则把数据报传送给表中所指明的下一跳路由器；
-- 若路由表中有到达网络 N 的路由，则把数据报传送给路由表中所指明的下一跳路由器；
-- 若路由表中有一个默认路由，则把数据报传送给路由表中所指明的默认路由器；
-- 报告转发分组出错。
-
-<div align="center"> <img src="../pics//1ab49e39-012b-4383-8284-26570987e3c4.jpg" width="800"/> </div><br>
-
-## 路由选择协议
-
-路由选择协议都是自适应的，能随着网络通信量和拓扑结构的变化而自适应地进行调整。
-
-互联网可以划分为许多较小的自治系统 AS，一个 AS 可以使用一种和别的 AS 不同的路由选择协议。
-
-可以把路由选择协议划分为两大类：
-
-- 自治系统内部的路由选择：RIP 和 OSPF
-- 自治系统间的路由选择：BGP
-
-### 1. 内部网关协议 RIP
-
-RIP 是一种基于距离向量的路由选择协议。距离是指跳数，直接相连的路由器跳数为 1。跳数最多为 15，超过 15 表示不可达。
-
-RIP 按固定的时间间隔仅和相邻路由器交换自己的路由表，经过若干次交换之后，所有路由器最终会知道到达本自治系统中任何一个网络的最短距离和下一跳路由器地址。
-
-距离向量算法：
-
-- 对地址为 X 的相邻路由器发来的 RIP 报文，先修改报文中的所有项目，把下一跳字段中的地址改为 X，并把所有的距离字段加 1；
-- 对修改后的 RIP 报文中的每一个项目，进行以下步骤：
- - 若原来的路由表中没有目的网络 N，则把该项目添加到路由表中；
- - 否则：若下一跳路由器地址是 X，则把收到的项目替换原来路由表中的项目；否则：若收到的项目中的距离 d 小于路由表中的距离，则进行更新（例如原始路由表项为 Net2, 5, P，新表项为 Net2, 4, X，则更新）；否则什么也不做。
-- 若 3 分钟还没有收到相邻路由器的更新路由表，则把该相邻路由器标为不可达，即把距离置为 16。
-
-RIP 协议实现简单，开销小。但是 RIP 能使用的最大距离为 15，限制了网络的规模。并且当网络出现故障时，要经过比较长的时间才能将此消息传送到所有路由器。
-
-### 2. 内部网关协议 OSPF
-
-开放最短路径优先 OSPF，是为了克服 RIP 的缺点而开发出来的。
-
-开放表示 OSPF 不受某一家厂商控制，而是公开发表的；最短路径优先表示使用了 Dijkstra 提出的最短路径算法 SPF。
-
-OSPF 具有以下特点：
-
-- 向本自治系统中的所有路由器发送信息，这种方法是洪泛法。
-- 发送的信息就是与相邻路由器的链路状态，链路状态包括与哪些路由器相连以及链路的度量，度量用费用、距离、时延、带宽等来表示。
-- 只有当链路状态发生变化时，路由器才会发送信息。
-
-所有路由器都具有全网的拓扑结构图，并且是一致的。相比于 RIP，OSPF 的更新过程收敛的很快。
-
-### 3. 外部网关协议 BGP
-
-BGP（Border Gateway Protocol，边界网关协议）
-
-AS 之间的路由选择很困难，主要是由于：
-
-- 互联网规模很大；
-- 各个 AS 内部使用不同的路由选择协议，无法准确定义路径的度量；
-- AS 之间的路由选择必须考虑有关的策略，比如有些 AS 不愿意让其它 AS 经过。
-
-BGP 只能寻找一条比较好的路由，而不是最佳路由。
-
-每个 AS 都必须配置 BGP 发言人，通过在两个相邻 BGP 发言人之间建立 TCP 连接来交换路由信息。
-
-<div align="center"> <img src="../pics//9cd0ae20-4fb5-4017-a000-f7d3a0eb3529.png" width="600"/> </div><br>
-
-# 五、运输层*
-
-网络层只把分组发送到目的主机，但是真正通信的并不是主机而是主机中的进程。运输层提供了进程间的逻辑通信，运输层向高层用户屏蔽了下面网络层的核心细节，使应用程序看起来像是在两个运输层实体之间有一条端到端的逻辑通信信道。
-
-## UDP 和 TCP 的特点
-
-- 用户数据报协议 UDP（User Datagram Protocol）是无连接的，尽最大可能交付，没有拥塞控制，面向报文（对于应用程序传下来的报文不合并也不拆分，只是添加 UDP 首部），支持一对一、一对多、多对一和多对多的交互通信。
-
-- 传输控制协议 TCP（Transmission Control Protocol）是面向连接的，提供可靠交付，有流量控制，拥塞控制，提供全双工通信，面向字节流（把应用层传下来的报文看成字节流，把字节流组织成大小不等的数据块），每一条 TCP 连接只能是点对点的（一对一）。
-
-## UDP 首部格式
-
-<div align="center"> <img src="../pics//d4c3a4a1-0846-46ec-9cc3-eaddfca71254.jpg" width="600"/> </div><br>
-
-首部字段只有 8 个字节，包括源端口、目的端口、长度、检验和。12 字节的伪首部是为了计算检验和临时添加的。
-
-## TCP 首部格式
-
-<div align="center"> <img src="../pics//55dc4e84-573d-4c13-a765-52ed1dd251f9.png" width="700"/> </div><br>
-
--  **序号**  ：用于对字节流进行编号，例如序号为 301，表示第一个字节的编号为 301，如果携带的数据长度为 100 字节，那么下一个报文段的序号应为 401。
-
--  **确认号**  ：期望收到的下一个报文段的序号。例如 B 正确收到 A 发送来的一个报文段，序号为 501，携带的数据长度为 200 字节，因此 B 期望下一个报文段的序号为 701，B 发送给 A 的确认报文段中确认号就为 701。
-
--  **数据偏移**  ：指的是数据部分距离报文段起始处的偏移量，实际上指的是首部的长度。
-
--  **确认 ACK**  ：当 ACK=1 时确认号字段有效，否则无效。TCP 规定，在连接建立后所有传送的报文段都必须把 ACK 置 1。
-
--  **同步 SYN**  ：在连接建立时用来同步序号。当 SYN=1，ACK=0 时表示这是一个连接请求报文段。若对方同意建立连接，则响应报文中 SYN=1，ACK=1。
-
--  **终止 FIN**  ：用来释放一个连接，当 FIN=1 时，表示此报文段的发送方的数据已发送完毕，并要求释放连接。
-
--  **窗口**  ：窗口值作为接收方让发送方设置其发送窗口的依据。之所以要有这个限制，是因为接收方的数据缓存空间是有限的。
-
-## TCP 的三次握手
-
-<div align="center"> <img src="../pics//e92d0ebc-7d46-413b-aec1-34a39602f787.png" width="600"/> </div><br>
-
-假设 A 为客户端，B 为服务器端。
-
-- 首先 B 处于 LISTEN（监听）状态，等待客户的连接请求。
-
-- A 向 B 发送连接请求报文段，SYN=1，ACK=0，选择一个初始的序号 x。
-
-- B 收到连接请求报文段，如果同意建立连接，则向 A 发送连接确认报文段，SYN=1，ACK=1，确认号为 x+1，同时也选择一个初始的序号 y。
-
-- A 收到 B 的连接确认报文段后，还要向 B 发出确认，确认号为 y+1，序号为 x+1。
-
-- B 收到 A 的确认后，连接建立。
-
-**三次握手的原因** 
-
-第三次握手是为了防止失效的连接请求到达服务器，让服务器错误打开连接。
-
-客户端发送的连接请求如果在网络中滞留，那么就会隔很长一段时间才能收到服务器端发回的连接确认。客户端等待一个超时重传时间之后，就会重新请求连接。但是这个滞留的连接请求最后还是会到达服务器，如果不进行三次握手，那么服务器就会打开两个连接。如果有第三次握手，客户端会忽略服务器之后发送的对滞留连接请求的连接确认，不进行第三次握手，因此就不会再次打开连接。
-
-## TCP 的四次挥手
-
-<div align="center"> <img src="../pics//f87afe72-c2df-4c12-ac03-9b8d581a8af8.jpg" width="600"/> </div><br>
-
-以下描述不讨论序号和确认号，因为序号和确认号的规则比较简单。并且不讨论 ACK，因为 ACK 在连接建立之后都为 1。
-
-- A 发送连接释放报文段，FIN=1。
-
-- B 收到之后发出确认，此时 TCP 属于半关闭状态，B 能向 A 发送数据但是 A 不能向 B 发送数据。
-
-- 当 B 不再需要连接时，发送连接释放请求报文段，FIN=1。
-
-- A 收到后发出确认，进入 TIME-WAIT 状态，等待 2 MSL（最大报文存活时间）后释放连接。
-
-- B 收到 A 的确认后释放连接。
-
-**四次挥手的原因** 
-
-客户端发送了 FIN 连接释放报文之后，服务器收到了这个报文，就进入了 CLOSE-WAIT 状态。这个状态是为了让服务器端发送还未传送完毕的数据，传送完毕之后，服务器会发送 FIN 连接释放报文。
-
-**TIME_WAIT** 
-
-客户端接收到服务器端的 FIN 报文后进入此状态，此时并不是直接进入 CLOSED 状态，还需要等待一个时间计时器设置的时间 2MSL。这么做有两个理由：
-
-- 确保最后一个确认报文段能够到达。如果 B 没收到 A 发送来的确认报文段，那么就会重新发送连接释放请求报文段，A 等待一段时间就是为了处理这种情况的发生。
-
-- 等待一段时间是为了让本连接持续时间内所产生的所有报文段都从网络中消失，使得下一个新的连接不会出现旧的连接请求报文段。
-
-## TCP 滑动窗口
-
-<div align="center"> <img src="../pics//a3253deb-8d21-40a1-aae4-7d178e4aa319.jpg" width="800"/> </div><br>
-
-窗口是缓存的一部分，用来暂时存放字节流。发送方和接收方各有一个窗口，接收方通过 TCP 报文段中的窗口字段告诉发送方自己的窗口大小，发送方根据这个值和其它信息设置自己的窗口大小。
-
-发送窗口内的字节都允许被发送，接收窗口内的字节都允许被接收。如果发送窗口左部的字节已经发送并且收到了确认，那么就将发送窗口向右滑动一定距离，直到左部第一个字节不是已发送并且已确认的状态；接收窗口的滑动类似，接收窗口左部字节已经发送确认并交付主机，就向右滑动接收窗口。
-
-接收窗口只会对窗口内最后一个按序到达的字节进行确认，例如接收窗口已经收到的字节为 {31, 34, 35}，其中 {31} 按序到达，而 {34, 35} 就不是，因此只对字节 31 进行确认。发送方得到一个字节的确认之后，就知道这个字节之前的所有字节都已经被接收。
-
-## TCP 可靠传输
-
-TCP 使用超时重传来实现可靠传输：如果一个已经发送的报文段在超时时间内没有收到确认，那么就重传这个报文段。
-
-一个报文段从发送再到接收到确认所经过的时间称为往返时间 RTT，加权平均往返时间 RTTs 计算如下：
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?RTTs=(1-a)*(RTTs)+a*RTT"/></div> <br>
-
-超时时间 RTO 应该略大于 RTTs，TCP 使用的超时时间计算如下：
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?RTO=RTTs+4*RTT_d"/></div> <br>
-
-其中 RTT<sub>d</sub> 为偏差。
-
-## TCP 流量控制
-
-流量控制是为了控制发送方发送速率，保证接收方来得及接收。
-
-接收方发送的确认报文中的窗口字段可以用来控制发送方窗口大小，从而影响发送方的发送速率。将窗口字段设置为 0，则发送方不能发送数据。
-
-## TCP 拥塞控制
-
-如果网络出现拥塞，分组将会丢失，此时发送方会继续重传，从而导致网络拥塞程度更高。因此当出现拥塞时，应当控制发送方的速率。这一点和流量控制很像，但是出发点不同。流量控制是为了让接收方能来得及接收，而拥塞控制是为了降低整个网络的拥塞程度。
-
-<div align="center"> <img src="../pics//51e2ed95-65b8-4ae9-8af3-65602d452a25.jpg" width="500"/> </div><br>
-
-TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、快重传、快恢复。
-
-发送方需要维护一个叫做拥塞窗口（cwnd）的状态变量，注意拥塞窗口与发送方窗口的区别：拥塞窗口只是一个状态变量，实际决定发送方能发送多少数据的是发送方窗口。
-
-为了便于讨论，做如下假设：
-
-- 接收方有足够大的接收缓存，因此不会发生流量控制；
-- 虽然 TCP 的窗口基于字节，但是这里设窗口的大小单位为报文段。
-
-<div align="center"> <img src="../pics//910f613f-514f-4534-87dd-9b4699d59d31.png" width="800"/> </div><br>
-
-### 1. 慢开始与拥塞避免
-
-发送的最初执行慢开始，令 cwnd=1，发送方只能发送 1 个报文段；当收到确认后，将 cwnd 加倍，因此之后发送方能够发送的报文段数量为：2、4、8 ...
-
-注意到慢开始每个轮次都将 cwnd 加倍，这样会让 cwnd 增长速度非常快，从而使得发送方发送的速度增长速度过快，网络拥塞的可能也就更高。设置一个慢开始门限 ssthresh，当 cwnd >= ssthresh 时，进入拥塞避免，每个轮次只将 cwnd 加 1。
-
-如果出现了超时，则令 ssthresh = cwnd/2，然后重新执行慢开始。
-
-### 2. 快重传与快恢复
-
-在接收方，要求每次接收到报文段都应该对最后一个已收到的有序报文段进行确认。例如已经接收到 M<sub>1</sub> 和 M<sub>2</sub>，此时收到 M<sub>4</sub>，应当发送对 M<sub>2</sub> 的确认。
-
-在发送方，如果收到三个重复确认，那么可以知道下一个报文段丢失，此时执行快重传，立即重传下一个报文段。例如收到三个 M<sub>2</sub>，则 M<sub>3</sub> 丢失，立即重传 M<sub>3</sub>。
-
-在这种情况下，只是丢失个别报文段，而不是网络拥塞。因此执行快恢复，令 ssthresh = cwnd/2 ，cwnd = ssthresh，注意到此时直接进入拥塞避免。慢开始和快恢复的快慢指的是 cwnd 的设定值，而不是 cwnd 的增长速率。慢开始 cwnd 设定为 1，而快恢复 cwnd 设定为 ssthresh。
-
-<div align="center"> <img src="../pics//f61b5419-c94a-4df1-8d4d-aed9ae8cc6d5.png" width="600"/> </div><br>
-
-# 六、应用层
-
-## 域名系统
-
-DNS 是一个分布式数据库，提供了主机名和 IP 地址之间的转换。这里的分布式数据库是指，每个站点只保留它自己的那部分数据。
-
-域名具有层次结构，从上到下依次为：根域名、顶级域名、第二级域名。
-
-<div align="center"> <img src="../pics//b54eeb16-0b0e-484c-be62-306f57c40d77.jpg"/> </div><br>
-
-DNS 可以使用 UDP 或者 TCP 进行传输，使用的端口号都为 53。大多数情况下 DNS 使用 UDP 进行传输，这就要求域名解析器和域名服务器都必须自己处理超时和重传来保证可靠性。在两种情况下会使用 TCP 进行传输：
-
-- 因为 UDP 最大只支持 512 字节的数据，如果返回的响应超过的 512 字节就改用 TCP 进行传输。
-- 区域传送是主域名服务器向辅助域名服务器传送变化的那部分数据，区域传送需要使用 TCP 进行传输。
-
-## 文件传送协议
-
-FTP 使用 TCP 进行连接，它需要两个连接来传送一个文件：
-
-- 控制连接：服务器以打开端口号 21 等待客户端的连接，客户端主动建立连接后，使用这个连接将客户端的命令传送给服务器，并传回服务器的应答。
-- 数据连接：用来传送一个文件。
-
-根据数据连接是否是服务器端主动建立，FTP 有主动和被动两种模式：
-
-- 主动模式：服务器端主动建立数据连接，其中服务器端的端口号为 20，客户端的端口号随机，但是必须大于 1024，因为 0\~1023 是熟知端口号。
-
-<div align="center"> <img src="../pics//03f47940-3843-4b51-9e42-5dcaff44858b.jpg"/> </div><br>
-
-- 被动模式：客户端主动建立数据连接，其中客户端的端口号由客户端自己指定，服务器端的端口号随机。
-
-<div align="center"> <img src="../pics//be5c2c61-86d2-4dba-a289-b48ea23219de.jpg"/> </div><br>
-
-主动模式要求客户端开放端口号给服务器端，需要去配置客户端的防火墙。被动模式只需要服务器端开放端口号即可，无需客户端配置防火墙。但是被动模式会导致服务器端的安全性减弱，因为开放了过多的端口号。
-
-## 远程登录协议
-
-TELNET 用于登录到远程主机上，并且远程主机上的输出也会返回。
-
-TELNET 可以适应许多计算机和操作系统的差异，例如不同操作系统系统的换行符定义。
-
-## 电子邮件协议
-
-一个电子邮件系统由三部分组成：用户代理、邮件服务器以及邮件协议。
-
-邮件协议包含发送协议和读取协议，发送协议常用 SMTP，读取协议常用 POP3 和 IMAP。
-
-<div align="center"> <img src="../pics//7b3efa99-d306-4982-8cfb-e7153c33aab4.png" width="700"/> </div><br>
-
-### 1. POP3
-
-POP3 的特点是只要用户从服务器上读取了邮件，就把该邮件删除。
-
-### 2. IMAP
-
-IMAP 协议中客户端和服务器上的邮件保持同步，如果不去手动删除邮件，那么服务器上的邮件也不会被删除。IMAP 这种做法可以让用户随时随地去访问服务器上的邮件。
-
-### 3. SMTP
-
-SMTP 只能发送 ASCII 码，而互联网邮件扩充 MIME 可以发送二进制文件。MIME 并没有改动或者取代 SMTP，而是增加邮件主体的结构，定义了非 ASCII 码的编码规则。
-
-<div align="center"> <img src="../pics//ed5522bb-3a60-481c-8654-43e7195a48fe.png" width=""/> </div><br>
-
-## 动态主机配置协议
-
-DHCP (Dynamic Host Configuration Protocol) 提供了即插即用的连网方式，用户不再需要去手动配置 IP 地址等信息。DHCP 配置的内容不仅是 IP 地址，还包括子网掩码、网关 IP 地址。
-
-DHCP 工作过程如下：
-
-1. 客户端发送 Discover 报文，该报文的目的地址为 255.255.255.255:67，源地址为 0.0.0.0:68，被放入 UDP 中，该报文被广播到同一个子网的所有主机上。
-2. DHCP 服务器收到 Discover 报文之后，发送 Offer 报文给客户端，该报文包含了客户端所需要的信息。因为客户端可能收到多个 DHCP 服务器提供的信息，因此客户端需要进行选择。
-3. 如果客户端选择了某个 DHCP 服务器提供的信息，那么就发送 Request 报文给该 DHCP 服务器。
-4. DHCP 服务器发送 Ack 报文，表示客户端此时可以使用提供给它的信息。
-
-<div align="center"> <img src="../pics//bf16c541-0717-473b-b75d-4115864f4fbf.jpg"/> </div><br>
-
-如果客户端和 DHCP 服务器不在同一个子网，就需要使用中继代理。
-
-## 常用端口
-
-|应用| 应用层协议 | 端口号 | 运输层协议 | 备注 |
-| :---: | :--: | :--: | :--: | :--:
-| 域名解析 | DNS | 53 | UDP/TCP | 长度超过 512 字节时使用 TCP |
-| 动态主机配置协议 | DHCP | 67/68 | UDP | |
-| 简单网络管理协议 | SNMP | 161/162 | UDP | |
-| 文件传送协议 | FTP | 20/21 | TCP | 控制连接 21，数据连接 20
-| 远程终端协议 | TELNET | 23 | TCP | |
-| 超文本传送协议 | HTTP | 80 | TCP | |
-| 简单邮件传送协议 | SMTP | 25 | TCP | |
-| 邮件读取协议 | POP3 | 110 | TCP | |
-| 网际报文存取协议 | IMAP | 143 | TCP | |
-
-## Web 页面请求过程
-
-### 1. DHCP 配置主机信息
-
-- 假设主机最开始没有 IP 地址以及其它信息，那么就需要先使用 DHCP 来获取。
-
-- 主机生成一个 DHCP 请求报文，并将这个报文放入具有目的端口 67 和源端口 68 的 UDP 报文段中。
-
-- 该报文段则被放入在一个具有广播 IP 目的地址(255.255.255.255) 和源 IP 地址（0.0.0.0）的 IP 数据报中。
-
-- 该数据报则被放置在 MAC 帧中，该帧具有目的地址 FF:FF:FF:FF:FF:FF，将广播到与交换机连接的所有设备。
-
-- 连接在交换机的 DHCP 服务器收到广播帧之后，不断地向上分解得到 IP 数据报、UDP 报文段、DHCP 请求报文，之后生成 DHCP ACK 报文，该报文包含以下信息：IP 地址、DNS 服务器的 IP 地址、默认网关路由器的 IP 地址和子网掩码。该报文被放入 UDP 报文段中，UDP 报文段有被放入 IP 数据报中，最后放入 MAC 帧中。
-
-- 该帧的目的地址是请求主机的 MAC 地址，因为交换机具有自学习能力，之前主机发送了广播帧之后就记录了 MAC 地址到其转发接口的交换表项，因此现在交换机就可以直接知道应该向哪个接口发送该帧。
-
-- 主机收到该帧后，不断分解得到 DHCP 报文。之后就配置它的 IP 地址、子网掩码和 DNS 服务器的 IP 地址，并在其 IP 转发表中安装默认网关。
-
-### 2. ARP 解析 MAC 地址
-
-- 主机通过浏览器生成一个 TCP 套接字，套接字向 HTTP 服务器发送 HTTP 请求。为了生成该套接字，主机需要知道网站的域名对应的 IP 地址。
-
-- 主机生成一个 DNS 查询报文，该报文具有 53 号端口，因为 DNS 服务器的端口号是 53。
-
-- 该 DNS 查询报文被放入目的地址为 DNS 服务器 IP 地址的 IP 数据报中。
-
-- 该 IP 数据报被放入一个以太网帧中，该帧将发送到网关路由器。
-
-- DHCP 过程只知道网关路由器的 IP 地址，为了获取网关路由器的 MAC 地址，需要使用 ARP 协议。
-
-- 主机生成一个包含目的地址为网关路由器 IP 地址的 ARP 查询报文，将该 ARP 查询报文放入一个具有广播目的地址（FF:FF:FF:FF:FF:FF）的以太网帧中，并向交换机发送该以太网帧，交换机将该帧转发给所有的连接设备，包括网关路由器。
-
-- 网关路由器接收到该帧后，不断向上分解得到 ARP 报文，发现其中的 IP 地址与其接口的 IP 地址匹配，因此就发送一个 ARP 回答报文，包含了它的 MAC 地址，发回给主机。
-
-### 3. DNS 解析域名
-
-- 知道了网关路由器的 MAC 地址之后，就可以继续 DNS 的解析过程了。
-
-- 网关路由器接收到包含 DNS 查询报文的以太网帧后，抽取出 IP 数据报，并根据转发表决定该 IP 数据报应该转发的路由器。
-
-- 因为路由器具有内部网关协议（RIP、OSPF）和外部网关协议（BGP）这两种路由选择协议，因此路由表中已经配置了网关路由器到达 DNS 服务器的路由表项。
-
-- 到达 DNS 服务器之后，DNS 服务器抽取出 DNS 查询报文，并在 DNS 数据库中查找待解析的域名。
-
-- 找到 DNS 记录之后，发送 DNS 回答报文，将该回答报文放入 UDP 报文段中，然后放入 IP 数据报中，通过路由器反向转发回网关路由器，并经过以太网交换机到达主机。
-
-### 4. HTTP 请求页面
-
-- 有了 HTTP 服务器的 IP 地址之后，主机就能够生成 TCP 套接字，该套接字将用于向 Web 服务器发送 HTTP GET 报文。
-
-- 在生成 TCP 套接字之前，必须先与 HTTP 服务器进行三次握手来建立连接。生成一个具有目的端口 80 的 TCP SYN 报文段，并向 HTTP 服务器发送该报文段。
-
-- HTTP 服务器收到该报文段之后，生成 TCP SYN ACK 报文段，发回给主机。
-
-- 连接建立之后，浏览器生成 HTTP GET 报文，并交付给 HTTP 服务器。
-
-- HTTP 服务器从 TCP 套接字读取 HTTP GET 报文，生成一个 HTTP 响应报文，将 Web 页面内容放入报文主体中，发回给主机。
-
-- 浏览器收到 HTTP 响应报文后，抽取出 Web 页面内容，之后进行渲染，显示 Web 页面。
-
-
-# 参考资料
-
-- 计算机网络, 谢希仁
-- JamesF.Kurose, KeithW.Ross, 库罗斯, 等. 计算机网络: 自顶向下方法 [M]. 机械工业出版社, 2014.
-- W.RichardStevens. TCP/IP 详解. 卷 1, 协议 [M]. 机械工业出版社, 2006.
-- [Active vs Passive FTP Mode: Which One is More Secure?](https://securitywing.com/active-vs-passive-ftp-mode/)
-- [Active and Passive FTP Transfers Defined  -  KB Article #1138](http://www.serv-u.com/kb/1138/active-and-passive-ftp-transfers-defined)
-- [How DHCP works and DHCP Interview Questions and Answers](http://webcache.googleusercontent.com/search?q=cache:http://anandgiria.blogspot.com/2013/09/windows-dhcp-interview-questions-and.html)
-- [What is process of DORA in DHCP?](https://www.quora.com/What-is-process-of-DORA-in-DHCP)
-- [What is DHCP Server ?](https://tecadmin.net/what-is-dhcp-server/)
-- [Tackling emissions targets in Tokyo](http://www.climatechangenews.com/2011/html/university-tokyo.html)
-- [What does my ISP know when I use Tor?](http://www.climatechangenews.com/2011/html/university-tokyo.html)
-- [Technology-Computer Networking[1]-Computer Networks and the Internet](http://www.linyibin.cn/2017/02/12/technology-ComputerNetworking-Internet/)
-- [P2P 网络概述.](http://slidesplayer.com/slide/11616167/)
-- [Circuit Switching (a) Circuit switching. (b) Packet switching.](http://slideplayer.com/slide/5115386/)
diff --git a/notes/重构.md b/notes/重构.md
index 9a3242ba..649f563a 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -1,1420 +1,4 @@
 <!-- GFM-TOC -->
-* [一、第一个案例](#一第一个案例)
-* [二、重构原则](#二重构原则)
-    * [定义](#定义)
-    * [为何重构](#为何重构)
-    * [三次法则](#三次法则)
-    * [间接层与重构](#间接层与重构)
-    * [修改接口](#修改接口)
-    * [何时不该重构](#何时不该重构)
-    * [重构与设计](#重构与设计)
-    * [重构与性能](#重构与性能)
-* [三、代码的坏味道](#三代码的坏味道)
-    * [1. 重复代码](#1-重复代码)
-    * [2. 过长函数](#2-过长函数)
-    * [3. 过大的类](#3-过大的类)
-    * [4. 过长的参数列表](#4-过长的参数列表)
-    * [5. 发散式变化](#5-发散式变化)
-    * [6. 散弹式修改](#6-散弹式修改)
-    * [7. 依恋情结](#7-依恋情结)
-    * [8. 数据泥团](#8-数据泥团)
-    * [9. 基本类型偏执](#9-基本类型偏执)
-    * [10. switch 惊悚现身](#10-switch-惊悚现身)
-    * [11. 平行继承体系](#11-平行继承体系)
-    * [12. 冗余类](#12-冗余类)
-    * [13. 夸夸其谈未来性](#13-夸夸其谈未来性)
-    * [14. 令人迷惑的暂时字段](#14-令人迷惑的暂时字段)
-    * [15. 过度耦合的消息链](#15-过度耦合的消息链)
-    * [16. 中间人](#16-中间人)
-    * [17. 狎昵关系](#17-狎昵关系)
-    * [18. 异曲同工的类](#18-异曲同工的类)
-    * [19. 不完美的类库](#19-不完美的类库)
-    * [20. 幼稚的数据类](#20-幼稚的数据类)
-    * [21. 被拒绝的馈赠](#21-被拒绝的馈赠)
-    * [22. 过多的注释](#22-过多的注释)
-* [四、构筑测试体系](#四构筑测试体系)
-* [五、重新组织函数](#五重新组织函数)
-    * [1. 提炼函数](#1-提炼函数)
-    * [2. 内联函数](#2-内联函数)
-    * [3. 内联临时变量](#3-内联临时变量)
-    * [4. 以查询取代临时变量](#4-以查询取代临时变量)
-    * [5. 引起解释变量](#5-引起解释变量)
-    * [6. 分解临时变量](#6-分解临时变量)
-    * [7. 移除对参数的赋值](#7-移除对参数的赋值)
-    * [8. 以函数对象取代函数](#8-以函数对象取代函数)
-    * [9. 替换算法](#9-替换算法)
-* [六、在对象之间搬移特性](#六在对象之间搬移特性)
-    * [1. 搬移函数](#1-搬移函数)
-    * [2. 搬移字段](#2-搬移字段)
-    * [3. 提炼类](#3-提炼类)
-    * [4. 将类内联化](#4-将类内联化)
-    * [5. 隐藏委托关系](#5-隐藏委托关系)
-    * [6. 移除中间人](#6-移除中间人)
-    * [7. 引入外加函数](#7-引入外加函数)
-    * [8. 引入本地扩展](#8-引入本地扩展)
-* [七、重新组织数据](#七重新组织数据)
-    * [1. 自封装字段](#1-自封装字段)
-    * [2. 以对象取代数据值](#2-以对象取代数据值)
-    * [3. 将值对象改成引用对象](#3-将值对象改成引用对象)
-    * [4. 将引用对象改为值对象](#4-将引用对象改为值对象)
-    * [5. 以对象取代数组](#5-以对象取代数组)
-    * [6. 赋值被监视数据](#6-赋值被监视数据)
-    * [7. 将单向关联改为双向关联](#7-将单向关联改为双向关联)
-    * [8. 将双向关联改为单向关联](#8-将双向关联改为单向关联)
-    * [9. 以字面常量取代魔法数](#9-以字面常量取代魔法数)
-    * [10. 封装字段](#10-封装字段)
-    * [11. 封装集合](#11-封装集合)
-    * [12. 以数据类取代记录](#12-以数据类取代记录)
-    * [13. 以类取代类型码](#13-以类取代类型码)
-    * [14. 以子类取代类型码](#14-以子类取代类型码)
-    * [15. 以 State/Strategy 取代类型码](#15-以-statestrategy-取代类型码)
-    * [16. 以字段取代子类](#16-以字段取代子类)
-* [八、简化条件表达式](#八简化条件表达式)
-    * [1. 分解条件表达式](#1-分解条件表达式)
-    * [2. 合并条件表达式](#2-合并条件表达式)
-    * [3. 合并重复的条件片段](#3-合并重复的条件片段)
-    * [4. 移除控制标记](#4-移除控制标记)
-    * [5. 以卫语句取代嵌套条件表达式](#5-以卫语句取代嵌套条件表达式)
-    * [6. 以多态取代条件表达式](#6-以多态取代条件表达式)
-    * [7. 引入 Null 对象](#7-引入-null-对象)
-    * [8. 引入断言](#8-引入断言)
-* [九、简化函数调用](#九简化函数调用)
-    * [1. 函数改名](#1-函数改名)
-    * [2. 添加参数](#2-添加参数)
-    * [3. 移除参数](#3-移除参数)
-    * [4. 将查询函数和修改函数分离](#4-将查询函数和修改函数分离)
-    * [5. 令函数携带参数](#5-令函数携带参数)
-    * [6. 以明确函数取代参数](#6-以明确函数取代参数)
-    * [7. 保持对象完整](#7-保持对象完整)
-    * [8. 以函数取代参数](#8-以函数取代参数)
-    * [9. 引入参数对象](#9-引入参数对象)
-    * [10. 移除设值函数](#10-移除设值函数)
-    * [11. 隐藏函数](#11-隐藏函数)
-    * [12. 以工厂函数取代构造函数](#12-以工厂函数取代构造函数)
-    * [13. 封装向下转型](#13-封装向下转型)
-    * [14. 以异常取代错误码](#14-以异常取代错误码)
-    * [15. 以测试取代异常](#15-以测试取代异常)
-* [十、处理概括关系](#十处理概括关系)
-    * [1. 字段上移](#1-字段上移)
-    * [2. 函数上移](#2-函数上移)
-    * [3. 构造函数本体上移](#3-构造函数本体上移)
-    * [4. 函数下移](#4-函数下移)
-    * [5. 字段下移](#5-字段下移)
-    * [6. 提炼子类](#6-提炼子类)
-    * [7. 提炼超类](#7-提炼超类)
-    * [8. 提炼接口](#8-提炼接口)
-    * [9. 折叠继承体系](#9-折叠继承体系)
-    * [10. 塑造模板函数](#10-塑造模板函数)
-    * [11. 以委托取代继承](#11-以委托取代继承)
-    * [12. 以继承取代委托](#12-以继承取代委托)
-* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、第一个案例
-
-如果你发现自己需要为程序添加一个特性，而代码结构使你无法很方便地达成目的，那就先重构这个程序。
-
-在重构前，需要先构建好可靠的测试环境，确保安全地重构。
-
-重构需要以微小的步伐修改程序，如果重构过程发生错误，很容易就能发现错误。
-
-**案例分析** 
-
-影片出租店应用程序，需要计算每位顾客的消费金额。
-
-包括三个类：Movie、Rental 和 Customer。
-
-<div align="center"> <img src="../pics//c2f0c8e2-da66-498c-a38f-e1176abee29e.png"/> </div><br>
-
-最开始的实现是把所有的计费代码都放在 Customer 类中。可以发现，该代码没有使用 Customer 类中的任何信息，更多的是使用 Rental 类的信息，因此第一个可以重构的点就是把具体计费的代码移到 Rental 类中，然后 Customer 类的 getTotalCharge() 方法只需要调用 Rental 类中的计费方法即可。
-
-```java
-class Customer {
-
-    private List<Rental> rentals = new ArrayList<>();
-
-    void addRental(Rental rental) {
-        rentals.add(rental);
-    }
-
-    double getTotalCharge() {
-        double totalCharge = 0.0;
-        for (Rental rental : rentals) {
-            switch (rental.getMovie().getMovieType()) {
-                case Movie.Type1:
-                    totalCharge += rental.getDaysRented();
-                    break;
-                case Movie.Type2:
-                    totalCharge += rental.getDaysRented() * 2;
-                    break;
-                case Movie.Type3:
-                    totalCharge += rental.getDaysRented() * 3;
-                    break;
-            }
-        }
-        return totalCharge;
-    }
-}
-```
-
-```java
-class Rental {
-    private int daysRented;
-
-    private Movie movie;
-
-    Rental(int daysRented, Movie movie) {
-        this.daysRented = daysRented;
-        this.movie = movie;
-    }
-
-    Movie getMovie() {
-        return movie;
-    }
-
-    int getDaysRented() {
-        return daysRented;
-    }
-}
-```
-
-```java
-class Movie {
-
-    static final int Type1 = 0, Type2 = 1, Type3 = 2;
-
-    private int type;
-
-    Movie(int type) {
-        this.type = type;
-    }
-
-    int getMovieType() {
-        return type;
-    }
-}
-```
-
-```java
-public class App {
-    public static void main(String[] args) {
-        Customer customer = new Customer();
-        Rental rental1 = new Rental(1, new Movie(Movie.Type1));
-        Rental rental2 = new Rental(2, new Movie(Movie.Type2));
-        customer.addRental(rental1);
-        customer.addRental(rental2);
-        System.out.println(customer.getTotalCharge());
-    }
-}
-```
-
-```html
-5
-```
-
-使用 switch 的准则是：只使用 switch 所在类的数据。解释如下：switch 使用的数据通常是一组相关的数据，例如 getTotalCharge() 代码使用了 Movie 的多种类别数据。当这组类别的数据发生改变时，例如增加 Movie 的类别或者修改一种 Movie 类别的计费方法，就需要修改 switch 代码。如果违反了准则，就会有多个地方的 switch 使用了这部分的数据，那么这些 swtich 都需要进行修改，这些代码可能遍布在各个地方，修改工作往往会很难进行。上面的实现违反了这一准则，因此需要重构。
-
-以下是继承 Movie 的多态解决方案，这种方案可以解决上述的 switch 问题，因为每种电影类别的计费方式都被放到了对应 Movie 子类中，当变化发生时，只需要去修改对应子类中的代码即可。
-
-<div align="center"> <img src="../pics//41026c79-dfc1-40f7-85ae-062910fd272b.png"/> </div><br>
-
-有一条设计原则指示应该多用组合少用继承，这是因为组合比继承具有更高的灵活性。例如上面的继承方案，一部电影要改变它的计费方式，就要改变它所属的类，但是对象所属的类在编译时期就确定了，无法在运行过程中改变。（运行时多态可以在运行过程中改变一个父类引用指向的子类对象，但是无法改变一个对象所属的类。）
-
-策略模式就是使用组合替代继承的一种解决方案。引入 Price 类，它有多种实现。Movie 组合了一个 Price 对象，并且在运行时可以改变组合的 Price 对象，从而使得它的计费方式发生改变。
-
-<div align="center"> <img src="../pics//8c0b3ae1-1087-46f4-8637-8d46b4ae659c.png"/> </div><br>
-
-重构后整体的类图和时序图如下：
-
-<div align="center"> <img src="../pics//5b910141-08b6-442d-a4bc-a1608458c636.png"/> </div><br>
-
-<div align="center"> <img src="../pics//3ca58a41-8794-49c1-992e-de5d579a50d1.png"/> </div><br>
-
-重构后的代码：
-
-```java
-class Customer {
-    private List<Rental> rentals = new ArrayList<>();
-
-    void addRental(Rental rental) {
-        rentals.add(rental);
-    }
-
-    double getTotalCharge() {
-        double totalCharge = 0.0;
-        for (Rental rental : rentals) {
-            totalCharge += rental.getCharge();
-        }
-        return totalCharge;
-    }
-}
-```
-
-```java
-class Rental {
-    private int daysRented;
-
-    private Movie movie;
-
-    Rental(int daysRented, Movie movie) {
-        this.daysRented = daysRented;
-        this.movie = movie;
-    }
-
-    double getCharge() {
-        return daysRented * movie.getCharge();
-    }
-}
-```
-
-```java
-interface Price {
-    double getCharge();
-}
-```
-
-```java
-class Price1 implements Price {
-    @Override
-    public double getCharge() {
-        return 1;
-    }
-}
-```
-
-```java
-class Price2 implements Price {
-    @Override
-    public double getCharge() {
-        return 2;
-    }
-}
-```
-
-```java
-package imp2;
-
-class Price3 implements Price {
-    @Override
-    public double getCharge() {
-        return 3;
-    }
-}
-```
-
-```java
-class Movie {
-
-    private Price price;
-
-    Movie(Price price) {
-        this.price = price;
-    }
-
-    double getCharge() {
-        return price.getCharge();
-    }
-}
-```
-
-```java
-class App {
-
-    public static void main(String[] args) {
-        Customer customer = new Customer();
-        Rental rental1 = new Rental(1, new Movie(new Price1()));
-        Rental rental2 = new Rental(2, new Movie(new Price2()));
-        customer.addRental(rental1);
-        customer.addRental(rental2);
-        System.out.println(customer.getTotalCharge());
-    }
-}
-```
-
-# 二、重构原则
-
-## 定义
-
-重构是对软件内部结构的一种调整，目的是在不改变软件可观察行为的前提下，提高其可理解性，降低其修改成本。
-
-## 为何重构
-
-- 改进软件设计
-- 使软件更容易理解
-- 帮助找到 Bug
-- 提高编程速度
-
-## 三次法则
-
-第一次做某件事时只管去做；第二次做类似事情时可以去做；第三次再做类似的事，就应该重构。
-
-## 间接层与重构
-
-计算机科学中的很多问题可以通过增加一个间接层来解决，间接层具有以下价值：
-
-- 允许逻辑共享
-- 分开解释意图和实现
-- 隔离变化
-- 封装条件逻辑
-
-重构可以理解为在适当的位置插入间接层以及在不需要时移除间接层。
-
-## 修改接口
-
-如果重构手法改变了已发布的接口，就必须维护新旧两个接口。可以保留旧接口，让旧接口去调用新接口，并且使用 Java 提供的 @deprecation 将旧接口标记为弃用。
-
-可见修改接口特别麻烦，因此除非真有必要，否则不要发布接口，并且不要过早发布接口。
-
-## 何时不该重构
-
-当现有代码过于混乱时，应当重写而不是重构。
-
-一个折中的办法是，将代码封装成一个个组件，然后对各个组件做重写或者重构的决定。
-
-## 重构与设计
-
-软件开发无法预先设计，因为开发过程有很多变化发生，在最开始不可能都把所有情况考虑进去。
-
-重构可以简化设计，重构在一个简单的设计上进行修修改改，当变化发生时，以一种灵活的方式去应对变化，进而带来更好的设计。
-
-## 重构与性能
-
-为了软代码更容易理解，重构可能会导致性能减低。
-
-在编写代码时，不用对性能过多关注，只有在最后性能优化阶段再考虑性能问题。
-
-应当只关注关键代码的性能，并且只有一小部分的代码是关键代码。
-
-# 三、代码的坏味道
-
-本章主要介绍一些不好的代码，也就是说这些代码应该被重构。
-
-## 1. 重复代码
-
-> Duplicated Code
-
-同一个类的两个函数有相同表达式，则用 Extract Method 提取出重复代码；
-
-两个互为兄弟的子类含有相同的表达式，先使用 Extract Method，然后把提取出来的函数 Pull Up Method 推入超类。
-
-如果只是部分相同，用 Extract Method 分离出相似部分和差异部分，然后使用 Form Template Method 这种模板方法设计模式。
-
-如果两个毫不相关的类出现重复代码，则使用 Extract Class 方法将重复代码提取到一个独立类中。
-
-## 2. 过长函数
-
-> Long Method
-
-函数应该尽可能小，因为小函数具有解释能力、共享能力、选择能力。
-
-分解长函数的原则：当需要用注释来说明一段代码时，就需要把这部分代码写入一个独立的函数中。
-
-Extract Method 会把很多参数和临时变量都当做参数，可以用 Replace Temp with Query 消除临时变量，Introduce Parameter Object 和 Preserve Whole Object 可以将过长的参数列变得更简洁。
-
-条件和循环语句往往也需要提取到新的函数中。
-
-## 3. 过大的类
-
-> Large Class
-
-应该尽可能让一个类只做一件事，而过大的类做了过多事情，需要使用 Extract Class 或 Extract Subclass。
-
-先确定客户端如何使用该类，然后运用 Extract Interface 为每一种使用方式提取出一个接口。
-
-## 4. 过长的参数列表
-
-> Long Parameter List
-
-太长的参数列表往往会造成前后不一致，不易使用。
-
-面向对象程序中，函数所需要的数据通常能在宿主类中找到。
-
-## 5. 发散式变化
-
-> Divergent Change
-
-设计原则：一个类应该只有一个引起改变的原因。也就是说，针对某一外界变化所有相应的修改，都只应该发生在单一类中。
-
-针对某种原因的变化，使用 Extract Class 将它提炼到一个类中。
-
-## 6. 散弹式修改
-
-> Shotgun Surgery
-
-一个变化引起多个类修改。
-
-使用 Move Method 和 Move Field 把所有需要修改的代码放到同一个类中。
-
-## 7. 依恋情结
-
-> Feature Envy
-
-一个函数对某个类的兴趣高于对自己所处类的兴趣，通常是过多访问其它类的数据，
-
-使用 Move Method 将它移到该去的地方，如果对多个类都有 Feature Envy，先用 Extract Method 提取出多个函数。
-
-## 8. 数据泥团
-
-> Data Clumps
-
-有些数据经常一起出现，比如两个类具有相同的字段、许多函数有相同的参数，这些绑定在一起出现的数据应该拥有属于它们自己的对象。
-
-使用 Extract Class 将它们放在一起。
-
-## 9. 基本类型偏执
-
-> Primitive Obsession
-
-使用类往往比使用基本类型更好，使用 Replace Data Value with Object 将数据值替换为对象。
-
-## 10. switch 惊悚现身
-
-> Switch Statements
-
-具体参见第一章的案例。
-
-## 11. 平行继承体系
-
-> Parallel Inheritance Hierarchies
-
-每当为某个类增加一个子类，必须也为另一个类相应增加一个子类。
-
-这种结果会带来一些重复性，消除重复性的一般策略：让一个继承体系的实例引用另一个继承体系的实例。
-
-## 12. 冗余类
-
-> Lazy Class
-
-如果一个类没有做足够多的工作，就应该消失。
-
-## 13. 夸夸其谈未来性
-
-> Speculative Generality
-
-有些内容是用来处理未来可能发生的变化，但是往往会造成系统难以理解和维护，并且预测未来可能发生的改变很可能和最开始的设想相反。因此，如果不是必要，就不要这么做。
-
-## 14. 令人迷惑的暂时字段
-
-> Temporary Field
-
-某个字段仅为某种特定情况而设，这样的代码不易理解，因为通常认为对象在所有时候都需要它的所有字段。
-
-把这种字段和特定情况的处理操作使用 Extract Class 提炼到一个独立类中。
-
-## 15. 过度耦合的消息链
-
-> Message Chains
-
-一个对象请求另一个对象，然后再向后者请求另一个对象，然后...，这就是消息链。采用这种方式，意味着客户代码将与对象间的关系紧密耦合。
-
-改用函数链，用函数委托另一个对象来处理。
-
-## 16. 中间人
-
->  Middle Man
-
-中间人负责处理委托给它的操作，如果一个类中有过多的函数都委托给其它类，那就是过度运用委托，应当 Remove Middle Man，直接与负责的对象打交道。
-
-## 17. 狎昵关系
-
-> Inappropriate Intimacy
-
-两个类多于亲密，花费太多时间去探讨彼此的 private 成分。
-
-## 18. 异曲同工的类
-
-> Alernative Classes with Different Interfaces
-
-两个函数做同一件事，却有着不同的签名。
-
-使用 Rename Method 根据它们的用途重新命名。
-
-## 19. 不完美的类库
-
-> Incomplete Library Class
-
-类库的设计者不可能设计出完美的类库，当我们需要对类库进行一些修改时，可以使用以下两种方法：如果只是修改一两个函数，使用 Introduce Foreign Method；如果要添加一大堆额外行为，使用 Introduce Local Extension。
-
-## 20. 幼稚的数据类
-
-> Data Class
-
-它只拥有一些数据字段，以及用于访问这些字段的函数，除此之外一无长物。
-
-找出字段使用的地方，然后把相应的操作移到 Data Class 中。
-
-## 21. 被拒绝的馈赠
-
-> Refused Bequest
-
-子类不想继承超类的所有函数和数据。
-
-为子类新建一个兄弟类，不需要的函数或数据使用 Push Down Method 和 Push Down Field 下推给那个兄弟。
-
-## 22. 过多的注释
-
-> Comments
-
-使用 Extract Method 提炼出需要注释的部分，然后用函数名来解释函数的行为。
-
-# 四、构筑测试体系
-
-Java 可以使用 Junit 进行单元测试。
-
-测试应该能够完全自动化，并能检查测试的结果。
-
-小步修改，频繁测试。
-
-单元测试的对象是类的方法，而功能测是以客户的角度保证软件正常运行。
-
-应当集中测试可能出错的边界条件。
-
-# 五、重新组织函数
-
-## 1. 提炼函数
-
-> Extract Method
-
-将这段代码放进一个独立函数中，并让函数名称解释该函数的用途。
-
-## 2. 内联函数
-
-> Inline Method
-
-一个函数的本体与名称同样清楚易懂。
-
-在函数调用点插入函数本体，然后移除该函数。
-
-## 3. 内联临时变量
-
-> Inline Temp
-
-一个临时变量，只被简单表达式赋值一次，而它妨碍了其它重构手法。
-
-将所有对该变量的引用替换为对它赋值的那个表达式自身。
-
-```java
-double basePrice = anOrder.basePrice();
-return basePrice > 1000;
-```
-
-```java
-return anOrder.basePrice() > 1000;
-```
-
-## 4. 以查询取代临时变量
-
-> Replace Temp with Query
-
-以临时变量保存某一表达式的运算结果，将这个表达式提炼到一个独立函数中，将所有对临时变量的引用点替换为对新函数的调用。
-
-Replace Temp with Query 往往是 Extract Method 之前必不可少的一个步骤，因为局部变量会使代码难以提炼。
-
-```java
-double basePrice = quantity * itemPrice;
-if (basePrice > 1000)
-    return basePrice * 0.95;
-else
-    return basePrice * 0.98;
-```
-
-```java
-if (basePrice() > 1000)
-    return basePrice() * 0.95;
-else
-    return basePrice() * 0.98;
-
-// ...
-double basePrice(){
-    return quantity * itemPrice;
-}
-```
-
-## 5. 引起解释变量
-
-> Introduce Explaining Variable
-
-将复杂表达式（或其中一部分）的结果放进一个临时变量， 以此变量名称来解释表达式用途。
-
-```java
-if ((platform.toUpperCase().indexOf("MAC") > -1) &&
-  (browser.toUpperCase().indexOf("IE") > -1) &&
-  wasInitialized() && resize > 0) {
-    // do something
-}
-```
-
-```java
-final boolean isMacOS = platform.toUpperCase().indexOf("MAC") > -1;
-final boolean isIEBrower = browser.toUpperCase().indexOf("IE") > -1;
-final boolean wasResized = resize > 0;
-
-if (isMacOS && isIEBrower && wasInitialized() && wasResized) {
-    // do something
-}
-```
-
-## 6. 分解临时变量
-
-> Split Temporary Variable
-
-某个临时变量被赋值超过一次，它既不是循环变量，也不是用于收集计算结果。
-
-针对每次赋值，创造一个独立、对应的临时变量，每个临时变量只承担一个责任。
-
-## 7. 移除对参数的赋值
-
->  Remove Assigments to Parameters
-
-以一个临时变量取代对该参数的赋值。
-
-```java
-int discount (int inputVal, int quentity, int yearToDate) {
-    if (inputVal > 50) inputVal -= 2;
-    ...
-}
-```
-
-```java
-int discount (int inputVal, int quentity, int yearToDate) {
-    int result = inputVal;
-    if (inputVal > 50) result -= 2;
-    ...
-}
-```
-
-## 8. 以函数对象取代函数
-
-> Replace Method with Method Object
-
-当对一个大型函数采用 Extract Method 时，由于包含了局部变量使得很难进行该操作。
-
-将这个函数放进一个单独对象中，如此一来局部变量就成了对象内的字段。然后可以在同一个对象中将这个大型函数分解为多个小型函数。
-
-## 9. 替换算法
-
-> Subsititute Algorithn
-
-# 六、在对象之间搬移特性
-
-## 1. 搬移函数
-
-> Move Method
-
-类中的某个函数与另一个类进行更多交流：调用后者或者被后者调用。
-
-将这个函数搬移到另一个类中。
-
-## 2. 搬移字段
-
-> Move Field
-
-类中的某个字段被另一个类更多地用到，这里的用到是指调用取值设值函数，应当把该字段移到另一个类中。
-
-## 3. 提炼类
-
-> Extract Class
-
-某个类做了应当由两个类做的事。
-
-应当建立一个新类，将相关的字段和函数从旧类搬移到新类。
-
-## 4. 将类内联化
-
-> Inline Class
-
-与 Extract Class 相反。
-
-## 5. 隐藏委托关系
-
-> Hide Delegate
-
-建立所需的函数，隐藏委托关系。
-
-```java
-class Person {
-    Department department;
-
-    public Department getDepartment() {
-        return department;
-    }
-}
-
-class Department {
-    private Person manager;
-
-    public Person getManager() {
-        return manager;
-    }
-}
-```
-
-如果客户希望知道某人的经理是谁，必须获得 Department 对象，这样就对客户揭露了 Department 的工作原理。
-
-```java
-Person manager = john.getDepartment().getManager();
-```
-
-通过为 Peron 建立一个函数来隐藏这种委托关系。
-
-```java
-public Person getManager() {
-    return department.getManager();
-}
-```
-
-## 6. 移除中间人
-
-> Remove Middle Man
-
-与 Hide Delegate 相反，本方法需要移除委托函数，让客户直接调用委托类。
-
-Hide Delegate 有很大好处，但是它的代价是：每当客户要使用受托类的新特性时，就必须在服务器端添加一个简单的委托函数。随着受委托的特性越来越多，服务器类完全变成了一个“中间人”。
-
-## 7. 引入外加函数
-
-> Introduce Foreign Method
-
-需要为提供服务的类添加一个函数，但是无法修改这个类。
-
-可以在客户类中建立一个函数，并以第一参数形式传入一个服务类的实例，让客户类组合服务器实例。
-
-## 8. 引入本地扩展
-
-> Introduce Local Extension
-
-和 Introduce Foreign Method 目的一样，但是 Introduce Local Extension 通过建立新的类来实现。有两种方式：子类或者包装类，子类就是通过继承实现，包装类就是通过组合实现。
-
-# 七、重新组织数据
-
-## 1. 自封装字段
-
-> Self Encapsulate Field
-
-为字段建立取值/设值函数，并用这些函数来访问字段。只有当子类想访问超类的一个字段，又想在子类中将对这个字段访问改为一个计算后的值，才使用这种方式，否则直接访问字段的方式简洁明了。
-
-## 2. 以对象取代数据值
-
-> Replace Data Value with Object
-
-在开发初期，往往会用简单的数据项表示简单的情况，但是随着开发的进行，一些简单数据项会具有一些特殊行为。比如一开始会把电话号码存成字符串，但是随后发现电话号码需要“格式化”、“抽取区号”之类的特殊行为。
-
-## 3. 将值对象改成引用对象
-
-> Change Value to Reference
-
-将彼此相等的实例替换为同一个对象。这就要用一个工厂来创建这种唯一对象，工厂类中需要保留一份已经创建对象的列表，当要创建一个对象时，先查找这份列表中是否已经存在该对象，如果存在，则返回列表中的这个对象；否则，新建一个对象，添加到列表中，并返回该对象。
-
-## 4. 将引用对象改为值对象
-
-> Change Reference to value
-
-以 Change Value to Reference 相反。值对象有个非常重要的特性：它是不可变的，不可变表示如果要改变这个对象，必须用一个新的对象来替换旧对象，而不是修改旧对象。
-
-需要为值对象实现 equals() 和 hashCode() 方法。
-
-## 5. 以对象取代数组
-
-> Replace Array with Object
-
-有一个数组，其中的元素各自代表不同的东西。
-
-以对象替换数组，对于数组中的每个元素，以一个字段来表示，这样方便操作，也更容易理解。
-
-## 6. 赋值被监视数据
-
-> Duplicate Observed Data
-
-一些领域数据置身于 GUI 控件中，而领域函数需要访问这些数据。
-
-将该数据赋值到一个领域对象中，建立一个 Oberver 模式，用于同步领域对象和 GUI 对象内的重复数据。
-
-<div align="center"> <img src="../pics//e024bd7e-fb4e-4239-9451-9a6227f50b00.jpg" width=""/> </div><br>
-
-## 7. 将单向关联改为双向关联
-
-> Change Unidirectional Association to Bidirectional
-
-当两个类都需要对方的特性时，可以使用双向关联。
-
-有两个类，分别为订单 Order 和客户 Customer，Order 引用了 Customer，Customer 也需要引用 Order 来查看其所有订单详情。
-
-```java
-class Order {
-    private Customer customer;
-    public void setCustomer(Customer customer) {
-        if (this.customer != null)
-            this.customer.removeOrder(this);
-        this.customer = customer;
-        this.customer.add(this);
-    }
-}
-```
-```java
-class Curstomer {
-    private Set<Order> orders = new HashSet<>();
-    public void removeOrder(Order order) {
-        orders.remove(order);
-    }
-    public void addOrder(Order order) {
-        orders.add(order);
-    }
-}
-```
-
-注意到，这里让 Curstomer 类来控制关联关系。有以下原则来决定哪个类来控制关联关系：如果某个对象是组成另一个对象的部件，那么由后者负责控制关联关系；如果是一对多关系，则由单一引用那一方来控制关联关系。
-
-## 8. 将双向关联改为单向关联
-
-> Change Bidirectional Association to Unidirectional
-
-和 Change Unidirectional Association to Bidirectiona 为反操作。
-
-双向关联维护成本高，并且也不易于理解。大量的双向连接很容易造成“僵尸对象”：某个对象本身已经死亡了，却保留在系统中，因为它的引用还没有全部完全清除。
-
-## 9. 以字面常量取代魔法数
-
-> Replace Magic Number with Symbolic Constant
-
-创建一个常量，根据其意义为它命名，并将字面常量换为这个常量。
-
-## 10. 封装字段
-
-> Encapsulate Field
-
-public 字段应当改为 private，并提供相应的访问函数。
-
-## 11. 封装集合
-
-> Encapsulate Collection
-
-函数返回集合的一个只读副本，并在这个类中提供添加/移除集合元素的函数。如果函数返回集合自身，会让用户得以修改集合内容而集合拥有者却一无所知。
-
-## 12. 以数据类取代记录
-
-> Replace Record with Data Class
-
-## 13. 以类取代类型码
-
-> Replace Type Code with Class
-
-类中有一个数值类型码，但它并不影响类的行为，就用一个新类替换该数值类型码。如果类型码出现在 switch 语句中，需要使用 Replace Conditional with Polymorphism 去掉 switch，首先必须运用 Replace Type Code with Subcalss 或 Replace Type Code with State/Strategy 去掉类型码。
-
-<div align="center"> <img src="../pics//27c2e0b3-8f95-453d-bedc-6398a8566ce9.jpg" width="500"/> </div><br>
-
-## 14. 以子类取代类型码
-
-> Replace Type Code with Subcalsses
-
-有一个不可变的类型码，它会影响类的行为，以子类取代这个类型码。
-
-<div align="center"> <img src="../pics//c41d3977-e0e7-4ee4-93e1-d84f1ae3e20e.jpg" width="500"/> </div><br>
-
-## 15. 以 State/Strategy 取代类型码
-
-> Replace Type Code with State/Strategy
-
-有一个可变的类型码，它会影响类的行为，以状态对象取代类型码。
-
-和 Replace Type Code with Subcalsses 的区别是 Replace Type Code with State/Strategy 的类型码是动态可变的，前者通过继承的方式来实现，后者通过组合的方式来实现。因为类型码可变，如果通过继承的方式，一旦一个对象的类型码改变，那么就要改变用新的对象来取代旧对象，而客户端难以改变新的对象。但是通过组合的方式，改变引用的状态类是很容易的。
-
-<div align="center"> <img src="../pics//81fd1d6f-a3b2-4160-9a0a-1f7cb50ba440.jpg" width="800"/> </div><br>
-
-## 16. 以字段取代子类
-
-> Replace Subclass with Fields
-
-各个子类的唯一差别只在“返回常量数据”的函数上。
-
-<div align="center"> <img src="../pics//f2e0cee9-ecdc-4a96-853f-d9f6a1ad6ad1.jpg" width="600"/> </div><br>
-
-# 八、简化条件表达式
-
-## 1. 分解条件表达式
-
-> Decompose Conditional
-
-对于一个复杂的条件语句，可以从 if、then、else 三个段落中分别提炼出独立函数。
-
-```java
-if (data.befor(SUMMER_START) || data.after(SUMMER_END))
-    charge = quantity * winterRate + winterServiceCharge;
-else
-    charge = quantity * summerRate;
-```
-
-```java
-if (notSummer(date))
-    charge = winterCharge(quantity);
-else
-    charge = summerCharge(quantity);
-```
-
-## 2. 合并条件表达式
-
-> Consolidate Conditional Expression
-
-有一系列条件测试，都得到相同结果。
-
-将这些测试合并为一个条件表达式，并将这个条件表达式提炼成为一个独立函数。
-
-```java
-double disabilityAmount() {
-    if (seniority < 2) return 0;
-    if (monthsDisabled > 12 ) return 0;
-    if (isPartTime) return 0;
-    // ...
-}
-```
-```java
-double disabilityAmount() {
-    if (isNotEligibleForDisability()) return 0;
-    // ...
-}
-```
-
-## 3. 合并重复的条件片段
-
-> Consolidate Duplicate Conditional Fragments
-
-在条件表达式的每个分支上有着相同的一段代码。
-
-将这段重复代码搬移到条件表达式之外。
-
-```java
-if (isSpecialDeal()) {
-    total = price * 0.95;
-    send();
-} else {
-    total = price * 0.98;
-    send();
-}
-```
-
-```java
-if (isSpecialDeal()) {
-    total = price * 0.95;
-} else {
-    total = price * 0.98;
-}
-send();
-```
-
-## 4. 移除控制标记
-
-> Remove Control Flag
-
-在一系列布尔表达式中，某个变量带有“控制标记”的作用。
-
-用 break 语句或 return 语句来取代控制标记。
-
-## 5. 以卫语句取代嵌套条件表达式
-
-> Replace Nested Conditional with Guard Clauses
-
-如果某个条件极其罕见，就应该单独检查该条件，并在该条件为真时立刻从函数中返回，这样的单独检查常常被称为“卫语句”（guard clauses）。
-
-条件表达式通常有两种表现形式。第一种形式是：所有分支都属于正常行为。第二种形式则是：条件表达式提供的答案中只有一种是正常行为，其他都是不常见的情况，可以使用卫语句表现所有特殊情况。
-
-```java
-double getPayAmount() {
-    double result;
-    if (isDead) result = deadAmount();
-    else {
-        if (isSeparated) result = separatedAmount();
-        else {
-            if (isRetired) result = retiredAmount();
-            else result = normalPayAmount();
-        };
-    }
-    return result;
-};
-```
-
-```java
-double getPayAmount() {
-    if (isDead) return deadAmount();
-    if (isSeparated) return separatedAmount();
-    if (isRetired) return retiredAmount();
-    return normalPayAmount();
-};
-```
-
-## 6. 以多态取代条件表达式
-
-> Replace Conditional with Polymorphism
-
-将这个条件表达式的每个分支放进一个子类内的覆写函数中，然后将原始函数声明为抽象函数。需要先使用 Replace Type Code with Subclass 或 Replace Type Code with State/Strategy 来建立继承结果。
-
-```java
-double getSpeed() {
-    switch (type) {
-        case EUROPEAN:
-            return getBaseSpeed();
-        case AFRICAN:
-            return getBaseSpeed()- getLoadFactor()* numberOfCoconuts;
-        case NORWEGIAN_BLUE:
-            return isNailed ? 0 : getBaseSpeed(voltage);
-    }
-    throw new RuntimeException("Should be unreachable");
-}
-```
-
-<div align="center"> <img src="../pics//1c8432c8-2552-457f-b117-1da36c697221.jpg" width="500"/> </div><br>
-
-
-## 7. 引入 Null 对象
-
-> Introduce Null Object
-
-将 null 值替换为 null 对象。这样做的好处在于，不需要询问对象是否为空，直接调用就行。
-
-```java
-if (customer == null) plan = BillingPlan.basic();
-else plan = customer.getPlan();
-```
-
-## 8. 引入断言
-
-> Introduce Assertion
-
-以断言明确表现某种假设。断言只能用于开发过程中，产品代码中不会有断言。
-
-```java
-double getExpenseLimit() {
-    // should have either expense limit or a primary project
-    return (expenseLimit != NULL_EXPENSE) ? expenseLimit :  primaryProject.getMemberExpenseLimit();
-}
-```
-
-```java
-double getExpenseLimit() {
-    Assert.isTrue (expenseLimit != NULL_EXPENSE || primaryProject != null);
-    return (expenseLimit != NULL_EXPENSE) ? expenseLimit :  primaryProject.getMemberExpenseLimit();
-}
-```
-
-# 九、简化函数调用
-
-## 1. 函数改名
-
-> Rename Method
-
-使函数名能解释函数的用途。
-
-## 2. 添加参数
-
-> Add Parameter
-
-使函数不需要通过调用获得某个信息。
-
-## 3. 移除参数
-
-> Remove Parameter
-
-与 Add Parameter 相反，改用调用的方式来获得某个信息。
-
-## 4. 将查询函数和修改函数分离
-
-> Separate Query from Modifier
-
-某个函数即返回对象状态值，又修改对象状态。
-
-应当建立两个不同的函数，其中一个负责查询，另一个负责修改。任何有返回值的函数，都不应该有看得到的副作用。
-
-```java
-getTotalOutstandingAndSetReadyForSummaries();
-```
-
-```java
-getTotalOutstanding();
-setReadyForSummaries();
-```
-
-## 5. 令函数携带参数
-
-> Parameterize Method
-
-若干函数做了类似的工作，但在函数本体中却包含了不同的值。
-
-建立单一函数，以参数表达那些不同的值。
-
-```java
-fivePercentRaise();
-tenPercentRaise();
-```
-```java
-raise(percentage);
-```
-
-## 6. 以明确函数取代参数
-
-> Replace Parameter with Explicit Methods
-
-有一个函数，完全取决于参数值而采取不同行为。
-
-针对该参数的每一个可能值，建立一个独立函数。
-
-```java
-void setValue(String name, int value){
-    if (name.equals("height")){
-        height = value;
-        return;
-    }
-    if (name.equals("width")){
-        width = value;
-        return;
-    }
-    Assert.shouldNeverReachHere();
-}
-```
-
-```java
-void setHeight(int arg){
-    height = arg;
-}
-void setWidth(int arg){
-    width = arg;
-}
-```
-
-## 7. 保持对象完整
-
-> Preserve Whole Object
-
-从某个对象中取出若干值，将它们作为某一次函数调用时的参数。
-
-改为传递整个对象。
-
-```java
-int low = daysTempRange().getLow();
-int high = daysTempRange().getHigh();
-withinPlan = plan.withinRange(low, high);
-```
-
-```java
-withinPlan = plan.withinRange(daysTempRange());
-```
-
-## 8. 以函数取代参数
-
-> Replace Parameter with Methods
-
-对象调用某个函数，并将所得结果作为参数，传递给另一个函数。而接受该参数的函数本身也能够调用前一个函数。
-
-让参数接收者去除该项参数，而是直接调用前一个函数。
-
-```java
-int basePrice = _quantity * _itemPrice;
-discountLevel = getDiscountLevel();
-double finalPrice = discountedPrice (basePrice, discountLevel);
-```
-
-```java
-int basePrice = _quantity * _itemPrice;
-double finalPrice = discountedPrice (basePrice);
-```
-
-## 9. 引入参数对象
-
-> Introduce Parameter Object
-
-某些参数总是很自然地同时出现，这些参数就是 Data Clumps。
-
-以一个对象取代这些参数。
-
-<div align="center"> <img src="../pics//08738dd0-ae8e-404a-ba78-a6b1b7d225b3.jpg" width="600"/> </div><br>
-
-## 10. 移除设值函数
-
-> Remove Setting Method
-
-类中的某个字段应该在对象创建时被设值，然后就不再改变。
-
-去掉该字段的所有设值函数，并将该字段设为 final。
-
-## 11. 隐藏函数
-
-> Hide Method
-
-有一个函数，从来没有被其他任何类用到。
-
-将这个函数修改为 private。
-
-## 12. 以工厂函数取代构造函数
-
-> Replace Constructor with Factory Method
-
-希望在创建对象时不仅仅是做简单的建构动作。
-
-将构造函数替换为工厂函数。
-
-## 13. 封装向下转型
-
-> Encapsulate Downcast
-
-某个函数返回的对象，需要由函数调用者执行向下转型（downcast）。
-
-将向下转型动作移到函数中。
-
-```java
-Object lastReading() {
-    return readings.lastElement();
-}
-```
-```java
-Reading lastReading() {
-    return (Reading)readings.lastElement();
-}
-```
-
-## 14. 以异常取代错误码
-
-> Replace Error Code with Exception
-
-某个函数返回一个特定的代码，用以表示某种错误情况。
-
-改用异常，异常将普通程序和错误处理分开，使代码更容易理解。
-
-## 15. 以测试取代异常
-
-> Replace Exception with Test
-
-面对一个调用者可以预先检查的条件，你抛出了一个异常。
-
-修改调用者，使它在调用函数之前先做检查。
-
-```java
-double getValueForPeriod(int periodNumber) {
-    try {
-        return values[periodNumber];
-    } catch (ArrayIndexOutOfBoundsException e) {
-        return 0;
-    }
-}
-```
-```java
-double getValueForPeriod(int periodNumber) {
-    if (periodNumber >= values.length) return 0;
-    return values[periodNumber];
-```
-
-# 十、处理概括关系
-
-## 1. 字段上移
-
-> Pull Up Field
-
-两个子类拥有相同的字段。
-
-将该字段移至超类。
-
-## 2. 函数上移
-
-> Pull Up Method
-
-有些函数，在各个子类中产生完全相同的结果。
-
-将该函数移至超类。
-
-## 3. 构造函数本体上移
-
-> Pull Up Constructor Body
-
-你在各个子类中拥有一些构造函数，它们的本体几乎完全一致。
-
-在超类中新建一个构造函数，并在子类构造函数中调用它。
-
-```java
-class Manager extends Employee...
-
-public Manager(String name, String id, int grade) {
-    this.name = name;
-    this.id = id;
-    this.grade = grade;
-}
-```
-
-```java
-public Manager(String name, String id, int grade) {
-    super(name, id);
-    this.grade = grade;
-}
-```
-
-## 4. 函数下移
-
-> Push Down Method
-
-超类中的某个函数只与部分子类有关。
-
-将这个函数移到相关的那些子类去。
-
-## 5. 字段下移
-
-> Push Down Field
-
-超类中的某个字段只被部分子类用到。
-
-将这个字段移到需要它的那些子类去。
-
-## 6. 提炼子类
-
-> Extract Subclass
-
-类中的某些特性只被某些实例用到。
-
-新建一个子类，将上面所说的那一部分特性移到子类中。
-
-## 7. 提炼超类
-
-> Extract Superclass
-
-两个类有相似特性。
-
-为这两个类建立一个超类，将相同特性移至超类。
-
-## 8. 提炼接口
-
-> Extract Interface
-
-若干客户使用类接口中的同一子集，或者两个类的接口有部分相同。
-
-将相同的子集提炼到一个独立接口中。
-
-## 9. 折叠继承体系
-
-> Collapse Hierarchy
-
-超类和子类之间无太大区别。
-
-将它们合为一体。
-
-## 10. 塑造模板函数
-
-> Form Template Method
-
-你有一些子类，其中相应的某些函数以相同顺序执行类似的操作，但各个操作的细节上有所不同。
-
-将这些操作分别放进独立函数中，并保持它们都有相同的签名，于是原函数也就变得相同了。然后将原函数上移至超类。(模板方法模式)
-
-## 11. 以委托取代继承
-
-> Replace Inheritance with Delegation
-
-某个子类只使用超类接口中的一部分，或是根本不需要继承而来的数据。
-
-在子类中新建一个字段用以保存超类，调整子类函数，令它改而委托超类，然后去掉两者之间的继承关系。
-
-## 12. 以继承取代委托
-
-> Replace Delegation with Inheritance
-
-你在两个类之间使用委托关系，并经常为整个接口编写许多极简单的委托函数。
-
-让委托类继承受托类。
-
-# 参考资料
-
-- MartinFowler, 福勒, 贝克, 等. 重构: 改善既有代码的设计 [M]. 电子工业出版社, 2011.
diff --git a/notes/集群.md b/notes/集群.md
index b36a0bfe..649f563a 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -1,194 +1,4 @@
 <!-- GFM-TOC -->
-* [一、负载均衡](#一负载均衡)
-    * [算法实现](#算法实现)
-    * [转发实现](#转发实现)
-* [二、集群下的 Session 管理](#二集群下的-session-管理)
-    * [Sticky Sessions](#sticky-sessions)
-    * [Session Replication](#session-replication)
-    * [Session Server](#session-server)
 <!-- GFM-TOC -->
 
 
-# 一、负载均衡
-
-集群中的应用服务器通常被设计成无状态，用户可以请求任何一个节点（应用服务器）。
-
-负载均衡器会根据集群中每个节点的负载情况，将用户请求转发到合适的节点上。
-
-负载均衡器可以用来实现高可用以及伸缩性：
-
-- 高可用：当某个节点故障时，负载均衡器不会将用户请求转发到该节点上，从而保证所有服务持续可用；
-- 伸缩性：可以很容易地添加移除节点。
-
-负载均衡运行过程包含两个部分：
-
-1. 根据负载均衡算法得到应该将请求转发到哪个节点上；
-2. 将请求进行转发；
-
-## 算法实现
-
-### 1. 轮询（Round Robin）
-
-轮询算法把每个请求轮流发送到每个服务器上。下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
-
-<div align="center"> <img src="../pics//2766d04f-7dad-42e4-99d1-60682c9d5c61.jpg"/> </div><br>
-
-该算法比较适合每个服务器的性能差不多的场景，如果有性能存在差异的情况下，那么性能较差的服务器可能无法承担过大的负载（下图的 Server 2）。
-
-<div align="center"> <img src="../pics//f7ecbb8d-bb8b-4d45-a3b7-f49425d6d83d.jpg"/> </div><br>
-
-### 2. 加权轮询（Weighted Round Robbin）
-
-加权轮询是在轮询的基础上，根据服务器的性能差异，为服务器赋予一定的权值，性能高的服务器分配更高的权值。例如下图中，服务器 1 被赋予的权值为 5，服务器 2 被赋予的权值为 1，那么 (1, 2, 3, 4, 5) 请求会被发送到服务器 1，(6) 请求会被发送到服务器 2。
-
-<div align="center"> <img src="../pics//211c60d4-75ca-4acd-8a4f-171458ed58b4.jpg"/> </div><br>
-
-### 3. 最少连接（least Connections）
-
-由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
-
-<div align="center"> <img src="../pics//3b0d1aa8-d0e0-46c2-8fd1-736bf08a11aa.jpg"/> </div><br>
-
-最少连接算法就是将请求发送给当前最少连接数的服务器上。例如下图中，服务器 1 当前连接数最小，那么新到来的请求 6 就会被发送到服务器 1 上。
-
-<div align="center"> <img src="../pics//1f4a7f10-52b2-4bd7-a67d-a9581d66dc62.jpg"/> </div><br>
-
-### 4. 加权最少连接（Weighted Least Connection）
-
-在最少连接的基础上，根据服务器的性能为每台服务器分配权重，再根据权重计算出每台服务器能处理的连接数。
-
-<div align="center"> <img src="../pics//44edefb7-4b58-4519-b8ee-4aca01697b78.jpg"/> </div><br>
-
-### 5. 随机算法（Random）
-
-把请求随机发送到服务器上。和轮询算法类似，该算法比较适合服务器性能差不多的场景。
-
-<div align="center"> <img src="../pics//0ee0f61b-c782-441e-bf34-665650198ae0.jpg"/> </div><br>
-
-### 6. 源地址哈希法 (IP Hash)
-
-源地址哈希通过对客户端 IP 计算哈希值之后，再对服务器数量进行取模运算得到目标服务器的序号。可以保证同一 IP 的客户端的请求会转发到同一台服务器上，可以用来实现会话粘滞（Sticky Session）
-
-<div align="center"> <img src="../pics//2018040302.jpg"/> </div><br>
-
-## 转发实现
-
-### 1. HTTP 重定向
-
-HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服务器的 IP 地址之后，将该地址写入 HTTP 重定向报文中，状态码为 302。客户端收到重定向报文之后，需要重新向服务器发起请求。
-
-缺点：
-
-- 需要两次请求，因此访问延迟比较高；
-- HTTP 负载均衡器处理能力有限，会限制集群的规模。
-
-该负载均衡转发的缺点比较明显，实际场景中很少使用它。
-
-<div align="center"> <img src="../pics//c5f611f0-fd5c-4158-9003-278141136e6e.jpg"/> </div><br>
-
-### 2. DNS 域名解析
-
-在 DNS 解析域名的同时使用负载均衡算法计算服务器地址。
-
-优点：
-
-- DNS 能够根据地理位置进行域名解析，返回离用户最近的服务器 IP 地址。
-
-缺点：
-
-- 由于 DNS 具有多级结构，每一级的域名记录都可能被缓存，当下线一台服务器需要修改 DNS 记录时，需要过很长一段时间才能生效；
-
-大型网站基本使用了 DNS 做为第一级负载均衡手段，然后在内部使用其它方式做第二级负载均衡。也就是说，域名解析的结果为内部的负载均衡服务器 IP 地址。
-
-<div align="center"> <img src="../pics//76a25fc8-a579-4d7c-974b-7640b57fbf39.jpg"/> </div><br>
-
-### 3. 反向代理服务器
-
-首先了解一下正向代理与反向代理的区别：
-
-- 正向代理：发生在客户端，是由用户主动发起的。比如翻墙，客户端通过主动访问代理服务器，让代理服务器获得需要的外网数据，然后转发回客户端；
-- 反向代理：发生在服务器端，用户不知道代理的存在。
-
-反向代理服务器位于源服务器前面，用户的请求需要先经过反向代理服务器才能到达源服务器。反向代理可以用来进行缓存、日志记录等，同时也可以用来做为负载均衡服务器。
-
-在这种负载均衡转发方式下，客户端不直接请求源服务器，因此源服务器不需要外部 IP 地址，而反向代理需要配置内部和外部两套 IP 地址。
-
-优点：
-
-- 与其它功能集成在一起，部署简单。
-
-缺点：
-
-- 所有请求和响应都需要经过反向代理服务器，它可能会成为性能瓶颈。
-
-### 4. 网络层
-
-负载均衡服务器在操作系统内核进程获取网络数据包，根据负载均衡算法计算源服务器的 IP 地址，并修改请求数据包的目的 IP 地址，最后进行转发。
-
-源服务器返回的响应也需要经过负载均衡服务器，通常是让负载均衡服务器同时作为集群的网关服务器来实现。
-
-优点：
-
-- 在内核进程中进行处理，性能比较高。
-
-缺点：
-
-- 和反向代理一样，所有的请求和相应都经过负载均衡服务器，会成为性能瓶颈。
-
-### 5. 链路层
-
-在链路层根据负载均衡算法计算源服务器的 MAC 地址，并修改请求数据包的目的 MAC 地址，并进行转发。
-
-通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要需要 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
-
-这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用 LVS（Linux Virtual Server）。
-
-参考：
-
-- [Comparing Load Balancing Algorithms](http://www.jscape.com/blog/load-balancing-algorithms)
-- [Redirection and Load Balancing](http://slideplayer.com/slide/6599069/#)
-
-# 二、集群下的 Session 管理
-
-一个用户的 Session 信息如果存储在一个服务器上，那么当负载均衡器把用户的下一个请求转发到另一个服务器，由于服务器没有用户的 Session 信息，那么该用户就需要重新进行登录等操作。
-
-## Sticky Sessions
-
-需要配置负载均衡器，使得一个用户的所有请求都路由到同一个服务器，这样就可以把用户的 Session 存放在该服务器中。
-
-缺点：
-
-- 当服务器宕机时，将丢失该服务器上的所有 Session。
-
-<div align="center"> <img src="../pics//MultiNode-StickySessions.jpg"/> </div><br>
-
-## Session Replication
-
-在服务器之间进行 Session 同步操作，每个服务器都有所有用户的 Session 信息，因此用户可以向任何一个服务器进行请求。
-
-缺点：
-
-- 占用过多内存；
-- 同步过程占用网络带宽以及服务器处理器时间。
-
-
-<div align="center"> <img src="../pics//MultiNode-SessionReplication.jpg"/> </div><br>
-
-## Session Server
-
-使用一个单独的服务器存储 Session 数据，可以使用 MySQL，也使用 Redis 或者 Memcached 这种内存型数据库。
-
-优点：
-
-- 为了使得大型网站具有伸缩性，集群中的应用服务器通常需要保持无状态，那么应用服务器不能存储用户的会话信息。Session Server 将用户的会话信息单独进行存储，从而保证了应用服务器的无状态。
-
-缺点：
-
-- 需要去实现存取 Session 的代码。
-
-<div align="center"> <img src="../pics//MultiNode-SpringSession.jpg"/> </div><br>
-
-参考：
-
-- [Session Management using Spring Session with JDBC DataStore](https://sivalabs.in/2018/02/session-management-using-spring-session-jdbc-datastore/)
-

From 19d6a24e7b237cc936ac5b7740ff29ddc71f0774 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 7 Aug 2018 12:40:02 +0800
Subject: [PATCH 073/172] auto commit

---
 notes/Java IO.md        |  23 ++++-
 notes/Java 容器.md      |   6 --
 notes/Redis.md          |  46 ++++++++-
 notes/SQL.md            |   6 --
 notes/分布式.md         |  27 +++++-
 notes/数据库系统原理.md |  41 +++++++-
 notes/正则表达式.md     |  14 ++-
 notes/系统设计基础.md   |  98 -------------------
 notes/计算机网络.md     |   6 --
 notes/重构.md           |   6 --
 notes/集群.md           | 208 ----------------------------------------
 11 files changed, 136 insertions(+), 345 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index 93b1c46f..1e3aedcd 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -1,9 +1,27 @@
 <!-- GFM-TOC -->
+* [一、概览](#一概览)
+* [二、磁盘操作](#二磁盘操作)
+* [三、字节操作](#三字节操作)
+* [四、字符操作](#四字符操作)
+* [五、对象操作](#五对象操作)
+* [六、网络操作](#六网络操作)
+    * [InetAddress](#inetaddress)
+    * [URL](#url)
+    * [Sockets](#sockets)
+    * [Datagram](#datagram)
+* [七、NIO](#七nio)
+    * [流与块](#流与块)
+    * [通道与缓冲区](#通道与缓冲区)
+    * [缓冲区状态变量](#缓冲区状态变量)
+    * [文件 NIO 实例](#文件-nio-实例)
+    * [选择器](#选择器)
+    * [套接字 NIO 实例](#套接字-nio-实例)
+    * [内存映射文件](#内存映射文件)
+    * [对比](#对比)
+* [八、参考资料](#八参考资料)
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
 # 一、概览
 
 Java 的 I/O 大概可以分成以下几类：
@@ -570,4 +588,3 @@ NIO 与普通 I/O 的区别主要有以下两点：
 - [NIO 与传统 IO 的区别](http://blog.csdn.net/shimiso/article/details/24990499)
 - [Decorator Design Pattern](http://stg-tud.github.io/sedc/Lecture/ws13-14/5.3-Decorator.html#mode=document)
 - [Socket Multicast](http://labojava.blogspot.com/2012/12/socket-multicast.html)
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index b5fbefc6..bf1921fc 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -1,9 +1,4 @@
 <!-- GFM-TOC -->
-<<<<<<< HEAD
-<!-- GFM-TOC -->
-
-
-=======
 * [一、概览](#一概览)
     * [Collection](#collection)
     * [Map](#map)
@@ -1152,4 +1147,3 @@ ListIterator <-- List
 - [Java 集合细节（二）：asList 的缺陷](http://wiki.jikexueyuan.com/project/java-enhancement/java-thirtysix.html)
 - [Java Collection Framework – The LinkedList Class](http://javaconceptoftheday.com/java-collection-framework-linkedlist-class/)
 
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/Redis.md b/notes/Redis.md
index d15b60af..0981abf0 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -1,9 +1,50 @@
 <!-- GFM-TOC -->
+* [一、概述](#一概述)
+* [二、数据类型](#二数据类型)
+    * [STRING](#string)
+    * [LIST](#list)
+    * [SET](#set)
+    * [HASH](#hash)
+    * [ZSET](#zset)
+* [三、数据结构](#三数据结构)
+    * [字典](#字典)
+    * [跳跃表](#跳跃表)
+* [四、使用场景](#四使用场景)
+    * [计数器](#计数器)
+    * [缓存](#缓存)
+    * [查找表](#查找表)
+    * [消息队列](#消息队列)
+    * [会话缓存](#会话缓存)
+    * [分布式锁实现](#分布式锁实现)
+    * [其它](#其它)
+* [五、Redis 与 Memcached](#五redis-与-memcached)
+    * [数据类型](#数据类型)
+    * [数据持久化](#数据持久化)
+    * [分布式](#分布式)
+    * [内存管理机制](#内存管理机制)
+* [六、键的过期时间](#六键的过期时间)
+* [七、数据淘汰策略](#七数据淘汰策略)
+* [八、持久化](#八持久化)
+    * [RDB 持久化](#rdb-持久化)
+    * [AOF 持久化](#aof-持久化)
+* [九、事务](#九事务)
+* [十、事件](#十事件)
+    * [文件事件](#文件事件)
+    * [时间事件](#时间事件)
+    * [事件的调度与执行](#事件的调度与执行)
+* [十一、复制](#十一复制)
+    * [连接过程](#连接过程)
+    * [主从链](#主从链)
+* [十二、Sentinel](#十二sentinel)
+* [十三、分片](#十三分片)
+* [十四、一个简单的论坛系统分析](#十四一个简单的论坛系统分析)
+    * [文章信息](#文章信息)
+    * [点赞功能](#点赞功能)
+    * [对文章进行排序](#对文章进行排序)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
 # 一、概述
 
 Redis 是速度非常快的非关系型（NoSQL）内存键值数据库，可以存储键和五种不同类型的值之间的映射。
@@ -566,4 +607,3 @@ Redis 没有关系型数据库中的表这一概念来将同种类型的数据
 - [Redis 应用场景](http://www.scienjus.com/redis-use-case/)
 - [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
 - [Using Redis as an LRU cache](https://redis.io/topics/lru-cache)
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/SQL.md b/notes/SQL.md
index 7ac49770..4a44aee8 100644
--- a/notes/SQL.md
+++ b/notes/SQL.md
@@ -1,9 +1,4 @@
 <!-- GFM-TOC -->
-<<<<<<< HEAD
-<!-- GFM-TOC -->
-
-
-=======
 * [一、基础](#一基础)
 * [二、创建表](#二创建表)
 * [三、修改表](#三修改表)
@@ -770,4 +765,3 @@ SET PASSWROD FOR myuser = Password('new_password');
 # 参考资料
 
 - BenForta. SQL 必知必会 [M]. 人民邮电出版社, 2013.
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/分布式.md b/notes/分布式.md
index d807bfc2..c048b43e 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -1,9 +1,31 @@
 <!-- GFM-TOC -->
+* [一、分布式锁](#一分布式锁)
+    * [数据库的唯一索引](#数据库的唯一索引)
+    * [Redis 的 SETNX 指令](#redis-的-setnx-指令)
+    * [Redis 的 RedLock 算法](#redis-的-redlock-算法)
+    * [Zookeeper 的有序节点](#zookeeper-的有序节点)
+* [二、分布式事务](#二分布式事务)
+    * [本地消息表](#本地消息表)
+    * [2PC](#2pc)
+* [三、CAP](#三cap)
+    * [一致性](#一致性)
+    * [可用性](#可用性)
+    * [分区容忍性](#分区容忍性)
+    * [权衡](#权衡)
+* [四、BASE](#四base)
+    * [基本可用](#基本可用)
+    * [软状态](#软状态)
+    * [最终一致性](#最终一致性)
+* [五、Paxos](#五paxos)
+    * [执行过程](#执行过程)
+    * [约束条件](#约束条件)
+* [五、Raft](#五raft)
+    * [单个 Candidate 的竞选](#单个-candidate-的竞选)
+    * [多个 Candidate 竞选](#多个-candidate-竞选)
+    * [日志复制](#日志复制)
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
 # 一、分布式锁
 
 在单机场景下，可以使用 Java 提供的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
@@ -333,4 +355,3 @@ Raft 主要是用来竞选主节点。
 
 - [Raft: Understandable Distributed Consensus](http://thesecretlivesofdata.com/raft)
 
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index 838e483b..151ffc0f 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -1,9 +1,45 @@
 <!-- GFM-TOC -->
+* [一、事务](#一事务)
+    * [概念](#概念)
+    * [ACID](#acid)
+    * [AUTOCOMMIT](#autocommit)
+* [二、并发一致性问题](#二并发一致性问题)
+    * [丢失修改](#丢失修改)
+    * [读脏数据](#读脏数据)
+    * [不可重复读](#不可重复读)
+    * [幻影读](#幻影读)
+* [三、封锁](#三封锁)
+    * [封锁粒度](#封锁粒度)
+    * [封锁类型](#封锁类型)
+    * [封锁协议](#封锁协议)
+    * [MySQL 隐式与显示锁定](#mysql-隐式与显示锁定)
+* [四、隔离级别](#四隔离级别)
+    * [未提交读（READ UNCOMMITTED）](#未提交读read-uncommitted)
+    * [提交读（READ COMMITTED）](#提交读read-committed)
+    * [可重复读（REPEATABLE READ）](#可重复读repeatable-read)
+    * [可串行化（SERIALIZABLE）](#可串行化serializable)
+* [五、多版本并发控制](#五多版本并发控制)
+    * [版本号](#版本号)
+    * [Undo 日志](#undo-日志)
+    * [实现过程](#实现过程)
+    * [快照读与当前读](#快照读与当前读)
+* [六、Next-Key Locks](#六next-key-locks)
+    * [Record Locks](#record-locks)
+    * [Gap Locks](#gap-locks)
+    * [Next-Key Locks](#next-key-locks)
+* [七、关系数据库设计理论](#七关系数据库设计理论)
+    * [函数依赖](#函数依赖)
+    * [异常](#异常)
+    * [范式](#范式)
+* [八、ER 图](#八er-图)
+    * [实体的三种联系](#实体的三种联系)
+    * [表示出现多次的关系](#表示出现多次的关系)
+    * [联系的多向性](#联系的多向性)
+    * [表示子类](#表示子类)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
 # 一、事务
 
 ## 概念
@@ -551,4 +587,3 @@ Entity-Relationship，有三个组成部分：实体、属性、联系。
 - [MySQL locking for the busy web developer](https://www.brightbox.com/blog/2013/10/31/on-mysql-locks/)
 - [浅入浅出 MySQL 和 InnoDB](https://draveness.me/mysql-innodb)
 - [Innodb 中的事务隔离级别和锁的关系](https://tech.meituan.com/innodb-lock.html)
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/正则表达式.md b/notes/正则表达式.md
index 6b563586..c713da63 100644
--- a/notes/正则表达式.md
+++ b/notes/正则表达式.md
@@ -1,9 +1,18 @@
 <!-- GFM-TOC -->
+* [一、概述](#一概述)
+* [二、匹配单个字符](#二匹配单个字符)
+* [三、匹配一组字符](#三匹配一组字符)
+* [四、使用元字符](#四使用元字符)
+* [五、重复匹配](#五重复匹配)
+* [六、位置匹配](#六位置匹配)
+* [七、使用子表达式](#七使用子表达式)
+* [八、回溯引用](#八回溯引用)
+* [九、前后查找](#九前后查找)
+* [十、嵌入条件](#十嵌入条件)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
 # 一、概述
 
 正则表达式用于文本内容的查找和替换。
@@ -377,4 +386,3 @@ aBCd
 # 参考资料
 
 - BenForta. 正则表达式必知必会 [M]. 人民邮电出版社, 2007.
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index 313eca83..649f563a 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -2,101 +2,3 @@
 <!-- GFM-TOC -->
 
 
-<<<<<<< HEAD
-=======
-# 一、性能
-
-## 性能指标
-
-### 1. 响应时间
-
-指从某个请求从发出到接收到响应消耗的时间。
-
-在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
-
-### 2. 吞吐量
-
-指系统在单位时间内可以处理的请求数量，通常使用每秒的请求数来衡量。
-
-### 3. 并发用户数
-
-指系统能同时处理的并发用户请求数量。
-
-在没有并发存在的系统中，请求被顺序执行，此时响应时间为吞吐量的倒数。例如系统支持的吞吐量为 100 req/s，那么平均响应时间应该为 0.01s。
-
-目前的大型系统都支持多线程来处理并发请求，多线程能够提高吞吐量以及缩短响应时间，主要有两个原因：
-
-- 多 CPU
-- IO 等待时间
-
-使用 IO 多路复用等方式，系统在等待一个 IO 操作完成的这段时间内不需要被阻塞，可以去处理其它请求。通过将这个等待时间利用起来，使得 CPU 利用率大大提高。
-
-并发用户数不是越高越好，因为如果并发用户数太高，系统来不及处理这么多的请求，会使得过多的请求需要等待，那么响应时间就会大大提高。
-
-## 性能优化
-
-### 1. 集群
-
-将多台服务器组成集群，使用负载均衡将请求转发到集群中，避免单一服务器的负载压力过大导致性能降低。
-
-### 2. 缓存
-
-缓存能够提高性能的原因如下：
-
-- 缓存数据通常位于内存等介质中，这种介质对于读操作特别快；
-- 缓存数据可以位于靠近用户的地理位置上；
-- 可以将计算结果进行缓存，从而避免重复计算。
-
-### 3. 异步
-
-某些流程可以将操作转换为消息，将消息发送到消息队列之后立即返回，之后这个操作会被异步处理。
-
-# 二、伸缩性
-
-指不断向集群中添加服务器来缓解不断上升的用户并发访问压力和不断增长的数据存储需求。
-
-## 伸缩性与性能
-
-如果系统存在性能问题，那么单个用户的请求总是很慢的；
-
-如果系统存在伸缩性问题，那么单个用户的请求可能会很快，但是在并发数很高的情况下系统会很慢。
-
-## 实现伸缩性
-
-应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
-
-关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
-
-对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
-
-# 三、扩展性
-
-指的是添加新功能时对现有系统的其它应用无影响，这就要求不同应用具备低耦合的特点。
-
-实现可扩展主要有两种方式：
-
-- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
-- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
-
-# 四、可用性
-
-## 冗余
-
-保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
-
-应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
-
-存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
-
-## 监控
-
-对 CPU、内存、磁盘、网络等系统负载信息进行监控，当某个数据达到一定阈值时通知运维人员，从而在系统发生故障之前及时发现问题。
-
-## 服务降级
-
-服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
-
-# 五、安全性
-
-要求系统的应对各种攻击手段时能够有可靠的应对措施。
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index 2297e981..e80ba7f8 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -1,9 +1,4 @@
 <!-- GFM-TOC -->
-<<<<<<< HEAD
-<!-- GFM-TOC -->
-
-
-=======
 * [一、概述](#一概述)
     * [网络的网络](#网络的网络)
     * [ISP](#isp)
@@ -905,4 +900,3 @@ IMAP 协议中客户端和服务器上的邮件保持同步，如果不去手动
 - [Technology-Computer Networking[1]-Computer Networks and the Internet](http://www.linyibin.cn/2017/02/12/technology-ComputerNetworking-Internet/)
 - [P2P 网络概述.](http://slidesplayer.com/slide/11616167/)
 - [Circuit Switching (a) Circuit switching. (b) Packet switching.](http://slideplayer.com/slide/5115386/)
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/重构.md b/notes/重构.md
index 04e21e5f..b5a834e0 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -1,9 +1,4 @@
 <!-- GFM-TOC -->
-<<<<<<< HEAD
-<!-- GFM-TOC -->
-
-
-=======
 * [一、第一个案例](#一第一个案例)
 * [二、重构原则](#二重构原则)
     * [定义](#定义)
@@ -1428,4 +1423,3 @@ public Manager(String name, String id, int grade) {
 # 参考资料
 
 - MartinFowler, 福勒, 贝克, 等. 重构: 改善既有代码的设计 [M]. 电子工业出版社, 2011.
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666
diff --git a/notes/集群.md b/notes/集群.md
index 82ddda76..649f563a 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -1,212 +1,4 @@
 <!-- GFM-TOC -->
-<<<<<<< HEAD
 <!-- GFM-TOC -->
 
 
-=======
-* [一、负载均衡](#一负载均衡)
-    * [算法实现](#算法实现)
-    * [转发实现](#转发实现)
-* [二、集群下的 Session 管理](#二集群下的-session-管理)
-    * [Sticky Session](#sticky-session)
-    * [Session Replication](#session-replication)
-    * [Session Server](#session-server)
-<!-- GFM-TOC -->
-
-
-# 一、负载均衡
-
-集群中的应用服务器通常被设计成无状态，用户可以请求任何一个节点（应用服务器）。
-
-负载均衡器会根据集群中每个节点的负载情况，将用户请求转发到合适的节点上。
-
-负载均衡器可以用来实现高可用以及伸缩性：
-
-- 高可用：当某个节点故障时，负载均衡器不会将用户请求转发到该节点上，从而保证所有服务持续可用；
-- 伸缩性：可以很容易地添加移除节点。
-
-负载均衡运行过程包含两个部分：
-
-1. 根据负载均衡算法得到请求转发的节点；
-2. 将请求进行转发；
-
-## 算法实现
-
-### 1. 轮询（Round Robin）
-
-轮询算法把每个请求轮流发送到每个服务器上。
-
-下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
-
-<div align="center"> <img src="../pics//2766d04f-7dad-42e4-99d1-60682c9d5c61.jpg"/> </div><br>
-
-该算法比较适合每个服务器的性能差不多的场景，如果有性能存在差异的情况下，那么性能较差的服务器可能无法承担过大的负载（下图的 Server 2）。
-
-<div align="center"> <img src="../pics//f7ecbb8d-bb8b-4d45-a3b7-f49425d6d83d.jpg"/> </div><br>
-
-### 2. 加权轮询（Weighted Round Robbin）
-
-加权轮询是在轮询的基础上，根据服务器的性能差异，为服务器赋予一定的权值，性能高的服务器分配更高的权值。
-
-例如下图中，服务器 1 被赋予的权值为 5，服务器 2 被赋予的权值为 1，那么 (1, 2, 3, 4, 5) 请求会被发送到服务器 1，(6) 请求会被发送到服务器 2。
-
-<div align="center"> <img src="../pics//211c60d4-75ca-4acd-8a4f-171458ed58b4.jpg"/> </div><br>
-
-### 3. 最少连接（least Connections）
-
-由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。
-
-例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
-
-<div align="center"> <img src="../pics//3b0d1aa8-d0e0-46c2-8fd1-736bf08a11aa.jpg"/> </div><br>
-
-最少连接算法就是将请求发送给当前最少连接数的服务器上。
-
-例如下图中，服务器 1 当前连接数最小，那么新到来的请求 6 就会被发送到服务器 1 上。
-
-<div align="center"> <img src="../pics//1f4a7f10-52b2-4bd7-a67d-a9581d66dc62.jpg"/> </div><br>
-
-### 4. 加权最少连接（Weighted Least Connection）
-
-在最少连接的基础上，根据服务器的性能为每台服务器分配权重，再根据权重计算出每台服务器能处理的连接数。
-
-<div align="center"> <img src="../pics//44edefb7-4b58-4519-b8ee-4aca01697b78.jpg"/> </div><br>
-
-### 5. 随机算法（Random）
-
-把请求随机发送到服务器上。
-
-和轮询算法类似，该算法比较适合服务器性能差不多的场景。
-
-<div align="center"> <img src="../pics//0ee0f61b-c782-441e-bf34-665650198ae0.jpg"/> </div><br>
-
-### 6. 源地址哈希法 (IP Hash)
-
-源地址哈希通过对客户端 IP 计算哈希值之后，再对服务器数量取模得到目标服务器的序号。
-
-可以保证同一 IP 的客户端的请求会转发到同一台服务器上，用来实现会话粘滞（Sticky Session）
-
-<div align="center"> <img src="../pics//2018040302.jpg"/> </div><br>
-
-## 转发实现
-
-### 1. HTTP 重定向
-
-HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服务器的 IP 地址之后，将该地址写入 HTTP 重定向报文中，状态码为 302。客户端收到重定向报文之后，需要重新向服务器发起请求。
-
-缺点：
-
-- 需要两次请求，因此访问延迟比较高；
-- HTTP 负载均衡器处理能力有限，会限制集群的规模。
-
-该负载均衡转发的缺点比较明显，实际场景中很少使用它。
-
-<div align="center"> <img src="../pics//c5f611f0-fd5c-4158-9003-278141136e6e.jpg"/> </div><br>
-
-### 2. DNS 域名解析
-
-在 DNS 解析域名的同时使用负载均衡算法计算服务器地址。
-
-优点：
-
-- DNS 能够根据地理位置进行域名解析，返回离用户最近的服务器 IP 地址。
-
-缺点：
-
-- 由于 DNS 具有多级结构，每一级的域名记录都可能被缓存，当下线一台服务器需要修改 DNS 记录时，需要过很长一段时间才能生效。
-
-大型网站基本使用了 DNS 做为第一级负载均衡手段，然后在内部使用其它方式做第二级负载均衡。也就是说，域名解析的结果为内部的负载均衡服务器 IP 地址。
-
-<div align="center"> <img src="../pics//76a25fc8-a579-4d7c-974b-7640b57fbf39.jpg"/> </div><br>
-
-### 3. 反向代理服务器
-
-首先了解一下正向代理与反向代理的区别：
-
-- 正向代理：发生在客户端，是由用户主动发起的。比如翻墙，客户端通过主动访问代理服务器，让代理服务器获得需要的外网数据，然后转发回客户端；
-- 反向代理：发生在服务器端，用户不知道代理的存在。
-
-反向代理服务器位于源服务器前面，用户的请求需要先经过反向代理服务器才能到达源服务器。反向代理可以用来进行缓存、日志记录等，同时也可以用来做为负载均衡服务器。
-
-在这种负载均衡转发方式下，客户端不直接请求源服务器，因此源服务器不需要外部 IP 地址，而反向代理需要配置内部和外部两套 IP 地址。
-
-优点：
-
-- 与其它功能集成在一起，部署简单。
-
-缺点：
-
-- 所有请求和响应都需要经过反向代理服务器，它可能会成为性能瓶颈。
-
-### 4. 网络层
-
-在操作系统内核进程获取网络数据包，根据负载均衡算法计算源服务器的 IP 地址，并修改请求数据包的目的 IP 地址，最后进行转发。
-
-源服务器返回的响应也需要经过负载均衡服务器，通常是让负载均衡服务器同时作为集群的网关服务器来实现。
-
-优点：
-
-- 在内核进程中进行处理，性能比较高。
-
-缺点：
-
-- 和反向代理一样，所有的请求和响应都经过负载均衡服务器，会成为性能瓶颈。
-
-### 5. 链路层
-
-在链路层根据负载均衡算法计算源服务器的 MAC 地址，并修改请求数据包的目的 MAC 地址，并进行转发。
-
-通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
-
-这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
-
-参考：
-
-- [Comparing Load Balancing Algorithms](http://www.jscape.com/blog/load-balancing-algorithms)
-- [Redirection and Load Balancing](http://slideplayer.com/slide/6599069/#)
-
-# 二、集群下的 Session 管理
-
-一个用户的 Session 信息如果存储在一个服务器上，那么当负载均衡器把用户的下一个请求转发到另一个服务器，由于服务器没有用户的 Session 信息，那么该用户就需要重新进行登录等操作。
-
-## Sticky Session
-
-需要配置负载均衡器，使得一个用户的所有请求都路由到同一个服务器，这样就可以把用户的 Session 存放在该服务器中。
-
-缺点：
-
-- 当服务器宕机时，将丢失该服务器上的所有 Session。
-
-<div align="center"> <img src="../pics//MultiNode-StickySessions.jpg"/> </div><br>
-
-## Session Replication
-
-在服务器之间进行 Session 同步操作，每个服务器都有所有用户的 Session 信息，因此用户可以向任何一个服务器进行请求。
-
-缺点：
-
-- 占用过多内存；
-- 同步过程占用网络带宽以及服务器处理器时间。
-
-
-<div align="center"> <img src="../pics//MultiNode-SessionReplication.jpg"/> </div><br>
-
-## Session Server
-
-使用一个单独的服务器存储 Session 数据，可以使用 MySQL，也使用 Redis 或者 Memcached 这种内存型数据库。
-
-优点：
-
-- 为了使得大型网站具有伸缩性，集群中的应用服务器通常需要保持无状态，那么应用服务器不能存储用户的会话信息。Session Server 将用户的会话信息单独进行存储，从而保证了应用服务器的无状态。
-
-缺点：
-
-- 需要去实现存取 Session 的代码。
-
-<div align="center"> <img src="../pics//MultiNode-SpringSession.jpg"/> </div><br>
-
-参考：
-
-- [Session Management using Spring Session with JDBC DataStore](https://sivalabs.in/2018/02/session-management-using-spring-session-jdbc-datastore/)
-
->>>>>>> 8c67483fe9ac599aa6f6358fb32a541d7e279666

From 0d34f6ec144f8207382711ccd42a417f6f8d6bf9 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 7 Aug 2018 17:52:24 +0800
Subject: [PATCH 074/172] auto commit

---
 notes/Leetcode 题解.md |   8 +-
 notes/Linux.md         |   9 +-
 notes/MySQL.md         |   2 +-
 notes/消息队列.md      |  76 ++++++++++++++++
 notes/系统设计基础.md  | 100 ++++++++++++++++++++
 notes/集群.md          | 202 +++++++++++++++++++++++++++++++++++++++++
 6 files changed, 391 insertions(+), 6 deletions(-)

diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 293f4328..2813ce5d 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -2371,7 +2371,9 @@ public List<Integer> diffWaysToCompute(String input) {
 
 题目描述：有 N 阶楼梯，每次可以上一阶或者两阶，求有多少种上楼梯的方法。
 
-定义一个数组 dp 存储上楼梯的方法数（为了方便讨论，数组下标从 1 开始），dp[i] 表示走到第 i 个楼梯的方法数目。第 i 个楼梯可以从第 i-1 和 i-2 个楼梯再走一步到达，走到第 i 个楼梯的方法数为走到第 i-1 和第 i-2 个楼梯的方法数之和。
+定义一个数组 dp 存储上楼梯的方法数（为了方便讨论，数组下标从 1 开始），dp[i] 表示走到第 i 个楼梯的方法数目。
+
+第 i 个楼梯可以从第 i-1 和 i-2 个楼梯再走一步到达，走到第 i 个楼梯的方法数为走到第 i-1 和第 i-2 个楼梯的方法数之和。
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]"/></div> <br>
 
@@ -2400,7 +2402,9 @@ public int climbStairs(int n) {
 
 题目描述：抢劫一排住户，但是不能抢邻近的住户，求最大抢劫量。
 
-定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以
+定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。
+
+由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2],dp[i-3])+nums[i]"/></div> <br>
 
diff --git a/notes/Linux.md b/notes/Linux.md
index ebcc347b..14d3b317 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -275,8 +275,6 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 
 ## 组成
 
-<div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
-
 最主要的几个组成部分如下：
 
 - inode：一个文件占用一个 inode，记录文件的属性，同时记录此文件的内容所在的 block 编号；
@@ -287,6 +285,9 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 - superblock：记录文件系统的整体信息，包括 inode 和 block 的总量、使用量、剩余量，以及文件系统的格式与相关信息等；
 - block bitmap：记录 block 是否被使用的位域；
 
+<div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
+
+
 ## 文件读取
 
 对于 Ext2 文件系统，当要读取一个文件的内容时，先在 inode 中去查找文件内容所在的所有 block，然后把所有 block 的内容读出来。
@@ -336,7 +337,9 @@ inode 中记录了文件内容所在的 block 编号，但是每个 block 非常
 
 ## 目录
 
-建立一个目录时，会分配一个 inode 与至少一个 block。block 记录的内容是目录下所有文件的 inode 编号以及文件名。可以看出文件的 inode 本身不记录文件名，文件名记录在目录中，因此新增文件、删除文件、更改文件名这些操作与目录的 w 权限有关。
+建立一个目录时，会分配一个 inode 与至少一个 block。block 记录的内容是目录下所有文件的 inode 编号以及文件名。
+
+可以看出文件的 inode 本身不记录文件名，文件名记录在目录中，因此新增文件、删除文件、更改文件名这些操作与目录的 w 权限有关。
 
 ## 日志
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index 1c575fae..690425db 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -139,7 +139,7 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 ### 2. 操作
 
-操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
+进行查找操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
 插入删除操作记录会破坏平衡树的平衡性，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作。
 
diff --git a/notes/消息队列.md b/notes/消息队列.md
index 649f563a..74155802 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -1,4 +1,80 @@
 <!-- GFM-TOC -->
+* [一、消息模型](#一消息模型)
+    * [点对点](#点对点)
+    * [发布/订阅](#发布订阅)
+* [二、使用场景](#二使用场景)
+    * [异步处理](#异步处理)
+    * [流量削锋](#流量削锋)
+    * [应用解耦](#应用解耦)
+* [三、可靠性](#三可靠性)
+    * [发送端的可靠性](#发送端的可靠性)
+    * [接收端的可靠性](#接收端的可靠性)
 <!-- GFM-TOC -->
 
 
+# 一、消息模型
+
+## 点对点
+
+消息生产者向消息队列中发送了一个消息之后，只能被一个消费者消费一次。
+
+<div align="center"> <img src="../pics//09b52bcb-88ba-4e36-8244-b375f16ad116.jpg"/> </div><br>
+
+## 发布/订阅
+
+消息生产者向频道发送一个消息之后，多个消费者可以从该频道订阅到这条消息并消费。
+
+<div align="center"> <img src="../pics//ddb5ff4c-4ada-46aa-9bf1-140bdb5e4676.jpg"/> </div><br>
+
+发布与订阅模式和观察者模式有以下不同：
+
+- 观察者模式中，观察者和主题都知道对方的存在；而在发布与订阅模式中，发布者与订阅者不知道对方的存在，它们之间通过频道进行通信。
+- 观察者模式是同步的，当事件触发时，主题会去调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息。
+
+<div align="center"> <img src="../pics//bee1ff1d-c80f-4b3c-b58c-7073a8896ab2.jpg"/> </div><br>
+
+参考：
+
+- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
+- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
+
+# 二、使用场景
+
+## 异步处理
+
+发送者将消息发送给消息队列之后，不需要同步等待消息接收者处理完毕，而是立即返回进行其它操作。消息接收者从消息队列中订阅消息之后异步处理。
+
+例如在注册流程中通常需要发送验证邮件来确保注册用户的身份合法，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
+
+只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成注册的话，就不能再使用消息队列。
+
+## 流量削锋
+
+在高并发的场景下，如果短时间有大量的请求到达会压垮服务器。
+
+可以将请求发送到消息队列中，服务器按照其处理能力从消息队列中订阅消息进行处理。
+
+## 应用解耦
+
+如果模块之间不直接进行调用，模块之间耦合度就会很低，那么修改一个模块或者新增一个模块对其它模块的影响会很小，从而实现可扩展性。
+
+通过使用消息队列，一个模块只需要向消息队列中发送消息，其它模块可以选择性地从消息队列中订阅消息从而完成调用。
+
+# 三、可靠性
+
+## 发送端的可靠性
+
+发送端完成操作后一定能将消息成功发送到消息队列中。
+
+实现方法：
+
+- 在本地数据库建一张消息表，将消息数据与业务数据保存在同一数据库实例里，这样就可以利用本地数据库的事务机制。事务提交成功后，将消息表中的消息转移到消息队列中，若转移消息成功则删除消息表中的数据，否则继续重传。
+
+## 接收端的可靠性
+
+接收端能够从消息中间件成功消费一次消息。
+
+实现方法：
+
+- 保证接收端处理消息的业务逻辑具有幂等性：只要具有幂等性，那么消费多少次消息，最后处理的结果都是一样的。
+- 保证消息具有唯一编号，并使用一张日志表来记录已经消费的消息编号。
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index 649f563a..40f0ca9d 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -1,4 +1,104 @@
 <!-- GFM-TOC -->
+* [一、性能](#一性能)
+* [二、伸缩性](#二伸缩性)
+* [三、扩展性](#三扩展性)
+* [四、可用性](#四可用性)
+* [五、安全性](#五安全性)
 <!-- GFM-TOC -->
 
 
+# 一、性能
+
+## 性能指标
+
+### 1. 响应时间
+
+指从某个请求从发出到接收到响应消耗的时间。
+
+在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
+
+### 2. 吞吐量
+
+指系统在单位时间内可以处理的请求数量，通常使用每秒的请求数来衡量。
+
+### 3. 并发用户数
+
+指系统能同时处理的并发用户请求数量。
+
+在没有并发存在的系统中，请求被顺序执行，此时响应时间为吞吐量的倒数。例如系统支持的吞吐量为 100 req/s，那么平均响应时间应该为 0.01s。
+
+目前的大型系统都支持多线程来处理并发请求，多线程能够提高吞吐量以及缩短响应时间，主要有两个原因：
+
+- 多 CPU
+- IO 等待时间
+
+使用 IO 多路复用等方式，系统在等待一个 IO 操作完成的这段时间内不需要被阻塞，可以去处理其它请求。通过将这个等待时间利用起来，使得 CPU 利用率大大提高。
+
+并发用户数不是越高越好，因为如果并发用户数太高，系统来不及处理这么多的请求，会使得过多的请求需要等待，那么响应时间就会大大提高。
+
+## 性能优化
+
+### 1. 集群
+
+将多台服务器组成集群，使用负载均衡将请求转发到集群中，避免单一服务器的负载压力过大导致性能降低。
+
+### 2. 缓存
+
+缓存能够提高性能的原因如下：
+
+- 缓存数据通常位于内存等介质中，这种介质对于读操作特别快；
+- 缓存数据可以位于靠近用户的地理位置上；
+- 可以将计算结果进行缓存，从而避免重复计算。
+
+### 3. 异步
+
+某些流程可以将操作转换为消息，将消息发送到消息队列之后立即返回，之后这个操作会被异步处理。
+
+# 二、伸缩性
+
+指不断向集群中添加服务器来缓解不断上升的用户并发访问压力和不断增长的数据存储需求。
+
+## 伸缩性与性能
+
+如果系统存在性能问题，那么单个用户的请求总是很慢的；
+
+如果系统存在伸缩性问题，那么单个用户的请求可能会很快，但是在并发数很高的情况下系统会很慢。
+
+## 实现伸缩性
+
+应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
+
+关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
+
+对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
+
+# 三、扩展性
+
+指的是添加新功能时对现有系统的其它应用无影响，这就要求不同应用具备低耦合的特点。
+
+实现可扩展主要有两种方式：
+
+- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
+- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
+
+# 四、可用性
+
+## 冗余
+
+保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
+
+应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
+
+存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
+
+## 监控
+
+对 CPU、内存、磁盘、网络等系统负载信息进行监控，当某个数据达到一定阈值时通知运维人员，从而在系统发生故障之前及时发现问题。
+
+## 服务降级
+
+服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
+
+# 五、安全性
+
+要求系统的应对各种攻击手段时能够有可靠的应对措施。
diff --git a/notes/集群.md b/notes/集群.md
index 649f563a..717ae637 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -1,4 +1,206 @@
 <!-- GFM-TOC -->
+* [一、负载均衡](#一负载均衡)
+    * [算法实现](#算法实现)
+    * [转发实现](#转发实现)
+* [二、集群下的 Session 管理](#二集群下的-session-管理)
+    * [Sticky Session](#sticky-session)
+    * [Session Replication](#session-replication)
+    * [Session Server](#session-server)
 <!-- GFM-TOC -->
 
 
+# 一、负载均衡
+
+集群中的应用服务器通常被设计成无状态，用户可以请求任何一个节点（应用服务器）。
+
+负载均衡器会根据集群中每个节点的负载情况，将用户请求转发到合适的节点上。
+
+负载均衡器可以用来实现高可用以及伸缩性：
+
+- 高可用：当某个节点故障时，负载均衡器不会将用户请求转发到该节点上，从而保证所有服务持续可用；
+- 伸缩性：可以很容易地添加移除节点。
+
+负载均衡运行过程包含两个部分：
+
+1. 根据负载均衡算法得到请求转发的节点；
+2. 将请求进行转发；
+
+## 算法实现
+
+### 1. 轮询（Round Robin）
+
+轮询算法把每个请求轮流发送到每个服务器上。
+
+下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
+
+<div align="center"> <img src="../pics//2766d04f-7dad-42e4-99d1-60682c9d5c61.jpg"/> </div><br>
+
+该算法比较适合每个服务器的性能差不多的场景，如果有性能存在差异的情况下，那么性能较差的服务器可能无法承担过大的负载（下图的 Server 2）。
+
+<div align="center"> <img src="../pics//f7ecbb8d-bb8b-4d45-a3b7-f49425d6d83d.jpg"/> </div><br>
+
+### 2. 加权轮询（Weighted Round Robbin）
+
+加权轮询是在轮询的基础上，根据服务器的性能差异，为服务器赋予一定的权值，性能高的服务器分配更高的权值。
+
+例如下图中，服务器 1 被赋予的权值为 5，服务器 2 被赋予的权值为 1，那么 (1, 2, 3, 4, 5) 请求会被发送到服务器 1，(6) 请求会被发送到服务器 2。
+
+<div align="center"> <img src="../pics//211c60d4-75ca-4acd-8a4f-171458ed58b4.jpg"/> </div><br>
+
+### 3. 最少连接（least Connections）
+
+由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。
+
+例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
+
+<div align="center"> <img src="../pics//3b0d1aa8-d0e0-46c2-8fd1-736bf08a11aa.jpg"/> </div><br>
+
+最少连接算法就是将请求发送给当前最少连接数的服务器上。
+
+例如下图中，服务器 1 当前连接数最小，那么新到来的请求 6 就会被发送到服务器 1 上。
+
+<div align="center"> <img src="../pics//1f4a7f10-52b2-4bd7-a67d-a9581d66dc62.jpg"/> </div><br>
+
+### 4. 加权最少连接（Weighted Least Connection）
+
+在最少连接的基础上，根据服务器的性能为每台服务器分配权重，再根据权重计算出每台服务器能处理的连接数。
+
+<div align="center"> <img src="../pics//44edefb7-4b58-4519-b8ee-4aca01697b78.jpg"/> </div><br>
+
+### 5. 随机算法（Random）
+
+把请求随机发送到服务器上。
+
+和轮询算法类似，该算法比较适合服务器性能差不多的场景。
+
+<div align="center"> <img src="../pics//0ee0f61b-c782-441e-bf34-665650198ae0.jpg"/> </div><br>
+
+### 6. 源地址哈希法 (IP Hash)
+
+源地址哈希通过对客户端 IP 计算哈希值之后，再对服务器数量取模得到目标服务器的序号。
+
+可以保证同一 IP 的客户端的请求会转发到同一台服务器上，用来实现会话粘滞（Sticky Session）
+
+<div align="center"> <img src="../pics//2018040302.jpg"/> </div><br>
+
+## 转发实现
+
+### 1. HTTP 重定向
+
+HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服务器的 IP 地址之后，将该地址写入 HTTP 重定向报文中，状态码为 302。客户端收到重定向报文之后，需要重新向服务器发起请求。
+
+缺点：
+
+- 需要两次请求，因此访问延迟比较高；
+- HTTP 负载均衡器处理能力有限，会限制集群的规模。
+
+该负载均衡转发的缺点比较明显，实际场景中很少使用它。
+
+<div align="center"> <img src="../pics//c5f611f0-fd5c-4158-9003-278141136e6e.jpg"/> </div><br>
+
+### 2. DNS 域名解析
+
+在 DNS 解析域名的同时使用负载均衡算法计算服务器地址。
+
+优点：
+
+- DNS 能够根据地理位置进行域名解析，返回离用户最近的服务器 IP 地址。
+
+缺点：
+
+- 由于 DNS 具有多级结构，每一级的域名记录都可能被缓存，当下线一台服务器需要修改 DNS 记录时，需要过很长一段时间才能生效。
+
+大型网站基本使用了 DNS 做为第一级负载均衡手段，然后在内部使用其它方式做第二级负载均衡。也就是说，域名解析的结果为内部的负载均衡服务器 IP 地址。
+
+<div align="center"> <img src="../pics//76a25fc8-a579-4d7c-974b-7640b57fbf39.jpg"/> </div><br>
+
+### 3. 反向代理服务器
+
+首先了解一下正向代理与反向代理的区别：
+
+- 正向代理：发生在客户端，是由用户主动发起的。比如翻墙，客户端通过主动访问代理服务器，让代理服务器获得需要的外网数据，然后转发回客户端；
+- 反向代理：发生在服务器端，用户不知道代理的存在。
+
+反向代理服务器位于源服务器前面，用户的请求需要先经过反向代理服务器才能到达源服务器。反向代理可以用来进行缓存、日志记录等，同时也可以用来做为负载均衡服务器。
+
+在这种负载均衡转发方式下，客户端不直接请求源服务器，因此源服务器不需要外部 IP 地址，而反向代理需要配置内部和外部两套 IP 地址。
+
+优点：
+
+- 与其它功能集成在一起，部署简单。
+
+缺点：
+
+- 所有请求和响应都需要经过反向代理服务器，它可能会成为性能瓶颈。
+
+### 4. 网络层
+
+在操作系统内核进程获取网络数据包，根据负载均衡算法计算源服务器的 IP 地址，并修改请求数据包的目的 IP 地址，最后进行转发。
+
+源服务器返回的响应也需要经过负载均衡服务器，通常是让负载均衡服务器同时作为集群的网关服务器来实现。
+
+优点：
+
+- 在内核进程中进行处理，性能比较高。
+
+缺点：
+
+- 和反向代理一样，所有的请求和响应都经过负载均衡服务器，会成为性能瓶颈。
+
+### 5. 链路层
+
+在链路层根据负载均衡算法计算源服务器的 MAC 地址，并修改请求数据包的目的 MAC 地址，并进行转发。
+
+通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
+
+这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
+
+参考：
+
+- [Comparing Load Balancing Algorithms](http://www.jscape.com/blog/load-balancing-algorithms)
+- [Redirection and Load Balancing](http://slideplayer.com/slide/6599069/#)
+
+# 二、集群下的 Session 管理
+
+一个用户的 Session 信息如果存储在一个服务器上，那么当负载均衡器把用户的下一个请求转发到另一个服务器，由于服务器没有用户的 Session 信息，那么该用户就需要重新进行登录等操作。
+
+## Sticky Session
+
+需要配置负载均衡器，使得一个用户的所有请求都路由到同一个服务器，这样就可以把用户的 Session 存放在该服务器中。
+
+缺点：
+
+- 当服务器宕机时，将丢失该服务器上的所有 Session。
+
+<div align="center"> <img src="../pics//MultiNode-StickySessions.jpg"/> </div><br>
+
+## Session Replication
+
+在服务器之间进行 Session 同步操作，每个服务器都有所有用户的 Session 信息，因此用户可以向任何一个服务器进行请求。
+
+缺点：
+
+- 占用过多内存；
+- 同步过程占用网络带宽以及服务器处理器时间。
+
+
+<div align="center"> <img src="../pics//MultiNode-SessionReplication.jpg"/> </div><br>
+
+## Session Server
+
+使用一个单独的服务器存储 Session 数据，可以使用 MySQL，也使用 Redis 或者 Memcached 这种内存型数据库。
+
+优点：
+
+- 为了使得大型网站具有伸缩性，集群中的应用服务器通常需要保持无状态，那么应用服务器不能存储用户的会话信息。Session Server 将用户的会话信息单独进行存储，从而保证了应用服务器的无状态。
+
+缺点：
+
+- 需要去实现存取 Session 的代码。
+
+<div align="center"> <img src="../pics//MultiNode-SpringSession.jpg"/> </div><br>
+
+参考：
+
+- [Session Management using Spring Session with JDBC DataStore](https://sivalabs.in/2018/02/session-management-using-spring-session-jdbc-datastore/)
+

From 38bb39be2b01bf77187624ecd7f03229325b264b Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Tue, 7 Aug 2018 18:26:21 +0800
Subject: [PATCH 075/172] =?UTF-8?q?repo=20=E9=93=BE=E6=8E=A5=E6=9C=89?=
 =?UTF-8?q?=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 other/leetcode 总结.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/other/leetcode 总结.md b/other/leetcode 总结.md
index a864b791..82ff61c7 100644
--- a/other/leetcode 总结.md	
+++ b/other/leetcode 总结.md	
@@ -5,7 +5,7 @@
 
 ## LeetCode 习题集合
 
-* [LeetCode 解题集合](https://github.com/apachecn/leetcode/docs/Leetcode_Solutions)
+* [LeetCode 解题集合](https://github.com/apachecn/LeetCode/tree/master/docs/Leetcode_Solutions)
 
 
 ## 模版要求
@@ -19,7 +19,7 @@
 
 > **案例模版**
 
-[模版：007. Reverse Integer 反转整数](https://github.com/apachecn/leetcode/docs/Leetcode_Solutions/007._Reverse_Integer.md)
+[模版：007. Reverse Integer 反转整数](https://github.com/apachecn/LeetCode/tree/master/docs/Leetcode_Solutions/007._Reverse_Integer.md)
 
 
 ## 项目贡献者

From fbc0a1e837d519b7a66b4efa2375496973b7a6a1 Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Wed, 8 Aug 2018 11:59:34 +0800
Subject: [PATCH 076/172] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=B8=80=E4=BA=9B?=
 =?UTF-8?q?=E6=95=B0=E6=8D=AE=E5=BA=93=E6=95=B0=E6=8D=AE=E8=AE=BF=E9=97=AE?=
 =?UTF-8?q?=E6=96=B9=E9=9D=A2=E7=9A=84=E4=BC=98=E5=8C=96=E6=96=B9=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/MySQL.md | 54 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 54 insertions(+)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 690425db..f2949801 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -304,6 +304,60 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 
 最有效的方式是使用索引来覆盖查询。
 
+### 3. 不要在列上使用函数和进行运算
+
+在列上使用函数和计算将导致索引失效而进行全表扫描。
+
+### 4. 尽量避免使用 != 或 not in 或 <> 等否定操作符
+
+尽量避免在 where 子句中使用 or 来连接条件，因为这会导致索引失效而进行全表扫描。
+
+### 5. 尽量避免使用 or 来连接条件
+
+尽量避免在 where 子句中使用 or 来连接条件，因为这会导致索引失效而进行全表扫描。
+
+### 6. 多个单列限制应该改成一个多列索引（复合索引）查询
+
+因为 MySQL 只能使用一个单列索引，所以为多个列创建单列索引并不能提高 MySQL 的查询性能。正确的做法是将这多个要限定条件的列单独做成一个多列索引（复合索引）。
+
+### 7. 查询时要按照复合索引的最左列开始查找
+
+查询条件中没有使用复合索引的第一个字段，索引是不会被使用的。复合索引遵守“最左前缀”原则，即在查询条件中使用了复合索引的第一个字段，索引才会被使用。
+
+假如我们只有一个复合索引 ```age_height_idx(age, height)```
+
+如果我们使用
+```mysql
+select * from user where height = 170;
+```
+那么复合索引 ```age_height_idx(age, height)```将不会被使用。
+
+### 8. 尽量不用范围查询
+
+如果查询中的某个列有范围查询，则其右边所有列都无法使用索引优化查找。
+
+比如说我们使用
+```mysql
+select * from users where age > 16 and age < 30 and height = 180
+```
+则索引中 age 右边所有列都无法使用索引优化查找。换句话说，```age_height_idx(age, height)``` 索引等价于 ```age_height_idx(age)```。
+
+### 9. 索引不要包含有NULL值的列
+
+只要列中包含有 NULL 值，它就不会被包含在索引中，复合索引中只要有一列含有 NULL 值，那么这一列对于此复合索引就是无效的。
+
+因此，在数据库设计时，除非有一个很特别的原因使用 NULL 值，不然尽量不要让字段的默认值为 NULL。
+
+### 10. 隐式转换可能带来不好的影响
+
+当 where 查询条件左右两侧类型不匹配的时候会发生隐式转换，隐式转换可能带来的不好影响就是导致索引失效而进行全表扫描。
+
+下面的案例中，brith_date 是字符串，然而匹配的是整数类型，从而发生隐式转换。
+
+```mysql
+select from users where brith_date = 19950610;
+```
+
 ## 重构查询方式
 
 ### 1. 切分大查询

From 30d980c04fb91efc5a348aef1faa92e313d16233 Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Wed, 8 Aug 2018 12:04:42 +0800
Subject: [PATCH 077/172] Update MySQL.md

---
 notes/MySQL.md | 4 ++++
 1 file changed, 4 insertions(+)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index f2949801..e49087bf 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -358,6 +358,10 @@ select * from users where age > 16 and age < 30 and height = 180
 select from users where brith_date = 19950610;
 ```
 
+### 11. like 语句的索引失效问题
+
+当我们使用 like “value%” 的方式时是会使用索引的，但是对于 like “%value%” 这样的方式，必定会执行全表查询，这在数据量小的表，不存在性能问题，但是对于海量数据，全表扫描是非常可怕的事情。所以，根据业务需求，考虑使用 ElasticSearch 或 Solr 是个不错的方案。
+
 ## 重构查询方式
 
 ### 1. 切分大查询

From 93a9fe11cb2b2c626206e5276132b1bc1961814a Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Wed, 8 Aug 2018 12:51:18 +0800
Subject: [PATCH 078/172] Update MySQL.md

---
 notes/MySQL.md | 2 ++
 1 file changed, 2 insertions(+)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index e49087bf..02dbe638 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -483,3 +483,5 @@ MySQL 读写分离能提高性能的原因在于：
 - [How to create unique row ID in sharded databases?](https://stackoverflow.com/questions/788829/how-to-create-unique-row-id-in-sharded-databases)
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
 - [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
+- [服务端指南 数据存储篇 | MySQL（04） 索引使用的注意事项](http://blog.720ui.com/2017/mysql_core_04_index_item/)
+- [mysql 聚簇索引 和聚簇索引 （二级索引）的 那些事](https://blog.csdn.net/bigtree_3721/article/details/51335479)

From 4b0db3f578c0414fe29847a7e70d56f557789745 Mon Sep 17 00:00:00 2001
From: Keqi Huang <keqih@smu.edu>
Date: Wed, 8 Aug 2018 14:10:19 +0800
Subject: [PATCH 079/172] =?UTF-8?q?Update=20=E6=95=B0=E6=8D=AE=E5=BA=93?=
 =?UTF-8?q?=E7=B3=BB=E7=BB=9F=E5=8E=9F=E7=90=86.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/数据库系统原理.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index 151ffc0f..6b8fcd86 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -389,11 +389,11 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 ## Gap Locks
 
-锁定一个范围内的索引，例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
+间隙锁锁住的是多行，是一个数据范围。间隙锁主要是为了防止出现幻读，但是它会把锁定范围扩大，有时候也会给我们带来麻烦。在数据库参数中，控制间隙锁的参数是 ```innodb_locks_unsafe_for_binlog```， 这个参数默认值是 OFF，也就是启用间隙锁，它是一个bool值，当值为 true 时表示 disable 间隙锁。那为了防止间隙锁是不是直接将 ```innodb_locaks_unsafe_for_binlog``` 设置为 ```true``` 就可以了呢？不一定！而且这个参数会影响到主从复制及灾难恢复，这个方法还尚待商量。
 
-```sql
-SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
-```
+间隙锁的出现主要集中在同一个事务中先 delete 后 insert 的情况下，当我们通过一个参数去删除一条记录的时候，如果参数在数据库中存在，那么这个时候产生的是普通行锁，锁住这个记录，然后删除，最后释放锁。如果这条记录不存在，问题就来了，数据库会扫描索引，发现这个记录不存在，这个时候的 delete 语句获取到的就是一个间隙锁，然后数据库会向左扫描扫到第一个比给定参数小的值，向右扫描扫描到第一个比给定参数大的值，然后以此为上下界限构建一个区间，进而锁住整个区间内的数据，一个特别容易出现死锁的间隙锁诞生了。
+
+通过修改 ```innodb_locaks_unsafe_for_binlog``` 参数来取消间隙锁从而达到避免这种情况的死锁的方式尚待商量，那就只有修改代码逻辑，存在才删除，尽量避免去删除不存在的记录。
 
 ## Next-Key Locks
 

From 9a1a7ef04780063398576266ee677f1ab86d0068 Mon Sep 17 00:00:00 2001
From: luocaodan <luocaodan@buaa.edu.cn>
Date: Wed, 8 Aug 2018 22:56:45 +0800
Subject: [PATCH 080/172] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87=20offer=20?=
 =?UTF-8?q?=E9=A2=98=E8=A7=A3.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/剑指 offer 题解.md | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 47d5d94e..5d9adbea 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1298,6 +1298,7 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
 
 ## 解题思路
 
+### 递归
 ```java
 public void Mirror(TreeNode root)
 {
@@ -1316,6 +1317,28 @@ private void swap(TreeNode root)
 }
 ```
 
+### 迭代
+```java
+public void Mirror(TreeNode root) {
+    if (root == null)
+        return;
+    Stack<TreeNode> stack = new Stack<>();
+    stack.push(root);
+    while (!stack.isEmpty()) {
+        TreeNode p = stack.pop();
+        if (p.left == null && p.right == null)
+            continue;
+        TreeNode left = p.left;
+        p.left = p.right;
+        p.right = left;
+        if (p.left != null)
+            stack.push(p.left);
+        if (p.right != null)
+            stack.push(p.right);
+    }
+}
+```
+
 # 28 对称的二叉树
 
 [NowCder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)

From c693786f509053f4d13fff4cf27082420a2de912 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Thu, 9 Aug 2018 22:57:33 +0800
Subject: [PATCH 081/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 2b4e688d..8dbf0507 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-qq-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->

From e7d09f99768866dcf830a5f2ddb27804f0f59521 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Thu, 9 Aug 2018 23:05:05 +0800
Subject: [PATCH 082/172] Update Group.md

---
 other/Group.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index fee7b8bf..60ff3946 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -12,4 +12,4 @@
 
 交流群不讨论政治，不讨论有争议性的话题，不发表仇视言论，不传播谣言，不发布广告（招聘信息之类的可以）。
 
-</br> <div align="center"><img src="group.jpg" width="300px"></div> </br>
+</br> <div align="center"><img src="group.png" width="300px"></div> </br>

From 0d7909db6d01bdfa0a0fd9abaee469b9ddd2e3a4 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Thu, 9 Aug 2018 23:05:46 +0800
Subject: [PATCH 083/172] Add files via upload

---
 other/group.png | Bin 0 -> 152701 bytes
 1 file changed, 0 insertions(+), 0 deletions(-)
 create mode 100644 other/group.png

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From dc6973b8e182a61fd17cb9fc23e66a75aa57d89b Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 9 Aug 2018 23:09:29 +0800
Subject: [PATCH 084/172] auto commit

---
 notes/HTTP.md            |   33 +-
 notes/Leetcode 题解.md   | 1971 +++++++++++++++++++-------------------
 notes/Linux.md           |    9 +-
 notes/MySQL.md           |    2 +-
 notes/剑指 offer 题解.md |  281 +++---
 notes/消息队列.md        |    6 +-
 notes/算法.md            |   30 +-
 notes/缓存.md            |   13 +-
 notes/设计模式.md        |   31 +-
 9 files changed, 1166 insertions(+), 1210 deletions(-)

diff --git a/notes/HTTP.md b/notes/HTTP.md
index 243dc283..537b24fe 100644
--- a/notes/HTTP.md
+++ b/notes/HTTP.md
@@ -45,7 +45,7 @@
     * [二进制分帧层](#二进制分帧层)
     * [服务端推送](#服务端推送)
     * [首部压缩](#首部压缩)
-* [八、GET 和 POST 的区别](#八get-和-post-的区别)
+* [八、GET 和 POST 比较](#八get-和-post-比较)
     * [作用](#作用)
     * [参数](#参数)
     * [安全](#安全)
@@ -61,13 +61,13 @@
 
 ## URL
 
-- URI（Uniform Resource Identifier，统一资源标识符）
-- URL（Uniform Resource Locator，统一资源定位符）
-- URN（Uniform Resource Name，统一资源名称），例如 urn:isbn:0-486-27557-4。
-
 URI 包含 URL 和 URN，目前 WEB 只有 URL 比较流行，所以见到的基本都是 URL。
 
-<div align="center"> <img src="../pics//f716427a-94f2-4875-9c86-98793cf5dcc3.jpg" width="400"/> </div><br>
+- URI（Uniform Resource Identifier，统一资源标识符）
+- URL（Uniform Resource Locator，统一资源定位符）
+- URN（Uniform Resource Name，统一资源名称）
+
+<div align="center"> <img src="../pics//urlnuri.jpg" width="600"/> </div><br>
 
 ## 请求和响应报文
 
@@ -197,7 +197,7 @@ CONNECT www.example.com:443 HTTP/1.1
 
 -  **204 No Content** ：请求已经成功处理，但是返回的响应报文不包含实体的主体部分。一般在只需要从客户端往服务器发送信息，而不需要返回数据时使用。
 
--  **206 Partial Content** ：表示客户端进行了范围请求。响应报文包含由 Content-Range 指定范围的实体内容。
+-  **206 Partial Content** ：表示客户端进行了范围请求，响应报文包含由 Content-Range 指定范围的实体内容。
 
 ## 3XX 重定向
 
@@ -219,7 +219,7 @@ CONNECT www.example.com:443 HTTP/1.1
 
 -  **401 Unauthorized** ：该状态码表示发送的请求需要有认证信息（BASIC 认证、DIGEST 认证）。如果之前已进行过一次请求，则表示用户认证失败。
 
--  **403 Forbidden** ：请求被拒绝，服务器端没有必要给出拒绝的详细理由。
+-  **403 Forbidden** ：请求被拒绝。
 
 -  **404 Not Found** 
 
@@ -331,7 +331,7 @@ Set-Cookie: tasty_cookie=strawberry
 [page content]
 ```
 
-客户端之后对同一个服务器发送请求时，会从浏览器中读出 Cookie 信息通过 Cookie 请求首部字段发送给服务器。
+客户端之后对同一个服务器发送请求时，会从浏览器中取出 Cookie 信息并通过 Cookie 请求首部字段发送给服务器。
 
 ```html
 GET /sample_page.html HTTP/1.1
@@ -382,9 +382,9 @@ Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径
 
 除了可以将用户信息通过 Cookie 存储在用户浏览器中，也可以利用 Session 存储在服务器端，存储在服务器端的信息更加安全。
 
-Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在内存型数据库中，比如 Redis，效率会更高。
+Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在 Redis 这种内存型数据库中，效率会更高。
 
-使用 Session 维护用户登录的过程如下：
+使用 Session 维护用户登录状态的过程如下：
 
 - 用户进行登录时，用户提交包含用户名和密码的表单，放入 HTTP 请求报文中；
 - 服务器验证该用户名和密码；
@@ -499,7 +499,7 @@ If-Modified-Since: Wed, 21 Oct 2015 07:28:00 GMT
 
 ### 1. 短连接与长连接
 
-当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源。如果每进行一次 HTTP 通信就要断开一次 TCP 连接，连接建立和断开的开销会很大。
+当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源。如果每进行一次 HTTP 通信就要新建一个 TCP 连接，那么开销会很大。
 
 长连接只需要建立一次 TCP 连接就能进行多次 HTTP 通信。
 
@@ -688,7 +688,7 @@ HTTPs 并不是新协议，而是让 HTTP 先和 SSL（Secure Sockets Layer）
 
 ### 3. HTTPs 采用的加密方式
 
-HTTPs 采用混合的加密机制，使用非对称密钥加密用于传输对称密钥来保证安全性，之后使用对称密钥加密进行通信来保证效率。（下图中的 Session Key 就是对称密钥）
+HTTPs 采用混合的加密机制，使用非对称密钥加密用于传输对称密钥来保证传输过程的安全性，之后使用对称密钥加密进行通信来保证通信过程的效率。（下图中的 Session Key 就是对称密钥）
 
 <div align="center"> <img src="../pics//How-HTTPS-Works.png" width="600"/> </div><br>
 
@@ -717,7 +717,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 ## HTTPs 的缺点
 
 - 因为需要进行加密解密等过程，因此速度会更慢；
-- 需要支付证书授权的高费用。
+- 需要支付证书授权的高额费用。
 
 ## 配置 HTTPs
 
@@ -727,7 +727,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 
 ## HTTP/1.x 缺陷
 
- HTTP/1.x 实现简单是以牺牲应用性能为代价的：
+ HTTP/1.x 实现简单是以牺牲性能为代价的：
 
 - 客户端需要使用多个连接才能实现并发和缩短延迟；
 - 不会压缩请求和响应首部，从而导致不必要的网络流量；
@@ -763,7 +763,7 @@ HTTP/2.0 要求客户端和服务器同时维护和更新一个包含之前见
 
 <div align="center"> <img src="../pics//_u4E0B_u8F7D.png" width="600"/> </div><br>
 
-# 八、GET 和 POST 的区别
+# 八、GET 和 POST 比较
 
 ## 作用
 
@@ -870,6 +870,7 @@ DELETE /idX/delete HTTP/1.1   -> Returns 404
 - [MDN : HTTP](https://developer.mozilla.org/en-US/docs/Web/HTTP)
 - [HTTP/2 简介](https://developers.google.com/web/fundamentals/performance/http2/?hl=zh-cn)
 - [htmlspecialchars](http://php.net/manual/zh/function.htmlspecialchars.php)
+- [Difference between file URI and URL in java](http://java2db.com/java-io/how-to-get-and-the-difference-between-file-uri-and-url-in-java)
 - [How to Fix SQL Injection Using Java PreparedStatement & CallableStatement](https://software-security.sans.org/developer-how-to/fix-sql-injection-in-java-using-prepared-callable-statement)
 - [浅谈 HTTP 中 Get 与 Post 的区别](https://www.cnblogs.com/hyddd/archive/2009/03/31/1426026.html)
 - [Are http:// and www really necessary?](https://www.webdancers.com/are-http-and-www-necesary/)
diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 293f4328..102fe9b6 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -33,10 +33,6 @@
         * [多数投票问题](#多数投票问题)
         * [其它](#其它)
 * [数据结构相关](#数据结构相关)
-    * [栈和队列](#栈和队列)
-    * [哈希表](#哈希表)
-    * [字符串](#字符串)
-    * [数组与矩阵](#数组与矩阵)
     * [链表](#链表)
     * [树](#树)
         * [递归](#递归)
@@ -44,6 +40,10 @@
         * [前中后序遍历](#前中后序遍历)
         * [BST](#bst)
         * [Trie](#trie)
+    * [栈和队列](#栈和队列)
+    * [哈希表](#哈希表)
+    * [字符串](#字符串)
+    * [数组与矩阵](#数组与矩阵)
     * [图](#图)
         * [二分图](#二分图)
         * [拓扑排序](#拓扑排序)
@@ -2371,7 +2371,9 @@ public List<Integer> diffWaysToCompute(String input) {
 
 题目描述：有 N 阶楼梯，每次可以上一阶或者两阶，求有多少种上楼梯的方法。
 
-定义一个数组 dp 存储上楼梯的方法数（为了方便讨论，数组下标从 1 开始），dp[i] 表示走到第 i 个楼梯的方法数目。第 i 个楼梯可以从第 i-1 和 i-2 个楼梯再走一步到达，走到第 i 个楼梯的方法数为走到第 i-1 和第 i-2 个楼梯的方法数之和。
+定义一个数组 dp 存储上楼梯的方法数（为了方便讨论，数组下标从 1 开始），dp[i] 表示走到第 i 个楼梯的方法数目。
+
+第 i 个楼梯可以从第 i-1 和 i-2 个楼梯再走一步到达，走到第 i 个楼梯的方法数为走到第 i-1 和第 i-2 个楼梯的方法数之和。
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]"/></div> <br>
 
@@ -2400,7 +2402,9 @@ public int climbStairs(int n) {
 
 题目描述：抢劫一排住户，但是不能抢邻近的住户，求最大抢劫量。
 
-定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以
+定义 dp 数组用来存储最大的抢劫量，其中 dp[i] 表示抢到第 i 个住户时的最大抢劫量。
+
+由于不能抢劫邻近住户，因此如果抢劫了第 i 个住户那么只能抢劫 i - 2 或者 i - 3 的住户，所以
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2],dp[i-3])+nums[i]"/></div> <br>
 
@@ -4032,973 +4036,6 @@ public int maximumProduct(int[] nums) {
 
 # 数据结构相关
 
-## 栈和队列
-
-**用栈实现队列** 
-
-[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
-
-```java
-class MyQueue {
-
-    private Stack<Integer> in = new Stack<>();
-    private Stack<Integer> out = new Stack<>();
-
-    public void push(int x) {
-        in.push(x);
-    }
-
-    public int pop() {
-        in2out();
-        return out.pop();
-    }
-
-    public int peek() {
-        in2out();
-        return out.peek();
-    }
-
-    private void in2out() {
-        if (out.isEmpty()) {
-            while (!in.isEmpty()) {
-                out.push(in.pop());
-            }
-        }
-    }
-
-    public boolean empty() {
-        return in.isEmpty() && out.isEmpty();
-    }
-}
-```
-
-**用队列实现栈** 
-
-[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
-
-在将一个元素 x 插入队列时，需要让除了 x 之外的所有元素出队列，再入队列。此时 x 在队首，第一个出队列，实现了后进先出顺序。
-
-```java
-class MyStack {
-
-    private Queue<Integer> queue;
-
-    public MyStack() {
-        queue = new LinkedList<>();
-    }
-
-    public void push(int x) {
-        queue.add(x);
-        int cnt = queue.size();
-        while (cnt-- > 1) {
-            queue.add(queue.poll());
-        }
-    }
-
-    public int pop() {
-        return queue.remove();
-    }
-
-    public int top() {
-        return queue.peek();
-    }
-
-    public boolean empty() {
-        return queue.isEmpty();
-    }
-}
-```
-
-**最小值栈** 
-
-[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
-
-```java
-class MinStack {
-
-    private Stack<Integer> dataStack;
-    private Stack<Integer> minStack;
-    private int min;
-
-    public MinStack() {
-        dataStack = new Stack<>();
-        minStack = new Stack<>();
-        min = Integer.MAX_VALUE;
-    }
-
-    public void push(int x) {
-        dataStack.add(x);
-        min = Math.min(min, x);
-        minStack.add(min);
-    }
-
-    public void pop() {
-        dataStack.pop();
-        minStack.pop();
-        min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
-    }
-
-    public int top() {
-        return dataStack.peek();
-    }
-
-    public int getMin() {
-        return minStack.peek();
-    }
-}
-```
-
-对于实现最小值队列问题，可以先将队列使用栈来实现，然后就将问题转换为最小值栈，这个问题出现在 编程之美：3.7。
-
-**用栈实现括号匹配** 
-
-[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
-
-```html
-"()[]{}"
-
-Output : true
-```
-
-```java
-public boolean isValid(String s) {
-    Stack<Character> stack = new Stack<>();
-    for (char c : s.toCharArray()) {
-        if (c == '(' || c == '{' || c == '[') {
-            stack.push(c);
-        } else {
-            if (stack.isEmpty()) {
-                return false;
-            }
-            char cStack = stack.pop();
-            boolean b1 = c == ')' && cStack != '(';
-            boolean b2 = c == ']' && cStack != '[';
-            boolean b3 = c == '}' && cStack != '{';
-            if (b1 || b2 || b3) {
-                return false;
-            }
-        }
-    }
-    return stack.isEmpty();
-}
-```
-
-**数组中元素与下一个比它大的元素之间的距离** 
-
-[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
-
-```html
-Input: [73, 74, 75, 71, 69, 72, 76, 73]
-Output: [1, 1, 4, 2, 1, 1, 0, 0]
-```
-
-在遍历数组时用 Stack 把数组中的数存起来，如果当前遍历的数比栈顶元素来的大，说明栈顶元素的下一个比它大的数就是当前元素。
-
-```java
-public int[] dailyTemperatures(int[] temperatures) {
-    int n = temperatures.length;
-    int[] dist = new int[n];
-    Stack<Integer> indexs = new Stack<>();
-    for (int curIndex = 0; curIndex < n; curIndex++) {
-        while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
-            int preIndex = indexs.pop();
-            dist[preIndex] = curIndex - preIndex;
-        }
-        indexs.add(curIndex);
-    }
-    return dist;
-}
-```
-
-**循环数组中比当前元素大的下一个元素** 
-
-[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
-
-```text
-Input: [1,2,1]
-Output: [2,-1,2]
-Explanation: The first 1's next greater number is 2; 
-The number 2 can't find next greater number; 
-The second 1's next greater number needs to search circularly, which is also 2.
-```
-
-与 739. Daily Temperatures (Medium) 不同的是，数组是循环数组，并且最后要求的不是距离而是下一个元素。
-
-```java
-public int[] nextGreaterElements(int[] nums) {
-    int n = nums.length;
-    int[] next = new int[n];
-    Arrays.fill(next, -1);
-    Stack<Integer> pre = new Stack<>();
-    for (int i = 0; i < n * 2; i++) {
-        int num = nums[i % n];
-        while (!pre.isEmpty() && nums[pre.peek()] < num) {
-            next[pre.pop()] = num;
-        }
-        if (i < n){
-            pre.push(i);
-        }
-    }
-    return next;
-}
-```
-
-## 哈希表
-
-哈希表使用 O(N) 空间复杂度存储数据，从而能够以 O(1) 时间复杂度求解问题。
-
-Java 中的  **HashSet**  用于存储一个集合，可以查找元素是否在集合中。
-
-如果元素有穷，并且范围不大，那么可以用一个布尔数组来存储一个元素是否存在。例如对于只有小写字符的元素，就可以用一个长度为 26 的布尔数组来存储一个字符集合，使得空间复杂度降低为 O(1)。
-
-Java 中的  **HashMap**  主要用于映射关系，从而把两个元素联系起来。
-
-在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中[Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)，利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
-
-HashMap 也可以用来对元素进行计数统计，此时键为元素，值为计数。和 HashSet 类似，如果元素有穷并且范围不大，可以用整型数组来进行统计。
-
-**数组中的两个数和为给定值** 
-
-[1. Two Sum (Easy)](https://leetcode.com/problems/two-sum/description/)
-
-可以先对数组进行排序，然后使用双指针方法或者二分查找方法。这样做的时间复杂度为 O(NlogN)，空间复杂度为 O(1)。
-
-用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i]，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
-
-```java
-public int[] twoSum(int[] nums, int target) {
-    HashMap<Integer, Integer> indexForNum = new HashMap<>();
-    for (int i = 0; i < nums.length; i++) {
-        if (indexForNum.containsKey(target - nums[i])) {
-            return new int[]{indexForNum.get(target - nums[i]), i};
-        } else {
-            indexForNum.put(nums[i], i);
-        }
-    }
-    return null;
-}
-```
-
-**判断数组是否含有重复元素** 
-
-[217. Contains Duplicate (Easy)](https://leetcode.com/problems/contains-duplicate/description/)
-
-```java
-public boolean containsDuplicate(int[] nums) {
-    Set<Integer> set = new HashSet<>();
-    for (int num : nums) {
-        set.add(num);
-    }
-    return set.size() < nums.length;
-}
-```
-
-**最长和谐序列** 
-
-[594. Longest Harmonious Subsequence (Easy)](https://leetcode.com/problems/longest-harmonious-subsequence/description/)
-
-```html
-Input: [1,3,2,2,5,2,3,7]
-Output: 5
-Explanation: The longest harmonious subsequence is [3,2,2,2,3].
-```
-
-和谐序列中最大数和最小数只差正好为 1，应该注意的是序列的元素不一定是数组的连续元素。
-
-```java
-public int findLHS(int[] nums) {
-    Map<Integer, Integer> countForNum = new HashMap<>();
-    for (int num : nums) {
-        countForNum.put(num, countForNum.getOrDefault(num, 0) + 1);
-    }
-    int longest = 0;
-    for (int num : countForNum.keySet()) {
-        if (countForNum.containsKey(num + 1)) {
-            longest = Math.max(longest, countForNum.get(num + 1) + countForNum.get(num));
-        }
-    }
-    return longest;
-}
-```
-
-**最长连续序列** 
-
-[128. Longest Consecutive Sequence (Hard)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
-
-```html
-Given [100, 4, 200, 1, 3, 2],
-The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
-```
-
-要求以 O(N) 的时间复杂度求解。
-
-```java
-public int longestConsecutive(int[] nums) {
-    Map<Integer, Integer> countForNum = new HashMap<>();
-    for (int num : nums) {
-        countForNum.put(num, 1);
-    }
-    for (int num : nums) {
-        forward(countForNum, num);
-    }
-    return maxCount(countForNum);
-}
-
-private int forward(Map<Integer, Integer> countForNum, int num) {
-    if (!countForNum.containsKey(num)) {
-        return 0;
-    }
-    int cnt = countForNum.get(num);
-    if (cnt > 1) {
-        return cnt;
-    }
-    cnt = forward(countForNum, num + 1) + 1;
-    countForNum.put(num, cnt);
-    return cnt;
-}
-
-private int maxCount(Map<Integer, Integer> countForNum) {
-    int max = 0;
-    for (int num : countForNum.keySet()) {
-        max = Math.max(max, countForNum.get(num));
-    }
-    return max;
-}
-```
-
-## 字符串
-
-**两个字符串包含的字符是否完全相同** 
-
-[242. Valid Anagram (Easy)](https://leetcode.com/problems/valid-anagram/description/)
-
-```html
-s = "anagram", t = "nagaram", return true.
-s = "rat", t = "car", return false.
-```
-
-字符串只包含小写字符，总共有 26 个小写字符。可以用 HashMap 来映射字符与出现次数。因为键的范围很小，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，然后比较两个字符串出现的字符数量是否相同。
-
-```java
-public boolean isAnagram(String s, String t) {
-    int[] cnts = new int[26];
-    for (char c : s.toCharArray()) {
-        cnts[c - 'a']++;
-    }
-    for (char c : t.toCharArray()) {
-        cnts[c - 'a']--;
-    }
-    for (int cnt : cnts) {
-        if (cnt != 0) {
-            return false;
-        }
-    }
-    return true;
-}
-```
-
-**计算一组字符集合可以组成的回文字符串的最大长度** 
-
-[409. Longest Palindrome (Easy)](https://leetcode.com/problems/longest-palindrome/description/)
-
-```html
-Input : "abccccdd"
-Output : 7
-Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
-```
-
-使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。
-
-因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
-
-```java
-public int longestPalindrome(String s) {
-    int[] cnts = new int[256];
-    for (char c : s.toCharArray()) {
-        cnts[c]++;
-    }
-    int palindrome = 0;
-    for (int cnt : cnts) {
-        palindrome += (cnt / 2) * 2;
-    }
-    if (palindrome < s.length()) {
-        palindrome++;   // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
-    }
-    return palindrome;
-}
-```
-
-**字符串同构** 
-
-[205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
-
-```html
-Given "egg", "add", return true.
-Given "foo", "bar", return false.
-Given "paper", "title", return true.
-```
-
-记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
-
-```java
-public boolean isIsomorphic(String s, String t) {
-    int[] preIndexOfS = new int[256];
-    int[] preIndexOfT = new int[256];
-    for (int i = 0; i < s.length(); i++) {
-        char sc = s.charAt(i), tc = t.charAt(i);
-        if (preIndexOfS[sc] != preIndexOfT[tc]) {
-            return false;
-        }
-        preIndexOfS[sc] = i + 1;
-        preIndexOfT[tc] = i + 1;
-    }
-    return true;
-}
-```
-
-**回文子字符串** 
-
-[647. Palindromic Substrings (Medium)](https://leetcode.com/problems/palindromic-substrings/description/)
-
-```html
-Input: "aaa"
-Output: 6
-Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
-```
-
-从字符串的某一位开始，尝试着去扩展子字符串。
-
-```java
-private int cnt = 0;
-
-public int countSubstrings(String s) {
-    for (int i = 0; i < s.length(); i++) {
-        extendSubstrings(s, i, i);     // 奇数长度
-        extendSubstrings(s, i, i + 1); // 偶数长度
-    }
-    return cnt;
-}
-
-private void extendSubstrings(String s, int start, int end) {
-    while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
-        start--;
-        end++;
-        cnt++;
-    }
-}
-```
-
-**判断一个整数是否是回文数** 
-
-[9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
-
-要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
-
-将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
-
-```java
-public boolean isPalindrome(int x) {
-    if (x == 0) {
-        return true;
-    }
-    if (x < 0 || x % 10 == 0) {
-        return false;
-    }
-    int right = 0;
-    while (x > right) {
-        right = right * 10 + x % 10;
-        x /= 10;
-    }
-    return x == right || x == right / 10;
-}
-```
-
-**统计二进制字符串中连续 1 和连续 0 数量相同的子字符串个数** 
-
-[696. Count Binary Substrings (Easy)](https://leetcode.com/problems/count-binary-substrings/description/)
-
-```html
-Input: "00110011"
-Output: 6
-Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
-```
-
-```java
-public int countBinarySubstrings(String s) {
-    int preLen = 0, curLen = 1, count = 0;
-    for (int i = 1; i < s.length(); i++) {
-        if (s.charAt(i) == s.charAt(i - 1)) {
-            curLen++;
-        } else {
-            preLen = curLen;
-            curLen = 1;
-        }
-
-        if (preLen >= curLen) {
-            count++;
-        }
-    }
-    return count;
-}
-```
-
-**字符串循环移位包含** 
-
-[编程之美：3.1](#)
-
-```html
-s1 = AABCD, s2 = CDAA
-Return : true
-```
-
-给定两个字符串 s1 和 s2，要求判定 s2 是否能够被 s1 做循环移位得到的字符串包含。
-
-s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2 是否是 s1s1 的子字符串即可。
-
-**字符串循环移位** 
-
-[编程之美：2.17](#)
-
-```html
-s = "abcd123" k = 3
-Return "123abcd"
-```
-
-将字符串向右循环移动 k 位。
-
-将 abcd123 中的 abcd 和 123 单独逆序，得到 dcba321，然后对整个字符串进行逆序，得到 123abcd。
-
-**字符串中单词的翻转** 
-
-[程序员代码面试指南](#)
-
-```html
-s = "I am a student"
-return "student a am I"
-```
-
-将每个单词逆序，然后将整个字符串逆序。
-
-## 数组与矩阵
-
-**把数组中的 0 移到末尾** 
-
-[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
-
-```html
-For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
-```
-
-```java
-public void moveZeroes(int[] nums) {
-    int idx = 0;
-    for (int num : nums) {
-        if (num != 0) {
-            nums[idx++] = num;
-        }
-    }
-    while (idx < nums.length) {
-        nums[idx++] = 0;
-    }
-}
-```
-
-**改变矩阵维度** 
-
-[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
-
-```html
-Input:
-nums =
-[[1,2],
- [3,4]]
-r = 1, c = 4
-
-Output:
-[[1,2,3,4]]
-
-Explanation:
-The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
-```
-
-```java
-public int[][] matrixReshape(int[][] nums, int r, int c) {
-    int m = nums.length, n = nums[0].length;
-    if (m * n != r * c) {
-        return nums;
-    }
-    int[][] reshapedNums = new int[r][c];
-    int index = 0;
-    for (int i = 0; i < r; i++) {
-        for (int j = 0; j < c; j++) {
-            reshapedNums[i][j] = nums[index / n][index % n];
-            index++;
-        }
-    }
-    return reshapedNums;
-}
-```
-
-**找出数组中最长的连续 1** 
-
-[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
-
-```java
-public int findMaxConsecutiveOnes(int[] nums) {
-    int max = 0, cur = 0;
-    for (int x : nums) {
-        cur = x == 0 ? 0 : cur + 1;
-        max = Math.max(max, cur);
-    }
-    return max;
-}
-```
-
-**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数** 
-
-[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
-
-```html
-Input: nums = [1,2,2,4]
-Output: [2,3]
-```
-
-```html
-Input: nums = [1,2,2,4]
-Output: [2,3]
-```
-
-最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
-
-主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。遍历数组，如果第 i 位上的元素不是 i + 1，那么一直交换第 i 位和 nums[i] - 1 位置上的元素。
-
-```java
-public int[] findErrorNums(int[] nums) {
-    for (int i = 0; i < nums.length; i++) {
-        while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
-            swap(nums, i, nums[i] - 1);
-        }
-    }
-    for (int i = 0; i < nums.length; i++) {
-        if (nums[i] != i + 1) {
-            return new int[]{nums[i], i + 1};
-        }
-    }
-    return null;
-}
-
-private void swap(int[] nums, int i, int j) {
-    int tmp = nums[i];
-    nums[i] = nums[j];
-    nums[j] = tmp;
-}
-```
-
-类似题目：
-
-- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
-- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
-
-**找出数组中重复的数，数组值在 [1, n] 之间** 
-
-[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
-
-要求不能修改数组，也不能使用额外的空间。
-
-二分查找解法：
-
-```java
-public int findDuplicate(int[] nums) {
-     int l = 1, h = nums.length - 1;
-     while (l <= h) {
-         int mid = l + (h - l) / 2;
-         int cnt = 0;
-         for (int i = 0; i < nums.length; i++) {
-             if (nums[i] <= mid) cnt++;
-         }
-         if (cnt > mid) h = mid - 1;
-         else l = mid + 1;
-     }
-     return l;
-}
-```
-
-双指针解法，类似于有环链表中找出环的入口：
-
-```java
-public int findDuplicate(int[] nums) {
-    int slow = nums[0], fast = nums[nums[0]];
-    while (slow != fast) {
-        slow = nums[slow];
-        fast = nums[nums[fast]];
-    }
-    fast = 0;
-    while (slow != fast) {
-        slow = nums[slow];
-        fast = nums[fast];
-    }
-    return slow;
-}
-```
-
-**有序矩阵查找** 
-
-[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
-
-```html
-[
-   [ 1,  5,  9],
-   [10, 11, 13],
-   [12, 13, 15]
-]
-```
-
-```java
-public boolean searchMatrix(int[][] matrix, int target) {
-    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
-    int m = matrix.length, n = matrix[0].length;
-    int row = 0, col = n - 1;
-    while (row < m && col >= 0) {
-        if (target == matrix[row][col]) return true;
-        else if (target < matrix[row][col]) col--;
-        else row++;
-    }
-    return false;
-}
-```
-
-**有序矩阵的 Kth Element** 
-
-[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
-
-```html
-matrix = [
-  [ 1,  5,  9],
-  [10, 11, 13],
-  [12, 13, 15]
-],
-k = 8,
-
-return 13.
-```
-
-解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
-
-二分查找解法：
-
-```java
-public int kthSmallest(int[][] matrix, int k) {
-    int m = matrix.length, n = matrix[0].length;
-    int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
-    while(lo <= hi) {
-        int mid = lo + (hi - lo) / 2;
-        int cnt = 0;
-        for(int i = 0; i < m; i++) {
-            for(int j = 0; j < n && matrix[i][j] <= mid; j++) {
-                cnt++;
-            }
-        }
-        if(cnt < k) lo = mid + 1;
-        else hi = mid - 1;
-    }
-    return lo;
-}
-```
-
-堆解法：
-
-```java
-public int kthSmallest(int[][] matrix, int k) {
-    int m = matrix.length, n = matrix[0].length;
-    PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
-    for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
-    for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
-        Tuple t = pq.poll();
-        if(t.x == m - 1) continue;
-        pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
-    }
-    return pq.poll().val;
-}
-
-class Tuple implements Comparable<Tuple> {
-    int x, y, val;
-    public Tuple(int x, int y, int val) {
-        this.x = x; this.y = y; this.val = val;
-    }
-
-    @Override
-    public int compareTo(Tuple that) {
-        return this.val - that.val;
-    }
-}
-```
-
-**数组相邻差值的个数** 
-
-[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
-
-```html
-Input: n = 3, k = 2
-Output: [1, 3, 2]
-Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
-```
-
-题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
-
-让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
-
-```java
-public int[] constructArray(int n, int k) {
-    int[] ret = new int[n];
-    ret[0] = 1;
-    for (int i = 1, interval = k; i <= k; i++, interval--) {
-        ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
-    }
-    for (int i = k + 1; i < n; i++) {
-        ret[i] = i + 1;
-    }
-    return ret;
-}
-```
-
-**数组的度** 
-
-[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
-
-```html
-Input: [1,2,2,3,1,4,2]
-Output: 6
-```
-
-题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
-
-```java
-public int findShortestSubArray(int[] nums) {
-    Map<Integer, Integer> numsCnt = new HashMap<>();
-    Map<Integer, Integer> numsLastIndex = new HashMap<>();
-    Map<Integer, Integer> numsFirstIndex = new HashMap<>();
-    for (int i = 0; i < nums.length; i++) {
-        int num = nums[i];
-        numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
-        numsLastIndex.put(num, i);
-        if (!numsFirstIndex.containsKey(num)) {
-            numsFirstIndex.put(num, i);
-        }
-    }
-    int maxCnt = 0;
-    for (int num : nums) {
-        maxCnt = Math.max(maxCnt, numsCnt.get(num));
-    }
-    int ret = nums.length;
-    for (int i = 0; i < nums.length; i++) {
-        int num = nums[i];
-        int cnt = numsCnt.get(num);
-        if (cnt != maxCnt) continue;
-        ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
-    }
-    return ret;
-}
-```
-
-**对角元素相等的矩阵** 
-
-[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
-
-```html
-1234
-5123
-9512
-
-In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
-```
-
-```java
-public boolean isToeplitzMatrix(int[][] matrix) {
-    for (int i = 0; i < matrix[0].length; i++) {
-        if (!check(matrix, matrix[0][i], 0, i)) {
-            return false;
-        }
-    }
-    for (int i = 0; i < matrix.length; i++) {
-        if (!check(matrix, matrix[i][0], i, 0)) {
-            return false;
-        }
-    }
-    return true;
-}
-
-private boolean check(int[][] matrix, int expectValue, int row, int col) {
-    if (row >= matrix.length || col >= matrix[0].length) {
-        return true;
-    }
-    if (matrix[row][col] != expectValue) {
-        return false;
-    }
-    return check(matrix, expectValue, row + 1, col + 1);
-}
-```
-
-**嵌套数组** 
-
-[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
-
-```html
-Input: A = [5,4,0,3,1,6,2]
-Output: 4
-Explanation:
-A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
-
-One of the longest S[K]:
-S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
-```
-
-题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
-
-```java
-public int arrayNesting(int[] nums) {
-    int max = 0;
-    for (int i = 0; i < nums.length; i++) {
-        int cnt = 0;
-        for (int j = i; nums[j] != -1; ) {
-            cnt++;
-            int t = nums[j];
-            nums[j] = -1; // 标记该位置已经被访问
-            j = t;
-
-        }
-        max = Math.max(max, cnt);
-    }
-    return max;
-}
-```
-
-**分隔数组** 
-
-[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
-
-```html
-Input: arr = [1,0,2,3,4]
-Output: 4
-Explanation:
-We can split into two chunks, such as [1, 0], [2, 3, 4].
-However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
-```
-
-题目描述：分隔数组，使得对每部分排序后数组就为有序。
-
-```java
-public int maxChunksToSorted(int[] arr) {
-    if (arr == null) return 0;
-    int ret = 0;
-    int right = arr[0];
-    for (int i = 0; i < arr.length; i++) {
-        right = Math.max(right, arr[i]);
-        if (right == i) ret++;
-    }
-    return ret;
-}
-```
-
 ## 链表
 
 链表是空节点，或者有一个值和一个指向下一个链表的指针，因此很多链表问题可以用递归来处理。
@@ -5032,7 +4069,10 @@ public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 }
 ```
 
-如果只是判断是否存在交点，那么就是另一个问题，即 [编程之美：3.6]() 的问题。有两种解法：把第一个链表的结尾连接到第二个链表的开头，看第二个链表是否存在环；或者直接比较两个链表的最后一个节点是否相同。
+如果只是判断是否存在交点，那么就是另一个问题，即 [编程之美 3.6]() 的问题。有两种解法：
+
+- 把第一个链表的结尾连接到第二个链表的开头，看第二个链表是否存在环；
+- 或者直接比较两个链表的最后一个节点是否相同。
 
 **链表反转** 
 
@@ -5097,7 +4137,7 @@ Given 1->1->2->3->3, return 1->2->3.
 
 ```java
 public ListNode deleteDuplicates(ListNode head) {
-    if(head == null || head.next == null) return head;
+    if (head == null || head.next == null) return head;
     head.next = deleteDuplicates(head.next);
     return head.val == head.next.val ? head.next : head;
 }
@@ -5419,8 +4459,8 @@ Input:
         3   2                     1   3
        /                           \   \
       5                             4   7
+
 Output:
-Merged tree:
          3
         / \
        4   5
@@ -5446,6 +4486,7 @@ public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
 
 ```html
 Given the below binary tree and sum = 22,
+
               5
              / \
             4   8
@@ -5453,10 +4494,11 @@ Given the below binary tree and sum = 22,
           11  13  4
          /  \      \
         7    2      1
+
 return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
 ```
 
-路径和定义为从 root 到 leaf 的所有节点的和
+路径和定义为从 root 到 leaf 的所有节点的和。
 
 ```java
 public boolean hasPathSum(TreeNode root, int sum) {
@@ -5848,7 +4890,7 @@ public List<Integer> preorderTraversal(TreeNode root) {
 
 [145. Binary Tree Postorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-postorder-traversal/description/)
 
-前序遍历为 root -> left -> right，后序遍历为 left -> right -> root，可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
+前序遍历为 root -> left -> right，后序遍历为 left -> right -> root。可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
 
 ```java
 public List<Integer> postorderTraversal(TreeNode root) {
@@ -6106,7 +5148,7 @@ public TreeNode sortedListToBST(ListNode head) {
     if (head == null) return null;
     if (head.next == null) return new TreeNode(head.val);
     ListNode preMid = preMid(head);
-    ListNode mid = preMid.next; 
+    ListNode mid = preMid.next;
     preMid.next = null;  // 断开链表
     TreeNode t = new TreeNode(mid.val);
     t.left = sortedListToBST(head);
@@ -6222,7 +5264,7 @@ private void inOrder(TreeNode node) {
 return [2].
 ```
 
-答案可能不止一个，也就是有多个值出现的次数一样多，并且是最大的。
+答案可能不止一个，也就是有多个值出现的次数一样多。
 
 ```java
 private int curCnt = 1;
@@ -6392,6 +5434,974 @@ class MapSum {
 }
 ```
 
+
+## 栈和队列
+
+**用栈实现队列** 
+
+[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
+
+栈的顺序为后进先出，而队列的顺序为先进先出。使用两个栈实现队列，一个元素需要经过两个栈才能出队列，在经过第一个栈时元素顺序被反转，经过第二个栈时再次被反转，此时就是先进先出顺序。
+
+```java
+class MyQueue {
+
+    private Stack<Integer> in = new Stack<>();
+    private Stack<Integer> out = new Stack<>();
+
+    public void push(int x) {
+        in.push(x);
+    }
+
+    public int pop() {
+        in2out();
+        return out.pop();
+    }
+
+    public int peek() {
+        in2out();
+        return out.peek();
+    }
+
+    private void in2out() {
+        if (out.isEmpty()) {
+            while (!in.isEmpty()) {
+                out.push(in.pop());
+            }
+        }
+    }
+
+    public boolean empty() {
+        return in.isEmpty() && out.isEmpty();
+    }
+}
+```
+
+**用队列实现栈** 
+
+[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
+
+在将一个元素 x 插入队列时，为了维护原来的后进先出顺序，需要让 x 插入队列首部。而队列的默认插入顺序是队列尾部，因此在将 x 插入队列尾部之后，需要让除了 x 之外的所有元素出队列，再入队列。
+
+```java
+class MyStack {
+
+    private Queue<Integer> queue;
+
+    public MyStack() {
+        queue = new LinkedList<>();
+    }
+
+    public void push(int x) {
+        queue.add(x);
+        int cnt = queue.size();
+        while (cnt-- > 1) {
+            queue.add(queue.poll());
+        }
+    }
+
+    public int pop() {
+        return queue.remove();
+    }
+
+    public int top() {
+        return queue.peek();
+    }
+
+    public boolean empty() {
+        return queue.isEmpty();
+    }
+}
+```
+
+**最小值栈** 
+
+[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
+
+```java
+class MinStack {
+
+    private Stack<Integer> dataStack;
+    private Stack<Integer> minStack;
+    private int min;
+
+    public MinStack() {
+        dataStack = new Stack<>();
+        minStack = new Stack<>();
+        min = Integer.MAX_VALUE;
+    }
+
+    public void push(int x) {
+        dataStack.add(x);
+        min = Math.min(min, x);
+        minStack.add(min);
+    }
+
+    public void pop() {
+        dataStack.pop();
+        minStack.pop();
+        min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
+    }
+
+    public int top() {
+        return dataStack.peek();
+    }
+
+    public int getMin() {
+        return minStack.peek();
+    }
+}
+```
+
+对于实现最小值队列问题，可以先将队列使用栈来实现，然后就将问题转换为最小值栈，这个问题出现在 编程之美：3.7。
+
+**用栈实现括号匹配** 
+
+[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
+
+```html
+"()[]{}"
+
+Output : true
+```
+
+```java
+public boolean isValid(String s) {
+    Stack<Character> stack = new Stack<>();
+    for (char c : s.toCharArray()) {
+        if (c == '(' || c == '{' || c == '[') {
+            stack.push(c);
+        } else {
+            if (stack.isEmpty()) {
+                return false;
+            }
+            char cStack = stack.pop();
+            boolean b1 = c == ')' && cStack != '(';
+            boolean b2 = c == ']' && cStack != '[';
+            boolean b3 = c == '}' && cStack != '{';
+            if (b1 || b2 || b3) {
+                return false;
+            }
+        }
+    }
+    return stack.isEmpty();
+}
+```
+
+**数组中元素与下一个比它大的元素之间的距离** 
+
+[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
+
+```html
+Input: [73, 74, 75, 71, 69, 72, 76, 73]
+Output: [1, 1, 4, 2, 1, 1, 0, 0]
+```
+
+在遍历数组时用栈把数组中的数存起来，如果当前遍历的数比栈顶元素来的大，说明栈顶元素的下一个比它大的数就是当前元素。
+
+```java
+public int[] dailyTemperatures(int[] temperatures) {
+    int n = temperatures.length;
+    int[] dist = new int[n];
+    Stack<Integer> indexs = new Stack<>();
+    for (int curIndex = 0; curIndex < n; curIndex++) {
+        while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
+            int preIndex = indexs.pop();
+            dist[preIndex] = curIndex - preIndex;
+        }
+        indexs.add(curIndex);
+    }
+    return dist;
+}
+```
+
+**循环数组中比当前元素大的下一个元素** 
+
+[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
+
+```text
+Input: [1,2,1]
+Output: [2,-1,2]
+Explanation: The first 1's next greater number is 2;
+The number 2 can't find next greater number;
+The second 1's next greater number needs to search circularly, which is also 2.
+```
+
+与 739. Daily Temperatures (Medium) 不同的是，数组是循环数组，并且最后要求的不是距离而是下一个元素。
+
+```java
+public int[] nextGreaterElements(int[] nums) {
+    int n = nums.length;
+    int[] next = new int[n];
+    Arrays.fill(next, -1);
+    Stack<Integer> pre = new Stack<>();
+    for (int i = 0; i < n * 2; i++) {
+        int num = nums[i % n];
+        while (!pre.isEmpty() && nums[pre.peek()] < num) {
+            next[pre.pop()] = num;
+        }
+        if (i < n){
+            pre.push(i);
+        }
+    }
+    return next;
+}
+```
+
+## 哈希表
+
+哈希表使用 O(N) 空间复杂度存储数据，并且以 O(1) 时间复杂度求解问题。
+
+- Java 中的  **HashSet**  用于存储一个集合，可以查找元素是否在集合中。如果元素有穷，并且范围不大，那么可以用一个布尔数组来存储一个元素是否存在。例如对于只有小写字符的元素，就可以用一个长度为 26 的布尔数组来存储一个字符集合，使得空间复杂度降低为 O(1)。
+
+- Java 中的  **HashMap**  主要用于映射关系，从而把两个元素联系起来。HashMap 也可以用来对元素进行计数统计，此时键为元素，值为计数。和 HashSet 类似，如果元素有穷并且范围不大，可以用整型数组来进行统计。在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中 [Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)，利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
+
+
+**数组中两个数的和为给定值** 
+
+[1. Two Sum (Easy)](https://leetcode.com/problems/two-sum/description/)
+
+可以先对数组进行排序，然后使用双指针方法或者二分查找方法。这样做的时间复杂度为 O(NlogN)，空间复杂度为 O(1)。
+
+用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i]，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
+
+```java
+public int[] twoSum(int[] nums, int target) {
+    HashMap<Integer, Integer> indexForNum = new HashMap<>();
+    for (int i = 0; i < nums.length; i++) {
+        if (indexForNum.containsKey(target - nums[i])) {
+            return new int[]{indexForNum.get(target - nums[i]), i};
+        } else {
+            indexForNum.put(nums[i], i);
+        }
+    }
+    return null;
+}
+```
+
+**判断数组是否含有重复元素** 
+
+[217. Contains Duplicate (Easy)](https://leetcode.com/problems/contains-duplicate/description/)
+
+```java
+public boolean containsDuplicate(int[] nums) {
+    Set<Integer> set = new HashSet<>();
+    for (int num : nums) {
+        set.add(num);
+    }
+    return set.size() < nums.length;
+}
+```
+
+**最长和谐序列** 
+
+[594. Longest Harmonious Subsequence (Easy)](https://leetcode.com/problems/longest-harmonious-subsequence/description/)
+
+```html
+Input: [1,3,2,2,5,2,3,7]
+Output: 5
+Explanation: The longest harmonious subsequence is [3,2,2,2,3].
+```
+
+和谐序列中最大数和最小数只差正好为 1，应该注意的是序列的元素不一定是数组的连续元素。
+
+```java
+public int findLHS(int[] nums) {
+    Map<Integer, Integer> countForNum = new HashMap<>();
+    for (int num : nums) {
+        countForNum.put(num, countForNum.getOrDefault(num, 0) + 1);
+    }
+    int longest = 0;
+    for (int num : countForNum.keySet()) {
+        if (countForNum.containsKey(num + 1)) {
+            longest = Math.max(longest, countForNum.get(num + 1) + countForNum.get(num));
+        }
+    }
+    return longest;
+}
+```
+
+**最长连续序列** 
+
+[128. Longest Consecutive Sequence (Hard)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
+
+```html
+Given [100, 4, 200, 1, 3, 2],
+The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
+```
+
+要求以 O(N) 的时间复杂度求解。
+
+```java
+public int longestConsecutive(int[] nums) {
+    Map<Integer, Integer> countForNum = new HashMap<>();
+    for (int num : nums) {
+        countForNum.put(num, 1);
+    }
+    for (int num : nums) {
+        forward(countForNum, num);
+    }
+    return maxCount(countForNum);
+}
+
+private int forward(Map<Integer, Integer> countForNum, int num) {
+    if (!countForNum.containsKey(num)) {
+        return 0;
+    }
+    int cnt = countForNum.get(num);
+    if (cnt > 1) {
+        return cnt;
+    }
+    cnt = forward(countForNum, num + 1) + 1;
+    countForNum.put(num, cnt);
+    return cnt;
+}
+
+private int maxCount(Map<Integer, Integer> countForNum) {
+    int max = 0;
+    for (int num : countForNum.keySet()) {
+        max = Math.max(max, countForNum.get(num));
+    }
+    return max;
+}
+```
+
+## 字符串
+
+**字符串循环移位包含** 
+
+[编程之美 3.1](#)
+
+```html
+s1 = AABCD, s2 = CDAA
+Return : true
+```
+
+给定两个字符串 s1 和 s2，要求判定 s2 是否能够被 s1 做循环移位得到的字符串包含。
+
+s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2 是否是 s1s1 的子字符串即可。
+
+**字符串循环移位** 
+
+[编程之美 2.17](#)
+
+```html
+s = "abcd123" k = 3
+Return "123abcd"
+```
+
+将字符串向右循环移动 k 位。
+
+将 abcd123 中的 abcd 和 123 单独翻转，得到 dcba321，然后对整个字符串进行翻转，得到 123abcd。
+
+**字符串中单词的翻转** 
+
+[程序员代码面试指南](#)
+
+```html
+s = "I am a student"
+Return "student a am I"
+```
+
+将每个单词翻转，然后将整个字符串翻转。
+
+**两个字符串包含的字符是否完全相同** 
+
+[242. Valid Anagram (Easy)](https://leetcode.com/problems/valid-anagram/description/)
+
+```html
+s = "anagram", t = "nagaram", return true.
+s = "rat", t = "car", return false.
+```
+
+可以用 HashMap 来映射字符与出现次数，然后比较两个字符串出现的字符数量是否相同。
+
+由于本题的字符串只包含 26 个小写字符，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，不再使用 HashMap。
+
+```java
+public boolean isAnagram(String s, String t) {
+    int[] cnts = new int[26];
+    for (char c : s.toCharArray()) {
+        cnts[c - 'a']++;
+    }
+    for (char c : t.toCharArray()) {
+        cnts[c - 'a']--;
+    }
+    for (int cnt : cnts) {
+        if (cnt != 0) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+**计算一组字符集合可以组成的回文字符串的最大长度** 
+
+[409. Longest Palindrome (Easy)](https://leetcode.com/problems/longest-palindrome/description/)
+
+```html
+Input : "abccccdd"
+Output : 7
+Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
+```
+
+使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。
+
+因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
+
+```java
+public int longestPalindrome(String s) {
+    int[] cnts = new int[256];
+    for (char c : s.toCharArray()) {
+        cnts[c]++;
+    }
+    int palindrome = 0;
+    for (int cnt : cnts) {
+        palindrome += (cnt / 2) * 2;
+    }
+    if (palindrome < s.length()) {
+        palindrome++;   // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
+    }
+    return palindrome;
+}
+```
+
+**字符串同构** 
+
+[205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
+
+```html
+Given "egg", "add", return true.
+Given "foo", "bar", return false.
+Given "paper", "title", return true.
+```
+
+记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
+
+```java
+public boolean isIsomorphic(String s, String t) {
+    int[] preIndexOfS = new int[256];
+    int[] preIndexOfT = new int[256];
+    for (int i = 0; i < s.length(); i++) {
+        char sc = s.charAt(i), tc = t.charAt(i);
+        if (preIndexOfS[sc] != preIndexOfT[tc]) {
+            return false;
+        }
+        preIndexOfS[sc] = i + 1;
+        preIndexOfT[tc] = i + 1;
+    }
+    return true;
+}
+```
+
+**回文子字符串个数** 
+
+[647. Palindromic Substrings (Medium)](https://leetcode.com/problems/palindromic-substrings/description/)
+
+```html
+Input: "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+```
+
+从字符串的某一位开始，尝试着去扩展子字符串。
+
+```java
+private int cnt = 0;
+
+public int countSubstrings(String s) {
+    for (int i = 0; i < s.length(); i++) {
+        extendSubstrings(s, i, i);     // 奇数长度
+        extendSubstrings(s, i, i + 1); // 偶数长度
+    }
+    return cnt;
+}
+
+private void extendSubstrings(String s, int start, int end) {
+    while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
+        start--;
+        end++;
+        cnt++;
+    }
+}
+```
+
+**判断一个整数是否是回文数** 
+
+[9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
+
+要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
+
+将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
+
+```java
+public boolean isPalindrome(int x) {
+    if (x == 0) {
+        return true;
+    }
+    if (x < 0 || x % 10 == 0) {
+        return false;
+    }
+    int right = 0;
+    while (x > right) {
+        right = right * 10 + x % 10;
+        x /= 10;
+    }
+    return x == right || x == right / 10;
+}
+```
+
+**统计二进制字符串中连续 1 和连续 0 数量相同的子字符串个数** 
+
+[696. Count Binary Substrings (Easy)](https://leetcode.com/problems/count-binary-substrings/description/)
+
+```html
+Input: "00110011"
+Output: 6
+Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
+```
+
+```java
+public int countBinarySubstrings(String s) {
+    int preLen = 0, curLen = 1, count = 0;
+    for (int i = 1; i < s.length(); i++) {
+        if (s.charAt(i) == s.charAt(i - 1)) {
+            curLen++;
+        } else {
+            preLen = curLen;
+            curLen = 1;
+        }
+
+        if (preLen >= curLen) {
+            count++;
+        }
+    }
+    return count;
+}
+```
+
+## 数组与矩阵
+
+**把数组中的 0 移到末尾** 
+
+[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
+
+```html
+For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
+```
+
+```java
+public void moveZeroes(int[] nums) {
+    int idx = 0;
+    for (int num : nums) {
+        if (num != 0) {
+            nums[idx++] = num;
+        }
+    }
+    while (idx < nums.length) {
+        nums[idx++] = 0;
+    }
+}
+```
+
+**改变矩阵维度** 
+
+[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
+
+```html
+Input:
+nums =
+[[1,2],
+ [3,4]]
+r = 1, c = 4
+
+Output:
+[[1,2,3,4]]
+
+Explanation:
+The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
+```
+
+```java
+public int[][] matrixReshape(int[][] nums, int r, int c) {
+    int m = nums.length, n = nums[0].length;
+    if (m * n != r * c) {
+        return nums;
+    }
+    int[][] reshapedNums = new int[r][c];
+    int index = 0;
+    for (int i = 0; i < r; i++) {
+        for (int j = 0; j < c; j++) {
+            reshapedNums[i][j] = nums[index / n][index % n];
+            index++;
+        }
+    }
+    return reshapedNums;
+}
+```
+
+**找出数组中最长的连续 1** 
+
+[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
+
+```java
+public int findMaxConsecutiveOnes(int[] nums) {
+    int max = 0, cur = 0;
+    for (int x : nums) {
+        cur = x == 0 ? 0 : cur + 1;
+        max = Math.max(max, cur);
+    }
+    return max;
+}
+```
+
+**有序矩阵查找** 
+
+[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
+
+```html
+[
+   [ 1,  5,  9],
+   [10, 11, 13],
+   [12, 13, 15]
+]
+```
+
+```java
+public boolean searchMatrix(int[][] matrix, int target) {
+    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
+    int m = matrix.length, n = matrix[0].length;
+    int row = 0, col = n - 1;
+    while (row < m && col >= 0) {
+        if (target == matrix[row][col]) return true;
+        else if (target < matrix[row][col]) col--;
+        else row++;
+    }
+    return false;
+}
+```
+
+**有序矩阵的 Kth Element** 
+
+[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
+
+```html
+matrix = [
+  [ 1,  5,  9],
+  [10, 11, 13],
+  [12, 13, 15]
+],
+k = 8,
+
+return 13.
+```
+
+解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
+
+二分查找解法：
+
+```java
+public int kthSmallest(int[][] matrix, int k) {
+    int m = matrix.length, n = matrix[0].length;
+    int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
+    while (lo <= hi) {
+        int mid = lo + (hi - lo) / 2;
+        int cnt = 0;
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
+                cnt++;
+            }
+        }
+        if (cnt < k) lo = mid + 1;
+        else hi = mid - 1;
+    }
+    return lo;
+}
+```
+
+堆解法：
+
+```java
+public int kthSmallest(int[][] matrix, int k) {
+    int m = matrix.length, n = matrix[0].length;
+    PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
+    for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
+    for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
+        Tuple t = pq.poll();
+        if(t.x == m - 1) continue;
+        pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
+    }
+    return pq.poll().val;
+}
+
+class Tuple implements Comparable<Tuple> {
+    int x, y, val;
+    public Tuple(int x, int y, int val) {
+        this.x = x; this.y = y; this.val = val;
+    }
+
+    @Override
+    public int compareTo(Tuple that) {
+        return this.val - that.val;
+    }
+}
+```
+
+**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数** 
+
+[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
+
+```html
+Input: nums = [1,2,2,4]
+Output: [2,3]
+```
+
+```html
+Input: nums = [1,2,2,4]
+Output: [2,3]
+```
+
+最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
+
+主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
+
+```java
+public int[] findErrorNums(int[] nums) {
+    for (int i = 0; i < nums.length; i++) {
+        while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
+            swap(nums, i, nums[i] - 1);
+        }
+    }
+    for (int i = 0; i < nums.length; i++) {
+        if (nums[i] != i + 1) {
+            return new int[]{nums[i], i + 1};
+        }
+    }
+    return null;
+}
+
+private void swap(int[] nums, int i, int j) {
+    int tmp = nums[i];
+    nums[i] = nums[j];
+    nums[j] = tmp;
+}
+```
+
+类似题目：
+
+- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
+- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
+
+**找出数组中重复的数，数组值在 [1, n] 之间** 
+
+[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
+
+要求不能修改数组，也不能使用额外的空间。
+
+二分查找解法：
+
+```java
+public int findDuplicate(int[] nums) {
+     int l = 1, h = nums.length - 1;
+     while (l <= h) {
+         int mid = l + (h - l) / 2;
+         int cnt = 0;
+         for (int i = 0; i < nums.length; i++) {
+             if (nums[i] <= mid) cnt++;
+         }
+         if (cnt > mid) h = mid - 1;
+         else l = mid + 1;
+     }
+     return l;
+}
+```
+
+双指针解法，类似于有环链表中找出环的入口：
+
+```java
+public int findDuplicate(int[] nums) {
+    int slow = nums[0], fast = nums[nums[0]];
+    while (slow != fast) {
+        slow = nums[slow];
+        fast = nums[nums[fast]];
+    }
+    fast = 0;
+    while (slow != fast) {
+        slow = nums[slow];
+        fast = nums[fast];
+    }
+    return slow;
+}
+```
+
+**数组相邻差值的个数** 
+
+[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
+
+```html
+Input: n = 3, k = 2
+Output: [1, 3, 2]
+Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
+```
+
+题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
+
+让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
+
+```java
+public int[] constructArray(int n, int k) {
+    int[] ret = new int[n];
+    ret[0] = 1;
+    for (int i = 1, interval = k; i <= k; i++, interval--) {
+        ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
+    }
+    for (int i = k + 1; i < n; i++) {
+        ret[i] = i + 1;
+    }
+    return ret;
+}
+```
+
+**数组的度** 
+
+[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
+
+```html
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
+
+```java
+public int findShortestSubArray(int[] nums) {
+    Map<Integer, Integer> numsCnt = new HashMap<>();
+    Map<Integer, Integer> numsLastIndex = new HashMap<>();
+    Map<Integer, Integer> numsFirstIndex = new HashMap<>();
+    for (int i = 0; i < nums.length; i++) {
+        int num = nums[i];
+        numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
+        numsLastIndex.put(num, i);
+        if (!numsFirstIndex.containsKey(num)) {
+            numsFirstIndex.put(num, i);
+        }
+    }
+    int maxCnt = 0;
+    for (int num : nums) {
+        maxCnt = Math.max(maxCnt, numsCnt.get(num));
+    }
+    int ret = nums.length;
+    for (int i = 0; i < nums.length; i++) {
+        int num = nums[i];
+        int cnt = numsCnt.get(num);
+        if (cnt != maxCnt) continue;
+        ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
+    }
+    return ret;
+}
+```
+
+**对角元素相等的矩阵** 
+
+[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
+
+```html
+1234
+5123
+9512
+
+In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
+```
+
+```java
+public boolean isToeplitzMatrix(int[][] matrix) {
+    for (int i = 0; i < matrix[0].length; i++) {
+        if (!check(matrix, matrix[0][i], 0, i)) {
+            return false;
+        }
+    }
+    for (int i = 0; i < matrix.length; i++) {
+        if (!check(matrix, matrix[i][0], i, 0)) {
+            return false;
+        }
+    }
+    return true;
+}
+
+private boolean check(int[][] matrix, int expectValue, int row, int col) {
+    if (row >= matrix.length || col >= matrix[0].length) {
+        return true;
+    }
+    if (matrix[row][col] != expectValue) {
+        return false;
+    }
+    return check(matrix, expectValue, row + 1, col + 1);
+}
+```
+
+**嵌套数组** 
+
+[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
+
+```html
+Input: A = [5,4,0,3,1,6,2]
+Output: 4
+Explanation:
+A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
+
+One of the longest S[K]:
+S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
+```
+
+题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
+
+```java
+public int arrayNesting(int[] nums) {
+    int max = 0;
+    for (int i = 0; i < nums.length; i++) {
+        int cnt = 0;
+        for (int j = i; nums[j] != -1; ) {
+            cnt++;
+            int t = nums[j];
+            nums[j] = -1; // 标记该位置已经被访问
+            j = t;
+
+        }
+        max = Math.max(max, cnt);
+    }
+    return max;
+}
+```
+
+**分隔数组** 
+
+[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
+
+```html
+Input: arr = [1,0,2,3,4]
+Output: 4
+Explanation:
+We can split into two chunks, such as [1, 0], [2, 3, 4].
+However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
+```
+
+题目描述：分隔数组，使得对每部分排序后数组就为有序。
+
+```java
+public int maxChunksToSorted(int[] arr) {
+    if (arr == null) return 0;
+    int ret = 0;
+    int right = arr[0];
+    for (int i = 0; i < arr.length; i++) {
+        right = Math.max(right, arr[i]);
+        if (right == i) ret++;
+    }
+    return ret;
+}
+```
+
+
 ## 图
 
 ### 二分图
@@ -7052,3 +7062,4 @@ public int[] countBits(int num) {
 - 何海涛, 软件工程师. 剑指 Offer: 名企面试官精讲典型编程题[M]. 电子工业出版社, 2014.
 - 《编程之美》小组. 编程之美[M]. 电子工业出版社, 2008.
 - 左程云. 程序员代码面试指南[M]. 电子工业出版社, 2015.
+
diff --git a/notes/Linux.md b/notes/Linux.md
index ebcc347b..14d3b317 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -275,8 +275,6 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 
 ## 组成
 
-<div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
-
 最主要的几个组成部分如下：
 
 - inode：一个文件占用一个 inode，记录文件的属性，同时记录此文件的内容所在的 block 编号；
@@ -287,6 +285,9 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 - superblock：记录文件系统的整体信息，包括 inode 和 block 的总量、使用量、剩余量，以及文件系统的格式与相关信息等；
 - block bitmap：记录 block 是否被使用的位域；
 
+<div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
+
+
 ## 文件读取
 
 对于 Ext2 文件系统，当要读取一个文件的内容时，先在 inode 中去查找文件内容所在的所有 block，然后把所有 block 的内容读出来。
@@ -336,7 +337,9 @@ inode 中记录了文件内容所在的 block 编号，但是每个 block 非常
 
 ## 目录
 
-建立一个目录时，会分配一个 inode 与至少一个 block。block 记录的内容是目录下所有文件的 inode 编号以及文件名。可以看出文件的 inode 本身不记录文件名，文件名记录在目录中，因此新增文件、删除文件、更改文件名这些操作与目录的 w 权限有关。
+建立一个目录时，会分配一个 inode 与至少一个 block。block 记录的内容是目录下所有文件的 inode 编号以及文件名。
+
+可以看出文件的 inode 本身不记录文件名，文件名记录在目录中，因此新增文件、删除文件、更改文件名这些操作与目录的 w 权限有关。
 
 ## 日志
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index 1c575fae..690425db 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -139,7 +139,7 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 ### 2. 操作
 
-操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
+进行查找操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
 插入删除操作记录会破坏平衡树的平衡性，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作。
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 47d5d94e..0c9bd1f9 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -633,11 +633,12 @@ public int RectCover(int n) {
 
 ## 解题思路
 
-当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；否则解在 [m + 1, h] 之间，令 l = m + 1。
+- 当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；
+- 否则解在 [m + 1, h] 之间，令 l = m + 1。
 
-因为 h 的赋值表达式为 h = m，因此循环体的循环条件应该为 l < h，详细解释请见 [Leetcode 题解](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md#%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE) 二分查找部分。
+因为 h 的赋值表达式为 h = m，因此循环体的循环条件应该为 l < h，详细解释请见 [Leetcode 题解](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md) 二分查找部分。
 
-但是如果出现 nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，因此需要切换到顺序查找。
+但是如果出现 nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，需要切换到顺序查找。
 
 复杂度：O(logN) + O(1)
 
@@ -1217,8 +1218,7 @@ public ListNode ReverseList(ListNode head) {
 ### 递归
 
 ```java
-public ListNode Merge(ListNode list1, ListNode list2)
-{
+public ListNode Merge(ListNode list1, ListNode list2) {
     if (list1 == null)
         return list2;
     if (list2 == null)
@@ -1236,8 +1236,7 @@ public ListNode Merge(ListNode list1, ListNode list2)
 ### 迭代
 
 ```java
-public ListNode Merge(ListNode list1, ListNode list2)
-{
+public ListNode Merge(ListNode list1, ListNode list2) {
     ListNode head = new ListNode(-1);
     ListNode cur = head;
     while (list1 != null && list2 != null) {
@@ -1269,15 +1268,13 @@ public ListNode Merge(ListNode list1, ListNode list2)
 ## 解题思路
 
 ```java
-public boolean HasSubtree(TreeNode root1, TreeNode root2)
-{
+public boolean HasSubtree(TreeNode root1, TreeNode root2) {
     if (root1 == null || root2 == null)
         return false;
     return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
 }
 
-private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
-{
+private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
     if (root2 == null)
         return true;
     if (root1 == null)
@@ -1298,9 +1295,10 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
 
 ## 解题思路
 
+### 递归
+
 ```java
-public void Mirror(TreeNode root)
-{
+public void Mirror(TreeNode root) {
     if (root == null)
         return;
     swap(root);
@@ -1308,14 +1306,36 @@ public void Mirror(TreeNode root)
     Mirror(root.right);
 }
 
-private void swap(TreeNode root)
-{
+private void swap(TreeNode root) {
     TreeNode t = root.left;
     root.left = root.right;
     root.right = t;
 }
 ```
 
+### 迭代
+
+```java
+public void Mirror(TreeNode root) {
+    Stack<TreeNode> stack = new Stack<>();
+    stack.push(root);
+    while (!stack.isEmpty()) {
+        TreeNode node = stack.pop();
+        if (node == null)
+            continue;
+        swap(node);
+        stack.push(node.left);
+        stack.push(node.right);
+    }
+}
+
+private void swap(TreeNode node) {
+    TreeNode t = node.left;
+    node.left = node.right;
+    node.right = t;
+}
+```
+
 # 28 对称的二叉树
 
 [NowCder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
@@ -1327,15 +1347,13 @@ private void swap(TreeNode root)
 ## 解题思路
 
 ```java
-boolean isSymmetrical(TreeNode pRoot)
-{
+boolean isSymmetrical(TreeNode pRoot) {
     if (pRoot == null)
         return true;
     return isSymmetrical(pRoot.left, pRoot.right);
 }
 
-boolean isSymmetrical(TreeNode t1, TreeNode t2)
-{
+boolean isSymmetrical(TreeNode t1, TreeNode t2) {
     if (t1 == null && t2 == null)
         return true;
     if (t1 == null || t2 == null)
@@ -1359,8 +1377,7 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2)
 ## 解题思路
 
 ```java
-public ArrayList<Integer> printMatrix(int[][] matrix)
-{
+public ArrayList<Integer> printMatrix(int[][] matrix) {
     ArrayList<Integer> ret = new ArrayList<>();
     int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
     while (r1 <= r2 && c1 <= c2) {
@@ -1394,25 +1411,21 @@ public ArrayList<Integer> printMatrix(int[][] matrix)
 private Stack<Integer> dataStack = new Stack<>();
 private Stack<Integer> minStack = new Stack<>();
 
-public void push(int node)
-{
+public void push(int node) {
     dataStack.push(node);
     minStack.push(minStack.isEmpty() ? node : Math.min(minStack.peek(), node));
 }
 
-public void pop()
-{
+public void pop() {
     dataStack.pop();
     minStack.pop();
 }
 
-public int top()
-{
+public int top() {
     return dataStack.peek();
 }
 
-public int min()
-{
+public int min() {
     return minStack.peek();
 }
 ```
@@ -1430,8 +1443,7 @@ public int min()
 使用一个栈来模拟压入弹出操作。
 
 ```java
-public boolean IsPopOrder(int[] pushSequence, int[] popSequence)
-{
+public boolean IsPopOrder(int[] pushSequence, int[] popSequence) {
     int n = pushSequence.length;
     Stack<Integer> stack = new Stack<>();
     for (int pushIndex = 0, popIndex = 0; pushIndex < n; pushIndex++) {
@@ -1464,8 +1476,7 @@ public boolean IsPopOrder(int[] pushSequence, int[] popSequence)
 不需要使用两个队列分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
 
 ```java
-public ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
-{
+public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
     Queue<TreeNode> queue = new LinkedList<>();
     ArrayList<Integer> ret = new ArrayList<>();
     queue.add(root);
@@ -1495,8 +1506,7 @@ public ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
 ## 解题思路
 
 ```java
-ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
-{
+ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     Queue<TreeNode> queue = new LinkedList<>();
     queue.add(pRoot);
@@ -1529,8 +1539,7 @@ ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
 ## 解题思路
 
 ```java
-public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
-{
+public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     Queue<TreeNode> queue = new LinkedList<>();
     queue.add(pRoot);
@@ -1571,15 +1580,13 @@ public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
 ## 解题思路
 
 ```java
-public boolean VerifySquenceOfBST(int[] sequence)
-{
+public boolean VerifySquenceOfBST(int[] sequence) {
     if (sequence == null || sequence.length == 0)
         return false;
     return verify(sequence, 0, sequence.length - 1);
 }
 
-private boolean verify(int[] sequence, int first, int last)
-{
+private boolean verify(int[] sequence, int first, int last) {
     if (last - first <= 1)
         return true;
     int rootVal = sequence[last];
@@ -1610,14 +1617,12 @@ private boolean verify(int[] sequence, int first, int last)
 ```java
 private ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
 
-public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target)
-{
+public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target) {
     backtracking(root, target, new ArrayList<>());
     return ret;
 }
 
-private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
-{
+private void backtracking(TreeNode node, int target, ArrayList<Integer> path) {
     if (node == null)
         return;
     path.add(node.val);
@@ -1641,8 +1646,7 @@ private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
 输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的 head。
 
 ```java
-public class RandomListNode
-{
+public class RandomListNode {
     int label;
     RandomListNode next = null;
     RandomListNode random = null;
@@ -1670,8 +1674,7 @@ public class RandomListNode
 <div align="center"> <img src="../pics//8f3b9519-d705-48fe-87ad-2e4052fc81d2.png" width="600"/> </div><br>
 
 ```java
-public RandomListNode Clone(RandomListNode pHead)
-{
+public RandomListNode Clone(RandomListNode pHead) {
     if (pHead == null)
         return null;
     // 插入新节点
@@ -1718,14 +1721,12 @@ public RandomListNode Clone(RandomListNode pHead)
 private TreeNode pre = null;
 private TreeNode head = null;
 
-public TreeNode Convert(TreeNode root)
-{
+public TreeNode Convert(TreeNode root) {
     inOrder(root);
     return head;
 }
 
-private void inOrder(TreeNode node)
-{
+private void inOrder(TreeNode node) {
     if (node == null)
         return;
     inOrder(node.left);
@@ -1752,21 +1753,18 @@ private void inOrder(TreeNode node)
 ```java
 private String deserializeStr;
 
-public String Serialize(TreeNode root)
-{
+public String Serialize(TreeNode root) {
     if (root == null)
         return "#";
     return root.val + " " + Serialize(root.left) + " " + Serialize(root.right);
 }
 
-public TreeNode Deserialize(String str)
-{
+public TreeNode Deserialize(String str) {
     deserializeStr = str;
     return Deserialize();
 }
 
-private TreeNode Deserialize()
-{
+private TreeNode Deserialize() {
     if (deserializeStr.length() == 0)
         return null;
     int index = deserializeStr.indexOf(" ");
@@ -1795,8 +1793,7 @@ private TreeNode Deserialize()
 ```java
 private ArrayList<String> ret = new ArrayList<>();
 
-public ArrayList<String> Permutation(String str)
-{
+public ArrayList<String> Permutation(String str) {
     if (str.length() == 0)
         return ret;
     char[] chars = str.toCharArray();
@@ -1805,8 +1802,7 @@ public ArrayList<String> Permutation(String str)
     return ret;
 }
 
-private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
-{
+private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s) {
     if (s.length() == chars.length) {
         ret.add(s.toString());
         return;
@@ -1836,8 +1832,7 @@ private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
 使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
 
 ```java
-public int MoreThanHalfNum_Solution(int[] nums)
-{
+public int MoreThanHalfNum_Solution(int[] nums) {
     int majority = nums[0];
     for (int i = 1, cnt = 1; i < nums.length; i++) {
         cnt = nums[i] == majority ? cnt + 1 : cnt - 1;
@@ -1868,8 +1863,7 @@ public int MoreThanHalfNum_Solution(int[] nums)
 快速排序的 partition() 方法，会返回一个整数 j 使得 a[l..j-1] 小于等于 a[j]，且 a[j+1..h] 大于等于 a[j]，此时 a[j] 就是数组的第 j 大元素。可以利用这个特性找出数组的第 K 个元素，这种找第 K 个元素的算法称为快速选择算法。
 
 ```java
-public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
-{
+public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
     ArrayList<Integer> ret = new ArrayList<>();
     if (k > nums.length || k <= 0)
         return ret;
@@ -1880,8 +1874,7 @@ public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
     return ret;
 }
 
-public void findKthSmallest(int[] nums, int k)
-{
+public void findKthSmallest(int[] nums, int k) {
     int l = 0, h = nums.length - 1;
     while (l < h) {
         int j = partition(nums, l, h);
@@ -1894,8 +1887,7 @@ public void findKthSmallest(int[] nums, int k)
     }
 }
 
-private int partition(int[] nums, int l, int h)
-{
+private int partition(int[] nums, int l, int h) {
     int p = nums[l];     /* 切分元素 */
     int i = l, j = h + 1;
     while (true) {
@@ -1909,8 +1901,7 @@ private int partition(int[] nums, int l, int h)
     return j;
 }
 
-private void swap(int[] nums, int i, int j)
-{
+private void swap(int[] nums, int i, int j) {
     int t = nums[i];
     nums[i] = nums[j];
     nums[j] = t;
@@ -1927,8 +1918,7 @@ private void swap(int[] nums, int i, int j)
 维护一个大小为 K 的最小堆过程如下：在添加一个元素之后，如果大顶堆的大小大于 K，那么需要将大顶堆的堆顶元素去除。
 
 ```java
-public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
-{
+public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
     if (k > nums.length || k <= 0)
         return new ArrayList<>();
     PriorityQueue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> o2 - o1);
@@ -1959,8 +1949,7 @@ private PriorityQueue<Integer> right = new PriorityQueue<>();
 /* 当前数据流读入的元素个数 */
 private int N = 0;
 
-public void Insert(Integer val)
-{
+public void Insert(Integer val) {
     /* 插入要保证两个堆存于平衡状态 */
     if (N % 2 == 0) {
         /* N 为偶数的情况下插入到右半边。
@@ -1975,8 +1964,7 @@ public void Insert(Integer val)
     N++;
 }
 
-public Double GetMedian()
-{
+public Double GetMedian() {
     if (N % 2 == 0)
         return (left.peek() + right.peek()) / 2.0;
     else
@@ -1998,16 +1986,14 @@ public Double GetMedian()
 private int[] cnts = new int[256];
 private Queue<Character> queue = new LinkedList<>();
 
-public void Insert(char ch)
-{
+public void Insert(char ch) {
     cnts[ch]++;
     queue.add(ch);
     while (!queue.isEmpty() && cnts[queue.peek()] > 1)
         queue.poll();
 }
 
-public char FirstAppearingOnce()
-{
+public char FirstAppearingOnce() {
     return queue.isEmpty() ? '#' : queue.peek();
 }
 ```
@@ -2023,8 +2009,7 @@ public char FirstAppearingOnce()
 ## 解题思路
 
 ```java
-public int FindGreatestSumOfSubArray(int[] nums)
-{
+public int FindGreatestSumOfSubArray(int[] nums) {
     if (nums == null || nums.length == 0)
         return 0;
     int greatestSum = Integer.MIN_VALUE;
@@ -2044,8 +2029,7 @@ public int FindGreatestSumOfSubArray(int[] nums)
 ## 解题思路
 
 ```java
-public int NumberOf1Between1AndN_Solution(int n)
-{
+public int NumberOf1Between1AndN_Solution(int n) {
     int cnt = 0;
     for (int m = 1; m <= n; m *= 10) {
         int a = n / m, b = n % m;
@@ -2066,8 +2050,7 @@ public int NumberOf1Between1AndN_Solution(int n)
 ## 解题思路
 
 ```java
-public int getDigitAtIndex(int index)
-{
+public int getDigitAtIndex(int index) {
     if (index < 0)
         return -1;
     int place = 1;  // 1 表示个位，2 表示 十位...
@@ -2085,8 +2068,7 @@ public int getDigitAtIndex(int index)
  * place 位数的数字组成的字符串长度
  * 10, 90, 900, ...
  */
-private int getAmountOfPlace(int place)
-{
+private int getAmountOfPlace(int place) {
     if (place == 1)
         return 10;
     return (int) Math.pow(10, place - 1) * 9;
@@ -2096,8 +2078,7 @@ private int getAmountOfPlace(int place)
  * place 位数的起始数字
  * 0, 10, 100, ...
  */
-private int getBeginNumberOfPlace(int place)
-{
+private int getBeginNumberOfPlace(int place) {
     if (place == 1)
         return 0;
     return (int) Math.pow(10, place - 1);
@@ -2106,8 +2087,7 @@ private int getBeginNumberOfPlace(int place)
 /**
  * 在 place 位数组成的字符串中，第 index 个数
  */
-private int getDigitAtIndex(int index, int place)
-{
+private int getDigitAtIndex(int index, int place) {
     int beginNumber = getBeginNumberOfPlace(place);
     int shiftNumber = index / place;
     String number = (beginNumber + shiftNumber) + "";
@@ -2129,8 +2109,7 @@ private int getDigitAtIndex(int index, int place)
 可以看成是一个排序问题，在比较两个字符串 S1 和 S2 的大小时，应该比较的是 S1+S2 和 S2+S1 的大小，如果 S1+S2 < S2+S1，那么应该把 S1 排在前面，否则应该把 S2 排在前面。
 
 ```java
-public String PrintMinNumber(int[] numbers)
-{
+public String PrintMinNumber(int[] numbers) {
     if (numbers == null || numbers.length == 0)
         return "";
     int n = numbers.length;
@@ -2156,8 +2135,7 @@ public String PrintMinNumber(int[] numbers)
 ## 解题思路
 
 ```java
-public int numDecodings(String s)
-{
+public int numDecodings(String s) {
     if (s == null || s.length() == 0)
         return 0;
     int n = s.length();
@@ -2200,8 +2178,7 @@ public int numDecodings(String s)
 应该用动态规划求解，而不是深度优先搜索，深度优先搜索过于复杂，不是最优解。
 
 ```java
-public int getMost(int[][] values)
-{
+public int getMost(int[][] values) {
     if (values == null || values.length == 0 || values[0].length == 0)
         return 0;
     int n = values[0].length;
@@ -2224,8 +2201,7 @@ public int getMost(int[][] values)
 ## 解题思路
 
 ```java
-public int longestSubStringWithoutDuplication(String str)
-{
+public int longestSubStringWithoutDuplication(String str) {
     int curLen = 0;
     int maxLen = 0;
     int[] preIndexs = new int[26];
@@ -2257,8 +2233,7 @@ public int longestSubStringWithoutDuplication(String str)
 ## 解题思路
 
 ```java
-public int GetUglyNumber_Solution(int N)
-{
+public int GetUglyNumber_Solution(int N) {
     if (N <= 6)
         return N;
     int i2 = 0, i3 = 0, i5 = 0;
@@ -2291,8 +2266,7 @@ public int GetUglyNumber_Solution(int N)
 最直观的解法是使用 HashMap 对出现次数进行统计，但是考虑到要统计的字符范围有限，因此可以使用整型数组代替 HashMap。
 
 ```java
-public int FirstNotRepeatingChar(String str)
-{
+public int FirstNotRepeatingChar(String str) {
     int[] cnts = new int[256];
     for (int i = 0; i < str.length(); i++)
         cnts[str.charAt(i)]++;
@@ -2306,8 +2280,7 @@ public int FirstNotRepeatingChar(String str)
 以上实现的空间复杂度还不是最优的。考虑到只需要找到只出现一次的字符，那么我们只需要统计的次数信息只有 0,1,更大，使用两个比特位就能存储这些信息。
 
 ```java
-public int FirstNotRepeatingChar2(String str)
-{
+public int FirstNotRepeatingChar2(String str) {
     BitSet bs1 = new BitSet(256);
     BitSet bs2 = new BitSet(256);
     for (char c : str.toCharArray()) {
@@ -2339,15 +2312,13 @@ public int FirstNotRepeatingChar2(String str)
 private long cnt = 0;
 private int[] tmp;  // 在这里创建辅助数组，而不是在 merge() 递归函数中创建
 
-public int InversePairs(int[] nums)
-{
+public int InversePairs(int[] nums) {
     tmp = new int[nums.length];
     mergeSort(nums, 0, nums.length - 1);
     return (int) (cnt % 1000000007);
 }
 
-private void mergeSort(int[] nums, int l, int h)
-{
+private void mergeSort(int[] nums, int l, int h) {
     if (h - l < 1)
         return;
     int m = l + (h - l) / 2;
@@ -2356,8 +2327,7 @@ private void mergeSort(int[] nums, int l, int h)
     merge(nums, l, m, h);
 }
 
-private void merge(int[] nums, int l, int m, int h)
-{
+private void merge(int[] nums, int l, int m, int h) {
     int i = l, j = m + 1, k = l;
     while (i <= m || j <= h) {
         if (i > m)
@@ -2392,8 +2362,7 @@ private void merge(int[] nums, int l, int m, int h)
 当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部重新开始访问链表 B；同样地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部重新开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。
 
 ```java
-public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2)
-{
+public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
     ListNode l1 = pHead1, l2 = pHead2;
     while (l1 != l2) {
         l1 = (l1 == null) ? pHead2 : l1.next;
@@ -2420,15 +2389,13 @@ Output:
 ## 解题思路
 
 ```java
-public int GetNumberOfK(int[] nums, int K)
-{
+public int GetNumberOfK(int[] nums, int K) {
     int first = binarySearch(nums, K);
     int last = binarySearch(nums, K + 1);
     return (first == nums.length || nums[first] != K) ? 0 : last - first;
 }
 
-private int binarySearch(int[] nums, int K)
-{
+private int binarySearch(int[] nums, int K) {
     int l = 0, h = nums.length;
     while (l < h) {
         int m = l + (h - l) / 2;
@@ -2453,14 +2420,12 @@ private int binarySearch(int[] nums, int K)
 private TreeNode ret;
 private int cnt = 0;
 
-public TreeNode KthNode(TreeNode pRoot, int k)
-{
+public TreeNode KthNode(TreeNode pRoot, int k) {
     inOrder(pRoot, k);
     return ret;
 }
 
-private void inOrder(TreeNode root, int k)
-{
+private void inOrder(TreeNode root, int k) {
     if (root == null || cnt >= k)
         return;
     inOrder(root.left, k);
@@ -2484,8 +2449,7 @@ private void inOrder(TreeNode root, int k)
 ## 解题思路
 
 ```java
-public int TreeDepth(TreeNode root)
-{
+public int TreeDepth(TreeNode root) {
     return root == null ? 0 : 1 + Math.max(TreeDepth(root.left), TreeDepth(root.right));
 }
 ```
@@ -2505,14 +2469,12 @@ public int TreeDepth(TreeNode root)
 ```java
 private boolean isBalanced = true;
 
-public boolean IsBalanced_Solution(TreeNode root)
-{
+public boolean IsBalanced_Solution(TreeNode root) {
     height(root);
     return isBalanced;
 }
 
-private int height(TreeNode root)
-{
+private int height(TreeNode root) {
     if (root == null || !isBalanced)
         return 0;
     int left = height(root.left);
@@ -2538,8 +2500,7 @@ private int height(TreeNode root)
 diff &= -diff 得到出 diff 最右侧不为 0 的位，也就是不存在重复的两个元素在位级表示上最右侧不同的那一位，利用这一位就可以将两个元素区分开来。
 
 ```java
-public void FindNumsAppearOnce(int[] nums, int num1[], int num2[])
-{
+public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
     int diff = 0;
     for (int num : nums)
         diff ^= num;
@@ -2570,8 +2531,7 @@ public void FindNumsAppearOnce(int[] nums, int num1[], int num2[])
 - 如果 sum < target，移动较小的元素，使 sum 变大一些。
 
 ```java
-public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum)
-{
+public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum) {
     int i = 0, j = array.length - 1;
     while (i < j) {
         int cur = array[i] + array[j];
@@ -2604,8 +2564,7 @@ public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum)
 ## 解题思路
 
 ```java
-public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum)
-{
+public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     int start = 1, end = 2;
     int curSum = 3;
@@ -2648,8 +2607,7 @@ public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum)
 正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
 
 ```java
-public String ReverseSentence(String str)
-{
+public String ReverseSentence(String str) {
     int n = str.length();
     char[] chars = str.toCharArray();
     int i = 0, j = 0;
@@ -2664,14 +2622,12 @@ public String ReverseSentence(String str)
     return new String(chars);
 }
 
-private void reverse(char[] c, int i, int j)
-{
+private void reverse(char[] c, int i, int j) {
     while (i < j)
         swap(c, i++, j--);
 }
 
-private void swap(char[] c, int i, int j)
-{
+private void swap(char[] c, int i, int j) {
     char t = c[i];
     c[i] = c[j];
     c[j] = t;
@@ -2691,8 +2647,7 @@ private void swap(char[] c, int i, int j)
 先将 "abc" 和 "XYZdef" 分别翻转，得到 "cbafedZYX"，然后再把整个字符串翻转得到 "XYZdefabc"。
 
 ```java
-public String LeftRotateString(String str, int n)
-{
+public String LeftRotateString(String str, int n) {
     if (n >= str.length())
         return str;
     char[] chars = str.toCharArray();
@@ -2702,14 +2657,12 @@ public String LeftRotateString(String str, int n)
     return new String(chars);
 }
 
-private void reverse(char[] chars, int i, int j)
-{
+private void reverse(char[] chars, int i, int j) {
     while (i < j)
         swap(chars, i++, j--);
 }
 
-private void swap(char[] chars, int i, int j)
-{
+private void swap(char[] chars, int i, int j) {
     char t = chars[i];
     chars[i] = chars[j];
     chars[j] = t;
@@ -2727,8 +2680,7 @@ private void swap(char[] chars, int i, int j)
 ## 解题思路
 
 ```java
-public ArrayList<Integer> maxInWindows(int[] num, int size)
-{
+public ArrayList<Integer> maxInWindows(int[] num, int size) {
     ArrayList<Integer> ret = new ArrayList<>();
     if (size > num.length || size < 1)
         return ret;
@@ -2762,8 +2714,7 @@ public ArrayList<Integer> maxInWindows(int[] num, int size)
 空间复杂度：O(N<sup>2</sup>)
 
 ```java
-public List<Map.Entry<Integer, Double>> dicesSum(int n)
-{
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
     final int face = 6;
     final int pointNum = face * n;
     long[][] dp = new long[n + 1][pointNum + 1];
@@ -2790,8 +2741,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n)
 空间复杂度：O(N)
 
 ```java
-public List<Map.Entry<Integer, Double>> dicesSum(int n)
-{
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
     final int face = 6;
     final int pointNum = face * n;
     long[][] dp = new long[2][pointNum + 1];
@@ -2829,8 +2779,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n)
 ## 解题思路
 
 ```java
-public boolean isContinuous(int[] nums)
-{
+public boolean isContinuous(int[] nums) {
     if (nums.length < 5)
         return false;
     Arrays.sort(nums);
@@ -2861,8 +2810,7 @@ public boolean isContinuous(int[] nums)
 约瑟夫环，圆圈长度为 n 的解可以看成长度为 n-1 的解再加上报数的长度 m。因为是圆圈，所以最后需要对 n 取余。
 
 ```java
-public int LastRemaining_Solution(int n, int m)
-{
+public int LastRemaining_Solution(int n, int m) {
     if (n == 0)     /* 特殊输入的处理 */
         return -1;
     if (n == 1)     /* 返回条件 */
@@ -2884,8 +2832,7 @@ public int LastRemaining_Solution(int n, int m)
 使用贪心策略，假设第 i 轮进行卖出操作，买入操作价格应该在 i 之前并且价格最低。
 
 ```java
-public int maxProfit(int[] prices)
-{
+public int maxProfit(int[] prices) {
     if (prices == null || prices.length == 0)
         return 0;
     int soFarMin = prices[0];
@@ -2915,8 +2862,7 @@ public int maxProfit(int[] prices)
 以下实现中，递归的返回条件为 n <= 0，取非后就是 n > 0，递归的主体部分为 sum += Sum_Solution(n - 1)，转换为条件语句后就是 (sum += Sum_Solution(n - 1)) > 0。
 
 ```java
-public int Sum_Solution(int n)
-{
+public int Sum_Solution(int n) {
     int sum = n;
     boolean b = (n > 0) && ((sum += Sum_Solution(n - 1)) > 0);
     return sum;
@@ -2938,8 +2884,7 @@ a ^ b 表示没有考虑进位的情况下两数的和，(a & b) << 1 就是进
 递归会终止的原因是 (a & b) << 1 最右边会多一个 0，那么继续递归，进位最右边的 0 会慢慢增多，最后进位会变为 0，递归终止。
 
 ```java
-public int Add(int a, int b)
-{
+public int Add(int a, int b) {
     return b == 0 ? a : Add(a ^ b, (a & b) << 1);
 }
 ```
@@ -2955,8 +2900,7 @@ public int Add(int a, int b)
 ## 解题思路
 
 ```java
-public int[] multiply(int[] A)
-{
+public int[] multiply(int[] A) {
     int n = A.length;
     int[] B = new int[n];
     for (int i = 0, product = 1; i < n; product *= A[i], i++)       /* 从左往右累乘 */
@@ -2988,8 +2932,7 @@ Output:
 ## 解题思路
 
 ```java
-public int StrToInt(String str)
-{
+public int StrToInt(String str) {
     if (str == null || str.length() == 0)
         return 0;
     boolean isNegative = str.charAt(0) == '-';
diff --git a/notes/消息队列.md b/notes/消息队列.md
index e28fcd3f..74155802 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -46,17 +46,17 @@
 
 例如在注册流程中通常需要发送验证邮件来确保注册用户的身份合法，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
 
-只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成操作的话，就不能再使用消息队列。
+只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成注册的话，就不能再使用消息队列。
 
 ## 流量削锋
 
-在高并发的场景下，如果短时间有大量的请求会压垮服务器。
+在高并发的场景下，如果短时间有大量的请求到达会压垮服务器。
 
 可以将请求发送到消息队列中，服务器按照其处理能力从消息队列中订阅消息进行处理。
 
 ## 应用解耦
 
-如果模块之间不直接进行调用，模块之间耦合度很低，那么修改一个模块或者新增一个模块对其它模块的影响会很小，从而实现可扩展性。
+如果模块之间不直接进行调用，模块之间耦合度就会很低，那么修改一个模块或者新增一个模块对其它模块的影响会很小，从而实现可扩展性。
 
 通过使用消息队列，一个模块只需要向消息队列中发送消息，其它模块可以选择性地从消息队列中订阅消息从而完成调用。
 
diff --git a/notes/算法.md b/notes/算法.md
index fb878742..cb5d4dbe 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -1623,10 +1623,10 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ## 2-3 查找树
 
-<div align="center"> <img src="../pics//ff396233-1bb1-4e74-8bc2-d7c90146f5dd.png" width="250"/> </div><br>
-
 2-3 查找树引入了 2- 节点和 3- 节点，目的是为了让树平衡。一颗完美平衡的 2-3 查找树的所有空链接到根节点的距离应该是相同的。
 
+<div align="center"> <img src="../pics//ff396233-1bb1-4e74-8bc2-d7c90146f5dd.png" width="250"/> </div><br>
+
 ### 1. 插入操作
 
 插入操作和 BST 的插入操作有很大区别，BST 的插入操作是先进行一次未命中的查找，然后再将节点插入到对应的空链接上。但是 2-3 查找树如果也这么做的话，那么就会破坏了平衡性。它是将新节点插入到叶子节点上。
@@ -2042,24 +2042,26 @@ public class SparseVector {
 
 ## 汉诺塔
 
+<div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
+
 这是一个经典的递归问题，分为三步求解：
 
-1. 将 n-1 个圆盘从 from -> buffer
-2. 将 1 个圆盘从 from -> to
-3. 将 n-1 个圆盘从 buffer -> to
+- 将 n-1 个圆盘从 from -> buffer
+
+<div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
+
+- 将 1 个圆盘从 from -> to
+
+<div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
+
+- 将 n-1 个圆盘从 buffer -> to
+
+<div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
 
 如果只有一个圆盘，那么只需要进行一次移动操作。
 
 从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然 a<sub>n</sub> = 2<sup>n</sup> - 1，n 个圆盘需要移动 2<sup>n</sup> - 1 次。
 
-<div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
-
 ```java
 public class Hanoi {
     public static void move(int n, String from, String buffer, String to) {
@@ -2105,7 +2107,7 @@ from H1 to H3
 
 生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到根节点，根节点代表的字符的编码就是这个路径编码。
 
-<div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png"/> </div><br>
+<div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png" width="400"/> </div><br>
 
 ```java
 public class Huffman {
diff --git a/notes/缓存.md b/notes/缓存.md
index e60373dd..449598ff 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -163,7 +163,11 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 分布式缓存
 
-使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存，本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。
+
+相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。
+
+不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
 
 ## 数据库缓存
 
@@ -171,7 +175,7 @@ MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查
 
 # 四、CDN
 
-内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源其它数据分发给用户。
+内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
 
 CDN 主要有以下优点：
 
@@ -190,7 +194,7 @@ CDN 主要有以下优点：
 
 ## 缓存穿透
 
-指的是对某个一定不存在的数据进行请求，该请求将会穿透缓存来到数据库。
+指的是对某个一定不存在的数据进行请求，该请求将会穿透缓存到达数据库。
 
 解决方案：
 
@@ -207,6 +211,7 @@ CDN 主要有以下优点：
 
 - 为了防止缓存在同一时间大面积过期导致的缓存雪崩，可以通过观察用户行为，合理设置缓存过期时间来实现；
 - 为了防止缓存服务器宕机出现的缓存雪崩，可以使用分布式缓存，分布式缓存中每一个节点只缓存部分的数据，当某个节点宕机时可以保证其它节点的缓存仍然可用。
+- 也可以在进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
 
 ## 缓存一致性
 
@@ -258,7 +263,7 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
 
-数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点号，从而使得数据分布也更加均匀。
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点好，从而使得数据分布也更加均匀。
 
 参考资料：
 
diff --git a/notes/设计模式.md b/notes/设计模式.md
index ae44213e..d48d7b36 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -1851,11 +1851,7 @@ No gumball dispensed
 
 ### 与状态模式的比较
 
-状态模式的类图和策略模式类似，并且都是能够动态改变对象的行为。
-
-但是状态模式是通过状态转移来改变 Context 所组合的 State 对象，而策略模式是通过 Context 本身的决策来改变组合的 Strategy 对象。
-
-所谓的状态转移，是指 Context 在运行过程中由于一些条件发生改变而使得 State 对象发生改变，注意必须要是在运行过程中。
+状态模式的类图和策略模式类似，并且都是能够动态改变对象的行为。但是状态模式是通过状态转移来改变 Context 所组合的 State 对象，而策略模式是通过 Context 本身的决策来改变组合的 Strategy 对象。所谓的状态转移，是指 Context 在运行过程中由于一些条件发生改变而使得 State 对象发生改变，注意必须要是在运行过程中。
 
 状态模式主要是用来解决状态转移的问题，当状态发生转移了，那么 Context 对象就会改变它的行为；而策略模式主要是用来封装一组可以互相替代的算法族，并且可以根据需要动态地去替换 Context 使用的算法。
 
@@ -1969,7 +1965,7 @@ public abstract class CaffeineBeverage {
 ```
 
 ```java
-public class Coffee extends CaffeineBeverage{
+public class Coffee extends CaffeineBeverage {
     @Override
     void brew() {
         System.out.println("Coffee.brew");
@@ -1983,7 +1979,7 @@ public class Coffee extends CaffeineBeverage{
 ```
 
 ```java
-public class Tea extends CaffeineBeverage{
+public class Tea extends CaffeineBeverage {
     @Override
     void brew() {
         System.out.println("Tea.brew");
@@ -2238,7 +2234,7 @@ Number of items:     6
 
 ### 意图
 
-使用什么都不做的空对象来替代 NULL。
+使用什么都不做的空对象来代替 NULL。
 
 一个方法返回 NULL，意味着方法的调用端需要去检查返回值是否是 NULL，这么做会导致非常多的冗余的检查代码。并且如果某一个调用端忘记了做这个检查返回值，而直接使用返回的对象，那么就有可能抛出空指针异常。
 
@@ -2393,7 +2389,7 @@ public abstract class TV {
 ```
 
 ```java
-public class Sony extends TV{
+public class Sony extends TV {
     @Override
     public void on() {
         System.out.println("Sony.on()");
@@ -2412,7 +2408,7 @@ public class Sony extends TV{
 ```
 
 ```java
-public class RCA extends TV{
+public class RCA extends TV {
     @Override
     public void on() {
         System.out.println("RCA.on()");
@@ -2551,9 +2547,6 @@ public abstract class Component {
 ```
 
 ```java
-import java.util.ArrayList;
-import java.util.List;
-
 public class Composite extends Component {
 
     private List<Component> child;
@@ -2659,7 +2652,7 @@ Composite:root
 
 ### 类图
 
-装饰者（Decorator）和具体组件（ConcreteComponent）都继承自组件（Component），具体组件的方法实现不需要依赖于其它对象，而装饰者组合了一个组件，这样它可以装饰其它装饰者或者具体组件。所谓装饰，就是把这个装饰者套在被装饰上，从而动态扩展被装饰者的功能。装饰者的方法有一部分是自己的，这属于它的功能，然后调用被装饰者的方法实现，从而也保留了被装饰者的功能。可以看到，具体组件应当是装饰层次的最低层，因为只有具体组件的方法实现不需要依赖于其它对象。
+装饰者（Decorator）和具体组件（ConcreteComponent）都继承自组件（Component），具体组件的方法实现不需要依赖于其它对象，而装饰者组合了一个组件，这样它可以装饰其它装饰者或者具体组件。所谓装饰，就是把这个装饰者套在被装饰者之上，从而动态扩展被装饰者的功能。装饰者的方法有一部分是自己的，这属于它的功能，然后调用被装饰者的方法实现，从而也保留了被装饰者的功能。可以看到，具体组件应当是装饰层次的最低层，因为只有具体组件的方法实现不需要依赖于其它对象。
 
 <div align="center"> <img src="../pics//137c593d-0a9e-47b8-a9e6-b71f540b82dd.png"/> </div><br>
 
@@ -2770,7 +2763,7 @@ public class Client {
 
 ### 实现
 
-观看电影需要操作很多电器，使用外观模式可以实现一键看电影功能。
+观看电影需要操作很多电器，使用外观模式实现一键看电影功能。
 
 ```java
 public class SubSystem {
@@ -2811,7 +2804,7 @@ public class Client {
 
 ### 设计原则
 
-最少知识原则：只和你的密友谈话。也就是客户对象所需要交互的对象应当尽可能少。
+最少知识原则：只和你的密友谈话。也就是说客户对象所需要交互的对象应当尽可能少。
 
 ## 6. 享元（Flyweight）
 
@@ -2822,8 +2815,8 @@ public class Client {
 ### 类图
 
 - Flyweight：享元对象
-- IntrinsicState：内部状态，相同的项元对象共享
-- ExtrinsicState：外部状态
+- IntrinsicState：内部状态，享元对象共享内部状态
+- ExtrinsicState：外部状态，每个享元对象的外部状态不同
 
 <div align="center"> <img src="../pics//d52270b4-9097-4667-9f18-f405fc661c99.png"/> </div><br>
 
@@ -2854,8 +2847,6 @@ public class ConcreteFlyweight implements Flyweight {
 ```
 
 ```java
-import java.util.HashMap;
-
 public class FlyweightFactory {
 
     private HashMap<String, Flyweight> flyweights = new HashMap<>();

From da479e25c6ef9f4268344e0c722903af34932d44 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 9 Aug 2018 23:10:49 +0800
Subject: [PATCH 085/172] auto commit

---
 notes/MySQL.md           | 60 ----------------------------------------
 notes/剑指 offer 题解.md | 25 -----------------
 notes/数据库系统原理.md  |  8 +++---
 3 files changed, 4 insertions(+), 89 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 02dbe638..690425db 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -304,64 +304,6 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 
 最有效的方式是使用索引来覆盖查询。
 
-### 3. 不要在列上使用函数和进行运算
-
-在列上使用函数和计算将导致索引失效而进行全表扫描。
-
-### 4. 尽量避免使用 != 或 not in 或 <> 等否定操作符
-
-尽量避免在 where 子句中使用 or 来连接条件，因为这会导致索引失效而进行全表扫描。
-
-### 5. 尽量避免使用 or 来连接条件
-
-尽量避免在 where 子句中使用 or 来连接条件，因为这会导致索引失效而进行全表扫描。
-
-### 6. 多个单列限制应该改成一个多列索引（复合索引）查询
-
-因为 MySQL 只能使用一个单列索引，所以为多个列创建单列索引并不能提高 MySQL 的查询性能。正确的做法是将这多个要限定条件的列单独做成一个多列索引（复合索引）。
-
-### 7. 查询时要按照复合索引的最左列开始查找
-
-查询条件中没有使用复合索引的第一个字段，索引是不会被使用的。复合索引遵守“最左前缀”原则，即在查询条件中使用了复合索引的第一个字段，索引才会被使用。
-
-假如我们只有一个复合索引 ```age_height_idx(age, height)```
-
-如果我们使用
-```mysql
-select * from user where height = 170;
-```
-那么复合索引 ```age_height_idx(age, height)```将不会被使用。
-
-### 8. 尽量不用范围查询
-
-如果查询中的某个列有范围查询，则其右边所有列都无法使用索引优化查找。
-
-比如说我们使用
-```mysql
-select * from users where age > 16 and age < 30 and height = 180
-```
-则索引中 age 右边所有列都无法使用索引优化查找。换句话说，```age_height_idx(age, height)``` 索引等价于 ```age_height_idx(age)```。
-
-### 9. 索引不要包含有NULL值的列
-
-只要列中包含有 NULL 值，它就不会被包含在索引中，复合索引中只要有一列含有 NULL 值，那么这一列对于此复合索引就是无效的。
-
-因此，在数据库设计时，除非有一个很特别的原因使用 NULL 值，不然尽量不要让字段的默认值为 NULL。
-
-### 10. 隐式转换可能带来不好的影响
-
-当 where 查询条件左右两侧类型不匹配的时候会发生隐式转换，隐式转换可能带来的不好影响就是导致索引失效而进行全表扫描。
-
-下面的案例中，brith_date 是字符串，然而匹配的是整数类型，从而发生隐式转换。
-
-```mysql
-select from users where brith_date = 19950610;
-```
-
-### 11. like 语句的索引失效问题
-
-当我们使用 like “value%” 的方式时是会使用索引的，但是对于 like “%value%” 这样的方式，必定会执行全表查询，这在数据量小的表，不存在性能问题，但是对于海量数据，全表扫描是非常可怕的事情。所以，根据业务需求，考虑使用 ElasticSearch 或 Solr 是个不错的方案。
-
 ## 重构查询方式
 
 ### 1. 切分大查询
@@ -483,5 +425,3 @@ MySQL 读写分离能提高性能的原因在于：
 - [How to create unique row ID in sharded databases?](https://stackoverflow.com/questions/788829/how-to-create-unique-row-id-in-sharded-databases)
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
 - [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
-- [服务端指南 数据存储篇 | MySQL（04） 索引使用的注意事项](http://blog.720ui.com/2017/mysql_core_04_index_item/)
-- [mysql 聚簇索引 和聚簇索引 （二级索引）的 那些事](https://blog.csdn.net/bigtree_3721/article/details/51335479)
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 00510fc8..0c9bd1f9 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1296,10 +1296,7 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
 ## 解题思路
 
 ### 递归
-<<<<<<< HEAD
 
-=======
->>>>>>> 0d7909db6d01bdfa0a0fd9abaee469b9ddd2e3a4
 ```java
 public void Mirror(TreeNode root) {
     if (root == null)
@@ -1317,7 +1314,6 @@ private void swap(TreeNode root) {
 ```
 
 ### 迭代
-<<<<<<< HEAD
 
 ```java
 public void Mirror(TreeNode root) {
@@ -1338,27 +1334,6 @@ private void swap(TreeNode node) {
     node.left = node.right;
     node.right = t;
 }
-=======
-```java
-public void Mirror(TreeNode root) {
-    if (root == null)
-        return;
-    Stack<TreeNode> stack = new Stack<>();
-    stack.push(root);
-    while (!stack.isEmpty()) {
-        TreeNode p = stack.pop();
-        if (p.left == null && p.right == null)
-            continue;
-        TreeNode left = p.left;
-        p.left = p.right;
-        p.right = left;
-        if (p.left != null)
-            stack.push(p.left);
-        if (p.right != null)
-            stack.push(p.right);
-    }
-}
->>>>>>> 0d7909db6d01bdfa0a0fd9abaee469b9ddd2e3a4
 ```
 
 # 28 对称的二叉树
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index 6b8fcd86..151ffc0f 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -389,11 +389,11 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 ## Gap Locks
 
-间隙锁锁住的是多行，是一个数据范围。间隙锁主要是为了防止出现幻读，但是它会把锁定范围扩大，有时候也会给我们带来麻烦。在数据库参数中，控制间隙锁的参数是 ```innodb_locks_unsafe_for_binlog```， 这个参数默认值是 OFF，也就是启用间隙锁，它是一个bool值，当值为 true 时表示 disable 间隙锁。那为了防止间隙锁是不是直接将 ```innodb_locaks_unsafe_for_binlog``` 设置为 ```true``` 就可以了呢？不一定！而且这个参数会影响到主从复制及灾难恢复，这个方法还尚待商量。
+锁定一个范围内的索引，例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
 
-间隙锁的出现主要集中在同一个事务中先 delete 后 insert 的情况下，当我们通过一个参数去删除一条记录的时候，如果参数在数据库中存在，那么这个时候产生的是普通行锁，锁住这个记录，然后删除，最后释放锁。如果这条记录不存在，问题就来了，数据库会扫描索引，发现这个记录不存在，这个时候的 delete 语句获取到的就是一个间隙锁，然后数据库会向左扫描扫到第一个比给定参数小的值，向右扫描扫描到第一个比给定参数大的值，然后以此为上下界限构建一个区间，进而锁住整个区间内的数据，一个特别容易出现死锁的间隙锁诞生了。
-
-通过修改 ```innodb_locaks_unsafe_for_binlog``` 参数来取消间隙锁从而达到避免这种情况的死锁的方式尚待商量，那就只有修改代码逻辑，存在才删除，尽量避免去删除不存在的记录。
+```sql
+SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
+```
 
 ## Next-Key Locks
 

From 02f03861376e9328e229f5a7264f8fde421d0395 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Thu, 9 Aug 2018 23:48:19 +0800
Subject: [PATCH 086/172] Update Group.md

---
 other/Group.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/other/Group.md b/other/Group.md
index 60ff3946..e4d54b22 100644
--- a/other/Group.md
+++ b/other/Group.md
@@ -1,6 +1,6 @@
 创建交流群的主要目的是为了给大家提供一个交流平台，方便大家在学习的过程中互相讨论。
 
-这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以帮助大家对知识的掌握程度。
+这个交流群不是一个笔者的问题回答群，我更希望大家能够愿意积极回答，我相信提问和回答的过程都可以提高大家对知识的掌握程度。
 
 因为笔者白天要上班，因此不能及时进行回复，大部分时间会处于潜水状态。
 

From e08c1f0e95d862edfbeef8cbd0c7a161430040ba Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 9 Aug 2018 23:49:33 +0800
Subject: [PATCH 087/172] auto commit

---
 notes/数据库系统原理.md | 30 ++++++++----------------------
 1 file changed, 8 insertions(+), 22 deletions(-)

diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index 151ffc0f..e7ef804b 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -383,13 +383,13 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 ## Record Locks
 
-锁定的对象是记录的索引，而不是记录本身。
+锁定一个记录上的索引，而不是记录本身。
 
 如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
 
 ## Gap Locks
 
-锁定一个范围内的索引，例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
+锁定索引之间的间隙，但是不包含索引本身。例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
 
 ```sql
 SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
@@ -397,28 +397,14 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 ## Next-Key Locks
 
-它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录，也锁定范围内的索引。在 user 中有以下记录：
+它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定范围内的索引。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
 
 ```sql
-|   id | last_name   | first_name   |   age |
-|------|-------------|--------------|-------|
-|    4 | stark       | tony         |    21 |
-|    1 | tom         | hiddleston   |    30 |
-|    3 | morgan      | freeman      |    40 |
-|    5 | jeff        | dean         |    50 |
-|    2 | donald      | trump        |    80 |
-+------|-------------|--------------|-------+
-```
-
-那么就需要锁定以下范围：
-
-```sql
-(-∞, 21]
-(21, 30]
-(30, 40]
-(40, 50]
-(50, 80]
-(80, ∞)
+(negative infinity, 10]
+(10, 11]
+(11, 13]
+(13, 20]
+(20, positive infinity)
 ```
 
 # 七、关系数据库设计理论

From 2015cf3e8594d22723d67e5d18801b71a8cb37ce Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 10 Aug 2018 16:41:20 +0800
Subject: [PATCH 088/172] Update README.md

---
 README.md | 66 +++++++++++++++++++++++++++----------------------------
 1 file changed, 33 insertions(+), 33 deletions(-)

diff --git a/README.md b/README.md
index 8dbf0507..f4c2b510 100644
--- a/README.md
+++ b/README.md
@@ -13,151 +13,151 @@
 
 ## 算法 :pencil2:
 
-> [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
+- [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
 目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
 
-> [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
+- [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
 
 对题目做了一个大致分类，并对每种题型的解题思路做了总结。
 
-> [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
+- [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
 
 排序、并查集、栈和队列、红黑树、散列表。
 
 ## 操作系统 :computer:
 
-> [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
+- [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
 进程管理、内存管理、设备管理、链接。
 
-> [Linux](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Linux.md)
+- [Linux](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Linux.md)
 
 基本实现原理以及基本操作。
 
 ## 网络 :cloud:
 
-> [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
+- [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
 物理层、链路层、网络层、运输层、应用层。
 
-> [HTTP](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/HTTP.md)
+- [HTTP](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/HTTP.md)
 
 方法、状态码、Cookie、缓存、连接管理、HTTPs、HTTP 2.0。
 
-> [Socket](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Socket.md)
+- [Socket](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Socket.md)
 
 I/O 模型、I/O 多路复用。
 
 ## 面向对象 :couple:
 
-> [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
+- [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
 实现了 Gof 的 23 种设计模式。
 
-> [面向对象思想](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/面向对象思想.md)
+- [面向对象思想](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/面向对象思想.md)
 
 三大原则（继承、封装、多态）、类图、设计原则。
 
 ## 数据库 :floppy_disk:
 
-> [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
+- [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
 事务、锁、隔离级别、MVCC、间隙锁、范式。
 
-> [SQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/SQL.md)
+- [SQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/SQL.md)
 
 SQL 基本语法。
 
-> [Leetcode-Database 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode-Database%20题解.md)
+- [Leetcode-Database 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode-Database%20题解.md)
 
 Leetcode 上数据库题目的解题记录。
 
-> [MySQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/MySQL.md)
+- [MySQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/MySQL.md)
 
 存储引擎、索引、查询优化、切分、复制。
 
-> [Redis](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Redis.md)
+- [Redis](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Redis.md)
 
 五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
 ## Java :coffee:
 
-> [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
+- [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
 不会涉及很多基本语法介绍，主要是一些实现原理以及关键特性。
 
-> [Java 容器](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20容器.md)
+- [Java 容器](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20容器.md)
 
 源码分析：ArrayList、Vector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
 
-> [Java 并发](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20并发.md)
+- [Java 并发](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20并发.md)
 
 线程使用方式、两种互斥同步方法、线程协作、JUC、线程安全、内存模型、锁优化。
 
-> [Java 虚拟机](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20虚拟机.md)
+- [Java 虚拟机](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20虚拟机.md)
 
 运行时数据区域、垃圾收集、类加载。
 
-> [Java I/O](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20IO.md)
+- [Java I/O](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20IO.md)
 
 NIO 的原理以及实例。
 
 ## 系统设计 :bulb:
 
-> [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
+- [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
 
 性能、伸缩性、扩展性、可用性、安全性
 
-> [分布式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/分布式.md)
+- [分布式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/分布式.md)
 
 分布式锁、分布式事务、CAP、BASE、Paxos、Raft
 
-> [集群](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/集群.md)
+- [集群](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/集群.md)
 
 负载均衡、Session 管理
 
-> [攻击技术](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/攻击技术.md)
+- [攻击技术](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/攻击技术.md)
 
 XSS、CSRF、SQL 注入、DDoS
 
-> [缓存](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/缓存.md)
+- [缓存](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/缓存.md)
 
 缓存特征、缓存位置、缓存问题、数据分布、一致性哈希、LRU、CDN
 
-> [消息队列](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/消息队列.md)
+- [消息队列](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/消息队列.md)
 
 消息处理模型、使用场景、可靠性
 
 ## 工具 :hammer:
 
-> [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
+- [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
 
 一些 Git 的使用和概念。
 
-> [Docker](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Docker.md)
+- [Docker](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Docker.md)
 
 Docker 基本原理。
 
-> [正则表达式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/正则表达式.md)
+- [正则表达式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/正则表达式.md)
 
 正则表达式基本语法。
 
-> [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
+- [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
 
 构建工具的基本概念、主流构建工具介绍。
 
 ## 编码实践 :speak_no_evil:
 
-> [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
+- [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
 
 参考 重构 改善既有代码的设计。
 
-> [代码可读性](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码可读性.md)
+- [代码可读性](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码可读性.md)
 
 参考 编写可读代码的艺术。
 
-> [代码风格规范](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码风格规范.md)
+- [代码风格规范](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码风格规范.md)
 
 Google 开源项目的代码风格规范。
 

From 8afe65df0c84979c576949f5e836a507ff7091d9 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 10 Aug 2018 16:42:23 +0800
Subject: [PATCH 089/172] Update README.md

---
 README.md | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/README.md b/README.md
index f4c2b510..728fb40c 100644
--- a/README.md
+++ b/README.md
@@ -15,25 +15,25 @@
 
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
-目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
+  目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
 
 - [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
 
-对题目做了一个大致分类，并对每种题型的解题思路做了总结。
+  对题目做了一个大致分类，并对每种题型的解题思路做了总结。
 
 - [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
 
-排序、并查集、栈和队列、红黑树、散列表。
+  排序、并查集、栈和队列、红黑树、散列表。
 
 ## 操作系统 :computer:
 
 - [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
-进程管理、内存管理、设备管理、链接。
+  进程管理、内存管理、设备管理、链接。
 
 - [Linux](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Linux.md)
 
-基本实现原理以及基本操作。
+  基本实现原理以及基本操作。
 
 ## 网络 :cloud:
 

From d2bafee9a193e05782c33d9ae88a253971ccb6bc Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 10 Aug 2018 16:43:23 +0800
Subject: [PATCH 090/172] Update README.md

---
 README.md | 56 +++++++++++++++++++++++++++----------------------------
 1 file changed, 28 insertions(+), 28 deletions(-)

diff --git a/README.md b/README.md
index 728fb40c..4de16ff4 100644
--- a/README.md
+++ b/README.md
@@ -39,127 +39,127 @@
 
 - [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
-物理层、链路层、网络层、运输层、应用层。
+  物理层、链路层、网络层、运输层、应用层。
 
 - [HTTP](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/HTTP.md)
 
-方法、状态码、Cookie、缓存、连接管理、HTTPs、HTTP 2.0。
+  方法、状态码、Cookie、缓存、连接管理、HTTPs、HTTP 2.0。
 
 - [Socket](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Socket.md)
 
-I/O 模型、I/O 多路复用。
+  I/O 模型、I/O 多路复用。
 
 ## 面向对象 :couple:
 
 - [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
-实现了 Gof 的 23 种设计模式。
+  实现了 Gof 的 23 种设计模式。
 
 - [面向对象思想](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/面向对象思想.md)
 
-三大原则（继承、封装、多态）、类图、设计原则。
+  三大原则（继承、封装、多态）、类图、设计原则。
 
 ## 数据库 :floppy_disk:
 
 - [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
-事务、锁、隔离级别、MVCC、间隙锁、范式。
+  事务、锁、隔离级别、MVCC、间隙锁、范式。
 
 - [SQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/SQL.md)
 
-SQL 基本语法。
+  SQL 基本语法。
 
 - [Leetcode-Database 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode-Database%20题解.md)
 
-Leetcode 上数据库题目的解题记录。
+  Leetcode 上数据库题目的解题记录。
 
 - [MySQL](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/MySQL.md)
 
-存储引擎、索引、查询优化、切分、复制。
+  存储引擎、索引、查询优化、切分、复制。
 
 - [Redis](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Redis.md)
 
-五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
+  五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
 ## Java :coffee:
 
 - [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
-不会涉及很多基本语法介绍，主要是一些实现原理以及关键特性。
+  不会涉及很多基本语法介绍，主要是一些实现原理以及关键特性。
 
 - [Java 容器](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20容器.md)
 
-源码分析：ArrayList、Vector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
+  源码分析：ArrayList、Vector、CopyOnWriteArrayList、LinkedList、HashMap、ConcurrentHashMap、LinkedHashMap、WeekHashMap。
 
 - [Java 并发](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20并发.md)
 
-线程使用方式、两种互斥同步方法、线程协作、JUC、线程安全、内存模型、锁优化。
+  线程使用方式、两种互斥同步方法、线程协作、JUC、线程安全、内存模型、锁优化。
 
 - [Java 虚拟机](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20虚拟机.md)
 
-运行时数据区域、垃圾收集、类加载。
+  运行时数据区域、垃圾收集、类加载。
 
 - [Java I/O](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20IO.md)
 
-NIO 的原理以及实例。
+  NIO 的原理以及实例。
 
 ## 系统设计 :bulb:
 
 - [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
 
-性能、伸缩性、扩展性、可用性、安全性
+  性能、伸缩性、扩展性、可用性、安全性
 
 - [分布式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/分布式.md)
 
-分布式锁、分布式事务、CAP、BASE、Paxos、Raft
+  分布式锁、分布式事务、CAP、BASE、Paxos、Raft
 
 - [集群](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/集群.md)
 
-负载均衡、Session 管理
+  负载均衡、Session 管理
 
 - [攻击技术](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/攻击技术.md)
 
-XSS、CSRF、SQL 注入、DDoS
+  XSS、CSRF、SQL 注入、DDoS
 
 - [缓存](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/缓存.md)
 
-缓存特征、缓存位置、缓存问题、数据分布、一致性哈希、LRU、CDN
+  缓存特征、缓存位置、缓存问题、数据分布、一致性哈希、LRU、CDN
 
 - [消息队列](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/消息队列.md)
 
-消息处理模型、使用场景、可靠性
+  消息处理模型、使用场景、可靠性
 
 ## 工具 :hammer:
 
 - [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
 
-一些 Git 的使用和概念。
+  一些 Git 的使用和概念。
 
 - [Docker](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Docker.md)
 
-Docker 基本原理。
+  Docker 基本原理。
 
 - [正则表达式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/正则表达式.md)
 
-正则表达式基本语法。
+  正则表达式基本语法。
 
 - [构建工具](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/构建工具.md)
 
-构建工具的基本概念、主流构建工具介绍。
+  构建工具的基本概念、主流构建工具介绍。
 
 ## 编码实践 :speak_no_evil:
 
 - [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
 
-参考 重构 改善既有代码的设计。
+  参考 重构 改善既有代码的设计。
 
 - [代码可读性](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码可读性.md)
 
-参考 编写可读代码的艺术。
+  参考 编写可读代码的艺术。
 
 - [代码风格规范](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/代码风格规范.md)
 
-Google 开源项目的代码风格规范。
+  Google 开源项目的代码风格规范。
 
 ## 后记 :memo:
 

From 305623c80f114d3bc4d04689902b9428d4c8a705 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 10 Aug 2018 16:49:08 +0800
Subject: [PATCH 091/172] Update README.md

---
 README.md | 18 +++++++++---------
 1 file changed, 9 insertions(+), 9 deletions(-)

diff --git a/README.md b/README.md
index 4de16ff4..42788e54 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 <div align="center">
     <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
-    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-qq-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
+    <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->
@@ -163,7 +163,7 @@
 
 ## 后记 :memo:
 
-**About**
+### About
 
 这个仓库是笔者的一个学习笔记，主要总结一些比较重要的知识点，希望对大家有所帮助。
 
@@ -171,7 +171,7 @@
 
 [BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)，这个书单是笔者至今看的一些比较好的技术书籍，虽然没有全都看完，但每本书多多少少都看了一部分。
 
-**How To Contribute**
+### How To Contribute
 
 笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
 
@@ -179,13 +179,13 @@
 
 欢迎在 Issue 中提交对本仓库的改进建议~
 
-**Authorization**
+### Authorization
 
 虽然没有加开源协议，但是允许非商业性使用。
 
 转载使用请注明出处，谢谢！
 
-**Typesetting**
+### Typesetting
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
 
@@ -193,7 +193,7 @@
 
 笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
 
-**Uploading**
+### Uploading
 
 笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
 
@@ -201,15 +201,15 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
-**Statement**
+### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)
 
-**Logo**
+### Logo
 
 Power by [logomakr](https://logomakr.com/).
 
-**Acknowledgements**
+### Acknowledgements
 
 感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与笔者联系。
 

From b6156c4247f05d29a65564f0e337545fde70fea1 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 10 Aug 2018 18:33:36 +0800
Subject: [PATCH 092/172] =?UTF-8?q?Update=20Leetcode=20=E9=A2=98=E8=A7=A3.?=
 =?UTF-8?q?md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Leetcode 题解.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 102fe9b6..d8e53838 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -57,9 +57,9 @@
 
 ## 贪心思想
 
-贪心思想保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
+保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
 
-**分配饼干** 
+**分配饼干**
 
 [455. Assign Cookies (Easy)](https://leetcode.com/problems/assign-cookies/description/)
 
@@ -74,7 +74,7 @@ You need to output 2.
 
 题目描述：每个孩子都有一个满足度，每个饼干都有一个大小，只有饼干的大小大于等于一个孩子的满足度，该孩子才会获得满足。求解最多可以获得满足的孩子数量。
 
-因为最小的孩子最容易得到满足，因此先满足最小孩子。给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因此贪心策略
+给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。
 
 证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，给该孩子分配第 n 个饼干，并且 m < n。我们可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
 

From 06b9804ec7f79f2340d9e97b4e4fc9b8a9adb726 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Fri, 10 Aug 2018 21:58:11 +0800
Subject: [PATCH 093/172] auto commit

---
 notes/Leetcode 题解.md          | 970 ++++++++++++++++----------------
 notes/Leetcode-Database 题解.md |  20 +-
 notes/设计模式.md               |  12 +-
 3 files changed, 508 insertions(+), 494 deletions(-)

diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 102fe9b6..44f3d9dc 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -1,18 +1,18 @@
 <!-- GFM-TOC -->
 * [算法思想](#算法思想)
-    * [贪心思想](#贪心思想)
     * [双指针](#双指针)
     * [排序](#排序)
         * [快速选择](#快速选择)
         * [堆排序](#堆排序)
         * [桶排序](#桶排序)
         * [荷兰国旗问题](#荷兰国旗问题)
+    * [贪心思想](#贪心思想)
     * [二分查找](#二分查找)
+    * [分治](#分治)
     * [搜索](#搜索)
         * [BFS](#bfs)
         * [DFS](#dfs)
         * [Backtracking](#backtracking)
-    * [分治](#分治)
     * [动态规划](#动态规划)
         * [斐波那契数列](#斐波那契数列)
         * [矩阵路径](#矩阵路径)
@@ -55,9 +55,449 @@
 
 # 算法思想
 
+## 双指针
+
+双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
+
+**有序数组的 Two Sum** 
+
+[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
+
+```html
+Input: numbers={2, 7, 11, 15}, target=9
+Output: index1=1, index2=2
+```
+
+题目描述：在有序数组中找出两个数，使它们的和为 target。
+
+使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
+
+- 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
+- 如果 sum > target，移动较大的元素，使 sum 变小一些；
+- 如果 sum < target，移动较小的元素，使 sum 变大一些。
+
+```java
+public int[] twoSum(int[] numbers, int target) {
+    int i = 0, j = numbers.length - 1;
+    while (i < j) {
+        int sum = numbers[i] + numbers[j];
+        if (sum == target) {
+            return new int[]{i + 1, j + 1};
+        } else if (sum < target) {
+            i++;
+        } else {
+            j--;
+        }
+    }
+    return null;
+}
+```
+
+**两数平方和** 
+
+[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
+
+```html
+Input: 5
+Output: True
+Explanation: 1 * 1 + 2 * 2 = 5
+```
+
+题目描述：判断一个数是否为两个数的平方和，例如 5 = 1<sup>2</sup> + 2<sup>2</sup>。
+
+```java
+public boolean judgeSquareSum(int c) {
+    int i = 0, j = (int) Math.sqrt(c);
+    while (i <= j) {
+        int powSum = i * i + j * j;
+        if (powSum == c) {
+            return true;
+        } else if (powSum > c) {
+            j--;
+        } else {
+            i++;
+        }
+    }
+    return false;
+}
+```
+
+**反转字符串中的元音字符** 
+
+[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
+
+```html
+Given s = "leetcode", return "leotcede".
+```
+
+使用双指针，指向待反转的两个元音字符，一个指针从头向尾遍历，一个指针从尾到头遍历。
+
+```java
+private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
+
+public String reverseVowels(String s) {
+    int i = 0, j = s.length() - 1;
+    char[] result = new char[s.length()];
+    while (i <= j) {
+        char ci = s.charAt(i);
+        char cj = s.charAt(j);
+        if (!vowels.contains(ci)) {
+            result[i++] = ci;
+        } else if (!vowels.contains(cj)) {
+            result[j--] = cj;
+        } else {
+            result[i++] = cj;
+            result[j--] = ci;
+        }
+    }
+    return new String(result);
+}
+```
+
+**回文字符串** 
+
+[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
+
+```html
+Input: "abca"
+Output: True
+Explanation: You could delete the character 'c'.
+```
+
+题目描述：可以删除一个字符，判断是否能构成回文字符串。
+
+```java
+public boolean validPalindrome(String s) {
+    int i = -1, j = s.length();
+    while (++i < --j) {
+        if (s.charAt(i) != s.charAt(j)) {
+            return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
+        }
+    }
+    return true;
+}
+
+private boolean isPalindrome(String s, int i, int j) {
+    while (i < j) {
+        if (s.charAt(i++) != s.charAt(j--)) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+**归并两个有序数组** 
+
+[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
+
+```html
+Input:
+nums1 = [1,2,3,0,0,0], m = 3
+nums2 = [2,5,6],       n = 3
+
+Output: [1,2,2,3,5,6]
+```
+
+题目描述：把归并结果存到第一个数组上。
+
+需要从尾开始遍历，否则在 nums1 上归并得到的值会覆盖还未进行归并比较的值。
+
+```java
+public void merge(int[] nums1, int m, int[] nums2, int n) {
+    int index1 = m - 1, index2 = n - 1;
+    int indexMerge = m + n - 1;
+    while (index1 >= 0 || index2 >= 0) {
+        if (index1 < 0) {
+            nums1[indexMerge--] = nums2[index2--];
+        } else if (index2 < 0) {
+            nums1[indexMerge--] = nums1[index1--];
+        } else if (nums1[index1] > nums2[index2]) {
+            nums1[indexMerge--] = nums1[index1--];
+        } else {
+            nums1[indexMerge--] = nums2[index2--];
+        }
+    }
+}
+```
+
+**判断链表是否存在环** 
+
+[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
+
+使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
+
+```java
+public boolean hasCycle(ListNode head) {
+    if (head == null) {
+        return false;
+    }
+    ListNode l1 = head, l2 = head.next;
+    while (l1 != null && l2 != null && l2.next != null) {
+        if (l1 == l2) {
+            return true;
+        }
+        l1 = l1.next;
+        l2 = l2.next.next;
+    }
+    return false;
+}
+```
+
+**最长子序列** 
+
+[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
+
+```
+Input:
+s = "abpcplea", d = ["ale","apple","monkey","plea"]
+
+Output:
+"apple"
+```
+
+题目描述：删除 s 中的一些字符，使得它构成字符串列表 d 中的一个字符串，找出能构成的最长字符串。如果有多个相同长度的结果，返回字典序的最大字符串。
+
+```java
+public String findLongestWord(String s, List<String> d) {
+    String longestWord = "";
+    for (String target : d) {
+        int l1 = longestWord.length(), l2 = target.length();
+        if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
+            continue;
+        }
+        if (isValid(s, target)) {
+            longestWord = target;
+        }
+    }
+    return longestWord;
+}
+
+private boolean isValid(String s, String target) {
+    int i = 0, j = 0;
+    while (i < s.length() && j < target.length()) {
+        if (s.charAt(i) == target.charAt(j)) {
+            j++;
+        }
+        i++;
+    }
+    return j == target.length();
+}
+```
+
+## 排序
+
+### 快速选择
+
+用于求解  **Kth Element**  问题，使用快速排序的 partition() 进行实现。
+
+需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
+
+### 堆排序
+
+用于求解  **TopK Elements**  问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。
+
+堆排序也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。
+
+快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。
+
+可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
+
+**Kth Element** 
+
+[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
+
+**排序** ：时间复杂度 O(NlogN)，空间复杂度 O(1)
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    Arrays.sort(nums);
+    return nums[nums.length - k];
+}
+```
+
+**堆排序** ：时间复杂度 O(NlogK)，空间复杂度 O(K)。
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
+    for (int val : nums) {
+        pq.add(val);
+        if (pq.size() > k) // 维护堆的大小为 K
+            pq.poll();
+    }
+    return pq.peek();
+}
+```
+
+**快速选择** ：时间复杂度 O(N)，空间复杂度 O(1)
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    k = nums.length - k;
+    int l = 0, h = nums.length - 1;
+    while (l < h) {
+        int j = partition(nums, l, h);
+        if (j == k) {
+            break;
+        } else if (j < k) {
+            l = j + 1;
+        } else {
+            h = j - 1;
+        }
+    }
+    return nums[k];
+}
+
+private int partition(int[] a, int l, int h) {
+    int i = l, j = h + 1;
+    while (true) {
+        while (a[++i] < a[l] && i < h) ;
+        while (a[--j] > a[l] && j > l) ;
+        if (i >= j) {
+            break;
+        }
+        swap(a, i, j);
+    }
+    swap(a, l, j);
+    return j;
+}
+
+private void swap(int[] a, int i, int j) {
+    int t = a[i];
+    a[i] = a[j];
+    a[j] = t;
+}
+```
+
+### 桶排序
+
+**出现频率最多的 k 个数** 
+
+[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
+
+```html
+Given [1,1,1,2,2,3] and k = 2, return [1,2].
+```
+
+设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。
+
+把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
+
+```java
+public List<Integer> topKFrequent(int[] nums, int k) {
+    Map<Integer, Integer> frequencyForNum = new HashMap<>();
+    for (int num : nums) {
+        frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
+    }
+    List<Integer>[] buckets = new ArrayList[nums.length + 1];
+    for (int key : frequencyForNum.keySet()) {
+        int frequency = frequencyForNum.get(key);
+        if (buckets[frequency] == null) {
+            buckets[frequency] = new ArrayList<>();
+        }
+        buckets[frequency].add(key);
+    }
+    List<Integer> topK = new ArrayList<>();
+    for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
+        if (buckets[i] != null) {
+            topK.addAll(buckets[i]);
+        }
+    }
+    return topK;
+}
+```
+
+**按照字符出现次数对字符串排序** 
+
+[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
+
+```html
+Input:
+"tree"
+
+Output:
+"eert"
+
+Explanation:
+'e' appears twice while 'r' and 't' both appear once.
+So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+```
+
+```java
+public String frequencySort(String s) {
+    Map<Character, Integer> frequencyForNum = new HashMap<>();
+    for (char c : s.toCharArray())
+        frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
+
+    List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
+    for (char c : frequencyForNum.keySet()) {
+        int f = frequencyForNum.get(c);
+        if (frequencyBucket[f] == null) {
+            frequencyBucket[f] = new ArrayList<>();
+        }
+        frequencyBucket[f].add(c);
+    }
+    StringBuilder str = new StringBuilder();
+    for (int i = frequencyBucket.length - 1; i >= 0; i--) {
+        if (frequencyBucket[i] == null) {
+            continue;
+        }
+        for (char c : frequencyBucket[i]) {
+            for (int j = 0; j < i; j++) {
+                str.append(c);
+            }
+        }
+    }
+    return str.toString();
+}
+```
+
+### 荷兰国旗问题
+
+荷兰国旗包含三种颜色：红、白、蓝。
+
+有三种颜色的球，算法的目标是将这三种球按颜色顺序正确地排列。
+
+它其实是三向切分快速排序的一种变种，在三向切分快速排序中，每次切分都将数组分成三个区间：小于切分元素、等于切分元素、大于切分元素，而该算法是将数组分成三个区间：等于红色、等于白色、等于蓝色。
+
+<div align="center"> <img src="../pics//3b49dd67-2c40-4b81-8ad2-7bbb1fe2fcbd.png"/> </div><br>
+
+**按颜色进行排序** 
+
+[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
+
+```html
+Input: [2,0,2,1,1,0]
+Output: [0,0,1,1,2,2]
+```
+
+题目描述：只有 0/1/2 三种颜色。
+
+```java
+public void sortColors(int[] nums) {
+    int zero = -1, one = 0, two = nums.length;
+    while (one < two) {
+        if (nums[one] == 0) {
+            swap(nums, ++zero, one++);
+        } else if (nums[one] == 2) {
+            swap(nums, --two, one);
+        } else {
+            ++one;
+        }
+    }
+}
+
+private void swap(int[] nums, int i, int j) {
+    int t = nums[i];
+    nums[i] = nums[j];
+    nums[j] = t;
+}
+```
+
 ## 贪心思想
 
-贪心思想保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
+保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
 
 **分配饼干** 
 
@@ -74,7 +514,7 @@ You need to output 2.
 
 题目描述：每个孩子都有一个满足度，每个饼干都有一个大小，只有饼干的大小大于等于一个孩子的满足度，该孩子才会获得满足。求解最多可以获得满足的孩子数量。
 
-因为最小的孩子最容易得到满足，因此先满足最小孩子。给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因此贪心策略
+给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。
 
 证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，给该孩子分配第 n 个饼干，并且 m < n。我们可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
 
@@ -362,434 +802,6 @@ public int maxProfit(int[] prices) {
 }
 ```
 
-## 双指针
-
-双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
-
-**有序数组的 Two Sum** 
-
-[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
-
-```html
-Input: numbers={2, 7, 11, 15}, target=9
-Output: index1=1, index2=2
-```
-
-题目描述：在有序数组中找出两个数，使它们的和为 target。
-
-使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
-
-如果两个指针指向元素的和 sum == target，那么得到要求的结果；如果 sum > target，移动较大的元素，使 sum 变小一些；如果 sum < target，移动较小的元素，使 sum 变大一些。
-
-```java
-public int[] twoSum(int[] numbers, int target) {
-    int i = 0, j = numbers.length - 1;
-    while (i < j) {
-        int sum = numbers[i] + numbers[j];
-        if (sum == target) {
-            return new int[]{i + 1, j + 1};
-        } else if (sum < target) {
-            i++;
-        } else {
-            j--;
-        }
-    }
-    return null;
-}
-```
-
-**两数平方和** 
-
-[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
-
-```html
-Input: 5
-Output: True
-Explanation: 1 * 1 + 2 * 2 = 5
-```
-
-题目描述：判断一个数是否为两个数的平方和，例如 5 = 1<sup>2</sup> + 2<sup>2</sup>。
-
-```java
-public boolean judgeSquareSum(int c) {
-    int i = 0, j = (int) Math.sqrt(c);
-    while (i <= j) {
-        int powSum = i * i + j * j;
-        if (powSum == c) {
-            return true;
-        } else if (powSum > c) {
-            j--;
-        } else {
-            i++;
-        }
-    }
-    return false;
-}
-```
-
-**反转字符串中的元音字符** 
-
-[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
-
-```html
-Given s = "leetcode", return "leotcede".
-```
-
-使用双指针，指向待反转的两个元音字符，一个指针从头向尾遍历，一个指针从尾到头遍历。
-
-```java
-private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
-
-public String reverseVowels(String s) {
-    int i = 0, j = s.length() - 1;
-    char[] result = new char[s.length()];
-    while (i <= j) {
-        char ci = s.charAt(i);
-        char cj = s.charAt(j);
-        if (!vowels.contains(ci)) {
-            result[i++] = ci;
-        } else if (!vowels.contains(cj)) {
-            result[j--] = cj;
-        } else {
-            result[i++] = cj;
-            result[j--] = ci;
-        }
-    }
-    return new String(result);
-}
-```
-
-**回文字符串** 
-
-[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
-
-```html
-Input: "abca"
-Output: True
-Explanation: You could delete the character 'c'.
-```
-
-题目描述：可以删除一个字符，判断是否能构成回文字符串。
-
-```java
-public boolean validPalindrome(String s) {
-    int i = -1, j = s.length();
-    while (++i < --j) {
-        if (s.charAt(i) != s.charAt(j)) {
-            return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
-        }
-    }
-    return true;
-}
-
-private boolean isPalindrome(String s, int i, int j) {
-    while (i < j) {
-        if (s.charAt(i++) != s.charAt(j--)) {
-            return false;
-        }
-    }
-    return true;
-}
-```
-
-**归并两个有序数组** 
-
-[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
-
-```html
-Input:
-nums1 = [1,2,3,0,0,0], m = 3
-nums2 = [2,5,6],       n = 3
-
-Output: [1,2,2,3,5,6]
-```
-
-题目描述：把归并结果存到第一个数组上。
-
-需要从尾开始遍历，否则在 nums1 上归并得到的值会覆盖还未进行归并比较的值
-
-```java
-public void merge(int[] nums1, int m, int[] nums2, int n) {
-    int index1 = m - 1, index2 = n - 1;
-    int indexMerge = m + n - 1;
-    while (index1 >= 0 || index2 >= 0) {
-        if (index1 < 0) {
-            nums1[indexMerge--] = nums2[index2--];
-        } else if (index2 < 0) {
-            nums1[indexMerge--] = nums1[index1--];
-        } else if (nums1[index1] > nums2[index2]) {
-            nums1[indexMerge--] = nums1[index1--];
-        } else {
-            nums1[indexMerge--] = nums2[index2--];
-        }
-    }
-}
-```
-
-**判断链表是否存在环** 
-
-[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
-
-使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
-
-```java
-public boolean hasCycle(ListNode head) {
-    if (head == null) {
-        return false;
-    }
-    ListNode l1 = head, l2 = head.next;
-    while (l1 != null && l2 != null && l2.next != null) {
-        if (l1 == l2) {
-            return true;
-        }
-        l1 = l1.next;
-        l2 = l2.next.next;
-    }
-    return false;
-}
-```
-
-**最长子序列** 
-
-[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
-
-```
-Input:
-s = "abpcplea", d = ["ale","apple","monkey","plea"]
-
-Output:
-"apple"
-```
-
-题目描述：删除 s 中的一些字符，使得它构成字符串列表 d 中的一个字符串，找出能构成的最长字符串。如果有多个相同长度的结果，返回按字典序排序的最大字符串。
-
-```java
-public String findLongestWord(String s, List<String> d) {
-    String longestWord = "";
-    for (String target : d) {
-        int l1 = longestWord.length(), l2 = target.length();
-        if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
-            continue;
-        }
-        if (isValid(s, target)) {
-            longestWord = target;
-        }
-    }
-    return longestWord;
-}
-
-private boolean isValid(String s, String target) {
-    int i = 0, j = 0;
-    while (i < s.length() && j < target.length()) {
-        if (s.charAt(i) == target.charAt(j)) {
-            j++;
-        }
-        i++;
-    }
-    return j == target.length();
-}
-```
-
-## 排序
-
-### 快速选择
-
-一般用于求解  **Kth Element**  问题，可以在 O(N) 时间复杂度，O(1) 空间复杂度完成求解工作。
-
-与快速排序一样，快速选择一般需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
-
-### 堆排序
-
-堆排序用于求解  **TopK Elements**  问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。当然它也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
-
-**Kth Element** 
-
-[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
-
-**排序** ：时间复杂度 O(NlogN)，空间复杂度 O(1)
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    Arrays.sort(nums);
-    return nums[nums.length - k];
-}
-```
-
-**堆排序** ：时间复杂度 O(NlogK)，空间复杂度 O(K)。
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
-    for (int val : nums) {
-        pq.add(val);
-        if (pq.size() > k) // 维护堆的大小为 K
-            pq.poll();
-    }
-    return pq.peek();
-}
-```
-
-**快速选择** ：时间复杂度 O(N)，空间复杂度 O(1)
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    k = nums.length - k;
-    int l = 0, h = nums.length - 1;
-    while (l < h) {
-        int j = partition(nums, l, h);
-        if (j == k) {
-            break;
-        } else if (j < k) {
-            l = j + 1;
-        } else {
-            h = j - 1;
-        }
-    }
-    return nums[k];
-}
-
-private int partition(int[] a, int l, int h) {
-    int i = l, j = h + 1;
-    while (true) {
-        while (a[++i] < a[l] && i < h) ;
-        while (a[--j] > a[l] && j > l) ;
-        if (i >= j) {
-            break;
-        }
-        swap(a, i, j);
-    }
-    swap(a, l, j);
-    return j;
-}
-
-private void swap(int[] a, int i, int j) {
-    int t = a[i];
-    a[i] = a[j];
-    a[j] = t;
-}
-```
-
-### 桶排序
-
-**出现频率最多的 k 个数** 
-
-[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
-
-```html
-Given [1,1,1,2,2,3] and k = 2, return [1,2].
-```
-
-设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
-
-```java
-public List<Integer> topKFrequent(int[] nums, int k) {
-    Map<Integer, Integer> frequencyForNum = new HashMap<>();
-    for (int num : nums) {
-        frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
-    }
-    List<Integer>[] buckets = new ArrayList[nums.length + 1];
-    for (int key : frequencyForNum.keySet()) {
-        int frequency = frequencyForNum.get(key);
-        if (buckets[frequency] == null) {
-            buckets[frequency] = new ArrayList<>();
-        }
-        buckets[frequency].add(key);
-    }
-    List<Integer> topK = new ArrayList<>();
-    for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
-        if (buckets[i] != null) {
-            topK.addAll(buckets[i]);
-        }
-    }
-    return topK;
-}
-```
-
-**按照字符出现次数对字符串排序** 
-
-[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
-
-```html
-Input:
-"tree"
-
-Output:
-"eert"
-
-Explanation:
-'e' appears twice while 'r' and 't' both appear once.
-So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
-```
-
-```java
-public String frequencySort(String s) {
-    Map<Character, Integer> frequencyForNum = new HashMap<>();
-    for (char c : s.toCharArray())
-        frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
-
-    List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
-    for (char c : frequencyForNum.keySet()) {
-        int f = frequencyForNum.get(c);
-        if (frequencyBucket[f] == null) {
-            frequencyBucket[f] = new ArrayList<>();
-        }
-        frequencyBucket[f].add(c);
-    }
-    StringBuilder str = new StringBuilder();
-    for (int i = frequencyBucket.length - 1; i >= 0; i--) {
-        if (frequencyBucket[i] == null) {
-            continue;
-        }
-        for (char c : frequencyBucket[i]) {
-            for (int j = 0; j < i; j++) {
-                str.append(c);
-            }
-        }
-    }
-    return str.toString();
-}
-```
-
-### 荷兰国旗问题
-
-荷兰国旗包含三种颜色：红、白、蓝。有这三种颜色的球，算法的目标是将这三种球按颜色顺序正确地排列。
-
-它其实是三向切分快速排序的一种变种，在三向切分快速排序中，每次切分都将数组分成三个区间：小于切分元素、等于切分元素、大于切分元素，而该算法是将数组分成三个区间：等于红色、等于白色、等于蓝色。
-
-<div align="center"> <img src="../pics//3b49dd67-2c40-4b81-8ad2-7bbb1fe2fcbd.png"/> </div><br>
-
-**按颜色进行排序** 
-
-[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
-
-```html
-Input: [2,0,2,1,1,0]
-Output: [0,0,1,1,2,2]
-```
-
-题目描述：只有 0/1/2 三种颜色。
-
-```java
-public void sortColors(int[] nums) {
-    int zero = -1, one = 0, two = nums.length;
-    while (one < two) {
-        if (nums[one] == 0) {
-            swap(nums, ++zero, one++);
-        } else if (nums[one] == 2) {
-            swap(nums, --two, one);
-        } else {
-            ++one;
-        }
-    }
-}
-
-private void swap(int[] nums, int i, int j) {
-    int t = nums[i];
-    nums[i] = nums[j];
-    nums[j] = t;
-}
-```
-
 ## 二分查找
 
 **正常实现** 
@@ -1065,6 +1077,53 @@ private int binarySearch(int[] nums, int target) {
 }
 ```
 
+## 分治
+
+**给表达式加括号** 
+
+[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
+
+```html
+Input: "2-1-1".
+
+((2-1)-1) = 0
+(2-(1-1)) = 2
+
+Output : [0, 2]
+```
+
+```java
+public List<Integer> diffWaysToCompute(String input) {
+    List<Integer> ways = new ArrayList<>();
+    for (int i = 0; i < input.length(); i++) {
+        char c = input.charAt(i);
+        if (c == '+' || c == '-' || c == '*') {
+            List<Integer> left = diffWaysToCompute(input.substring(0, i));
+            List<Integer> right = diffWaysToCompute(input.substring(i + 1));
+            for (int l : left) {
+                for (int r : right) {
+                    switch (c) {
+                        case '+':
+                            ways.add(l + r);
+                            break;
+                        case '-':
+                            ways.add(l - r);
+                            break;
+                        case '*':
+                            ways.add(l * r);
+                            break;
+                    }
+                }
+            }
+        }
+    }
+    if (ways.size() == 0) {
+        ways.add(Integer.valueOf(input));
+    }
+    return ways;
+}
+```
+
 ## 搜索
 
 深度优先搜索和广度优先搜索广泛运用于树和图中，但是它们的应用远远不止如此。
@@ -2312,53 +2371,6 @@ private void backtracking(int row) {
 }
 ```
 
-## 分治
-
-**给表达式加括号** 
-
-[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
-
-```html
-Input: "2-1-1".
-
-((2-1)-1) = 0
-(2-(1-1)) = 2
-
-Output : [0, 2]
-```
-
-```java
-public List<Integer> diffWaysToCompute(String input) {
-    List<Integer> ways = new ArrayList<>();
-    for (int i = 0; i < input.length(); i++) {
-        char c = input.charAt(i);
-        if (c == '+' || c == '-' || c == '*') {
-            List<Integer> left = diffWaysToCompute(input.substring(0, i));
-            List<Integer> right = diffWaysToCompute(input.substring(i + 1));
-            for (int l : left) {
-                for (int r : right) {
-                    switch (c) {
-                        case '+':
-                            ways.add(l + r);
-                            break;
-                        case '-':
-                            ways.add(l - r);
-                            break;
-                        case '*':
-                            ways.add(l * r);
-                            break;
-                    }
-                }
-            }
-        }
-    }
-    if (ways.size() == 0) {
-        ways.add(Integer.valueOf(input));
-    }
-    return ways;
-}
-```
-
 ## 动态规划
 
 递归和动态规划都是将原问题拆成多个子问题然后求解，他们之间最本质的区别是，动态规划保存了子问题的解，避免重复计算。
diff --git a/notes/Leetcode-Database 题解.md b/notes/Leetcode-Database 题解.md
index 5ab726c5..5b4a3d00 100644
--- a/notes/Leetcode-Database 题解.md	
+++ b/notes/Leetcode-Database 题解.md	
@@ -461,7 +461,7 @@ Employee 表：
 +----+-------+--------+-----------+
 ```
 
-查找所有员工，他们的薪资大于其经理薪资。
+查找薪资大于其经理薪资的员工信息。
 
 ## SQL Schema
 
@@ -924,27 +924,27 @@ VALUES
 ```sql
 SELECT
     s1.id - 1 AS id,
-    s1.student 
+    s1.student
 FROM
-    seat s1 
+    seat s1
 WHERE
     s1.id MOD 2 = 0 UNION
 SELECT
     s2.id + 1 AS id,
-    s2.student 
+    s2.student
 FROM
-    seat s2 
+    seat s2
 WHERE
-    s2.id MOD 2 = 1 
+    s2.id MOD 2 = 1
     AND s2.id != ( SELECT max( s3.id ) FROM seat s3 ) UNION
 SELECT
     s4.id AS id,
-    s4.student 
+    s4.student
 FROM
-    seat s4 
+    seat s4
 WHERE
-    s4.id MOD 2 = 1 
-    AND s4.id = ( SELECT max( s5.id ) FROM seat s5 ) 
+    s4.id MOD 2 = 1
+    AND s4.id = ( SELECT max( s5.id ) FROM seat s5 )
 ORDER BY
     id;
 ```
diff --git a/notes/设计模式.md b/notes/设计模式.md
index d48d7b36..b8f4dc9b 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -98,7 +98,9 @@ public static synchronized Singleton getUniqueInstance() {
 
 （三）饿汉式-线程安全
 
-线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。但是直接实例化的方式也丢失了延迟实例化带来的节约资源的优势。
+线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。
+
+但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
 
 ```java
 private static Singleton uniqueInstance = new Singleton();
@@ -106,7 +108,7 @@ private static Singleton uniqueInstance = new Singleton();
 
 （四）双重校验锁-线程安全
 
-uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行。也就是说，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
+uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
 
 双重校验锁先判断 uniqueInstance 是否已经被实例化，如果没有被实例化，那么才对实例化语句进行加锁。
 
@@ -131,7 +133,7 @@ public class Singleton {
 }
 ```
 
-考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程同时执行 if 语句，那么两个线程就会同时进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，也就是说会进行两次实例化，从而产生了两个实例。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
+考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程同时执行 if 语句，那么两个线程就会同时进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，那么就会进行两次实例化，从而产生了两个实例。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
 
 ```java
 if (uniqueInstance == null) {
@@ -157,7 +159,7 @@ uniqueInstance 采用 volatile 关键字修饰也是很有必要的。`uniqueIns
 
 这种方式不仅具有延迟初始化的好处，而且由虚拟机提供了对线程安全的支持。
 
-```source-java
+```java
 public class Singleton {
 
     private Singleton() {
@@ -299,7 +301,7 @@ public class Client {
 
 ### 意图
 
-定义了一个创建对象的接口，但由子类决定要实例化哪个类。工厂方法把实例化推迟到子类。
+定义了一个创建对象的接口，但由子类决定要实例化哪个类。工厂方法把实例化操作推迟到子类。
 
 ### 类图
 

From bcc266ff354dae6b5fdff7771c4f438399fa4c4c Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Fri, 10 Aug 2018 22:00:21 +0800
Subject: [PATCH 094/172] auto commit

---
 notes/Leetcode 题解.md | 10 ----------
 1 file changed, 10 deletions(-)

diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index aebe9d72..44f3d9dc 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -57,15 +57,9 @@
 
 ## 双指针
 
-<<<<<<< HEAD
 双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
 
 **有序数组的 Two Sum** 
-=======
-保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
-
-**分配饼干**
->>>>>>> b6156c4247f05d29a65564f0e337545fde70fea1
 
 [Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
 
@@ -76,11 +70,7 @@ Output: index1=1, index2=2
 
 题目描述：在有序数组中找出两个数，使它们的和为 target。
 
-<<<<<<< HEAD
 使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
-=======
-给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。
->>>>>>> b6156c4247f05d29a65564f0e337545fde70fea1
 
 - 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
 - 如果 sum > target，移动较大的元素，使 sum 变小一些；

From f857a0c99675fbe676c54749ccb631ee71b7e324 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E4=BC=9F?= <18056981502@163.com>
Date: Sun, 12 Aug 2018 22:19:41 +0800
Subject: [PATCH 095/172] List <|.. LinkeList  to  List <|.. LinkedList

---
 notes/Java 容器.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index bf1921fc..bd4ecfdb 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -1080,7 +1080,7 @@ Set <|.. LinkedHashSet
 SortSet <|.. TreeSet
 List <|.. ArrayList
 List <|.. Vector
-List <|.. LinkeList
+List <|.. LinkedList
 Queue <|.. LinkedList
 Queue <|.. PriorityQueue
 

From 9a4f6ea86cffff577870849dcdd5d724295937e9 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 12 Aug 2018 22:53:28 +0800
Subject: [PATCH 096/172] auto commit

---
 notes/HTTP.md                            |   33 +-
 notes/Java 容器.md                       |   61 +-
 notes/Leetcode 题解.md                   | 2933 +++++++++++-----------
 notes/Leetcode-Database 题解.md          |   20 +-
 notes/剑指 offer 题解.md                 |  281 +--
 notes/攻击技术.md                        |    8 +-
 notes/数据库系统原理.md                  |   30 +-
 notes/算法.md                            |   30 +-
 notes/系统设计基础.md                    |    2 +-
 notes/设计模式.md                        |   43 +-
 pics/VP4n3i8m34Ntd28NQ4_0KCJ2q044Oez.png |  Bin 0 -> 21504 bytes
 11 files changed, 1719 insertions(+), 1722 deletions(-)
 create mode 100644 pics/VP4n3i8m34Ntd28NQ4_0KCJ2q044Oez.png

diff --git a/notes/HTTP.md b/notes/HTTP.md
index 243dc283..537b24fe 100644
--- a/notes/HTTP.md
+++ b/notes/HTTP.md
@@ -45,7 +45,7 @@
     * [二进制分帧层](#二进制分帧层)
     * [服务端推送](#服务端推送)
     * [首部压缩](#首部压缩)
-* [八、GET 和 POST 的区别](#八get-和-post-的区别)
+* [八、GET 和 POST 比较](#八get-和-post-比较)
     * [作用](#作用)
     * [参数](#参数)
     * [安全](#安全)
@@ -61,13 +61,13 @@
 
 ## URL
 
-- URI（Uniform Resource Identifier，统一资源标识符）
-- URL（Uniform Resource Locator，统一资源定位符）
-- URN（Uniform Resource Name，统一资源名称），例如 urn:isbn:0-486-27557-4。
-
 URI 包含 URL 和 URN，目前 WEB 只有 URL 比较流行，所以见到的基本都是 URL。
 
-<div align="center"> <img src="../pics//f716427a-94f2-4875-9c86-98793cf5dcc3.jpg" width="400"/> </div><br>
+- URI（Uniform Resource Identifier，统一资源标识符）
+- URL（Uniform Resource Locator，统一资源定位符）
+- URN（Uniform Resource Name，统一资源名称）
+
+<div align="center"> <img src="../pics//urlnuri.jpg" width="600"/> </div><br>
 
 ## 请求和响应报文
 
@@ -197,7 +197,7 @@ CONNECT www.example.com:443 HTTP/1.1
 
 -  **204 No Content** ：请求已经成功处理，但是返回的响应报文不包含实体的主体部分。一般在只需要从客户端往服务器发送信息，而不需要返回数据时使用。
 
--  **206 Partial Content** ：表示客户端进行了范围请求。响应报文包含由 Content-Range 指定范围的实体内容。
+-  **206 Partial Content** ：表示客户端进行了范围请求，响应报文包含由 Content-Range 指定范围的实体内容。
 
 ## 3XX 重定向
 
@@ -219,7 +219,7 @@ CONNECT www.example.com:443 HTTP/1.1
 
 -  **401 Unauthorized** ：该状态码表示发送的请求需要有认证信息（BASIC 认证、DIGEST 认证）。如果之前已进行过一次请求，则表示用户认证失败。
 
--  **403 Forbidden** ：请求被拒绝，服务器端没有必要给出拒绝的详细理由。
+-  **403 Forbidden** ：请求被拒绝。
 
 -  **404 Not Found** 
 
@@ -331,7 +331,7 @@ Set-Cookie: tasty_cookie=strawberry
 [page content]
 ```
 
-客户端之后对同一个服务器发送请求时，会从浏览器中读出 Cookie 信息通过 Cookie 请求首部字段发送给服务器。
+客户端之后对同一个服务器发送请求时，会从浏览器中取出 Cookie 信息并通过 Cookie 请求首部字段发送给服务器。
 
 ```html
 GET /sample_page.html HTTP/1.1
@@ -382,9 +382,9 @@ Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径
 
 除了可以将用户信息通过 Cookie 存储在用户浏览器中，也可以利用 Session 存储在服务器端，存储在服务器端的信息更加安全。
 
-Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在内存型数据库中，比如 Redis，效率会更高。
+Session 可以存储在服务器上的文件、数据库或者内存中。也可以将 Session 存储在 Redis 这种内存型数据库中，效率会更高。
 
-使用 Session 维护用户登录的过程如下：
+使用 Session 维护用户登录状态的过程如下：
 
 - 用户进行登录时，用户提交包含用户名和密码的表单，放入 HTTP 请求报文中；
 - 服务器验证该用户名和密码；
@@ -499,7 +499,7 @@ If-Modified-Since: Wed, 21 Oct 2015 07:28:00 GMT
 
 ### 1. 短连接与长连接
 
-当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源。如果每进行一次 HTTP 通信就要断开一次 TCP 连接，连接建立和断开的开销会很大。
+当浏览器访问一个包含多张图片的 HTML 页面时，除了请求访问 HTML 页面资源，还会请求图片资源。如果每进行一次 HTTP 通信就要新建一个 TCP 连接，那么开销会很大。
 
 长连接只需要建立一次 TCP 连接就能进行多次 HTTP 通信。
 
@@ -688,7 +688,7 @@ HTTPs 并不是新协议，而是让 HTTP 先和 SSL（Secure Sockets Layer）
 
 ### 3. HTTPs 采用的加密方式
 
-HTTPs 采用混合的加密机制，使用非对称密钥加密用于传输对称密钥来保证安全性，之后使用对称密钥加密进行通信来保证效率。（下图中的 Session Key 就是对称密钥）
+HTTPs 采用混合的加密机制，使用非对称密钥加密用于传输对称密钥来保证传输过程的安全性，之后使用对称密钥加密进行通信来保证通信过程的效率。（下图中的 Session Key 就是对称密钥）
 
 <div align="center"> <img src="../pics//How-HTTPS-Works.png" width="600"/> </div><br>
 
@@ -717,7 +717,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 ## HTTPs 的缺点
 
 - 因为需要进行加密解密等过程，因此速度会更慢；
-- 需要支付证书授权的高费用。
+- 需要支付证书授权的高额费用。
 
 ## 配置 HTTPs
 
@@ -727,7 +727,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 
 ## HTTP/1.x 缺陷
 
- HTTP/1.x 实现简单是以牺牲应用性能为代价的：
+ HTTP/1.x 实现简单是以牺牲性能为代价的：
 
 - 客户端需要使用多个连接才能实现并发和缩短延迟；
 - 不会压缩请求和响应首部，从而导致不必要的网络流量；
@@ -763,7 +763,7 @@ HTTP/2.0 要求客户端和服务器同时维护和更新一个包含之前见
 
 <div align="center"> <img src="../pics//_u4E0B_u8F7D.png" width="600"/> </div><br>
 
-# 八、GET 和 POST 的区别
+# 八、GET 和 POST 比较
 
 ## 作用
 
@@ -870,6 +870,7 @@ DELETE /idX/delete HTTP/1.1   -> Returns 404
 - [MDN : HTTP](https://developer.mozilla.org/en-US/docs/Web/HTTP)
 - [HTTP/2 简介](https://developers.google.com/web/fundamentals/performance/http2/?hl=zh-cn)
 - [htmlspecialchars](http://php.net/manual/zh/function.htmlspecialchars.php)
+- [Difference between file URI and URL in java](http://java2db.com/java-io/how-to-get-and-the-difference-between-file-uri-and-url-in-java)
 - [How to Fix SQL Injection Using Java PreparedStatement & CallableStatement](https://software-security.sans.org/developer-how-to/fix-sql-injection-in-java-using-prepared-callable-statement)
 - [浅谈 HTTP 中 Get 与 Post 的区别](https://www.cnblogs.com/hyddd/archive/2009/03/31/1426026.html)
 - [Are http:// and www really necessary?](https://www.webdancers.com/are-http-and-www-necesary/)
diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index bf1921fc..af118e4c 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -25,7 +25,7 @@
 
 ## Collection
 
-<div align="center"> <img src="../pics//NP4z3i8m38Ntd28NQ4_0KCJ2q044Oez.png"/> </div><br>
+<div align="center"> <img src="../pics//VP4n3i8m34Ntd28NQ4_0KCJ2q044Oez.png"/> </div><br>
 
 ### 1. Set
 
@@ -129,12 +129,67 @@ private static final int DEFAULT_CAPACITY = 10;
 
 ArrayList 基于数组实现，并且具有动态扩容特性，因此保存元素的数组不一定都会被使用，那么就没必要全部进行序列化。
 
-保存元素的数组 elementData 使用 transient 修饰，该关键字声明数组默认不会被序列化。ArrayList 重写了 writeObject() 和 readObject() 来控制只序列化数组中有元素填充那部分内容。
+保存元素的数组 elementData 使用 transient 修饰，该关键字声明数组默认不会被序列化。
 
 ```java
 transient Object[] elementData; // non-private to simplify nested class access
 ```
 
+ArrayList 实现了 writeObject() 和 readObject() 来控制只序列化数组中有元素填充那部分内容。
+
+```java
+private void readObject(java.io.ObjectInputStream s)
+    throws java.io.IOException, ClassNotFoundException {
+    elementData = EMPTY_ELEMENTDATA;
+
+    // Read in size, and any hidden stuff
+    s.defaultReadObject();
+
+    // Read in capacity
+    s.readInt(); // ignored
+
+    if (size > 0) {
+        // be like clone(), allocate array based upon size not capacity
+        ensureCapacityInternal(size);
+
+        Object[] a = elementData;
+        // Read in all elements in the proper order.
+        for (int i=0; i<size; i++) {
+            a[i] = s.readObject();
+        }
+    }
+}
+```
+
+```java
+private void writeObject(java.io.ObjectOutputStream s)
+    throws java.io.IOException{
+    // Write out element count, and any hidden stuff
+    int expectedModCount = modCount;
+    s.defaultWriteObject();
+
+    // Write out size as capacity for behavioural compatibility with clone()
+    s.writeInt(size);
+
+    // Write out all elements in the proper order.
+    for (int i=0; i<size; i++) {
+        s.writeObject(elementData[i]);
+    }
+
+    if (modCount != expectedModCount) {
+        throw new ConcurrentModificationException();
+    }
+}
+```
+
+序列化时需要使用 ObjectOutputStream 的 writeObject() 将对象转换为字节流并输出。而 writeObject() 方法在传入的对象存在 writeObject() 的时候会去反射调用该对象的 writeObject() 来实现序列化。反序列化使用的是 ObjectInputStream 的 readObject() 方法，原理类似。
+
+```java
+ArrayList list = new ArrayList();
+ObjectOutputStream oos = new ObjectOutputStream(new FileOutputStream(file));
+oos.writeObject(list);
+```
+
 ### 3. 扩容
 
 添加元素时使用 ensureCapacityInternal() 方法来保证容量足够，如果不够时，需要使用 grow() 方法进行扩容，新容量的大小为 `oldCapacity + (oldCapacity >> 1)`，也就是旧容量的 1.5 倍。
@@ -1080,7 +1135,7 @@ Set <|.. LinkedHashSet
 SortSet <|.. TreeSet
 List <|.. ArrayList
 List <|.. Vector
-List <|.. LinkeList
+List <|.. LinkedList
 Queue <|.. LinkedList
 Queue <|.. PriorityQueue
 
diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 2813ce5d..44f3d9dc 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -1,18 +1,18 @@
 <!-- GFM-TOC -->
 * [算法思想](#算法思想)
-    * [贪心思想](#贪心思想)
     * [双指针](#双指针)
     * [排序](#排序)
         * [快速选择](#快速选择)
         * [堆排序](#堆排序)
         * [桶排序](#桶排序)
         * [荷兰国旗问题](#荷兰国旗问题)
+    * [贪心思想](#贪心思想)
     * [二分查找](#二分查找)
+    * [分治](#分治)
     * [搜索](#搜索)
         * [BFS](#bfs)
         * [DFS](#dfs)
         * [Backtracking](#backtracking)
-    * [分治](#分治)
     * [动态规划](#动态规划)
         * [斐波那契数列](#斐波那契数列)
         * [矩阵路径](#矩阵路径)
@@ -33,10 +33,6 @@
         * [多数投票问题](#多数投票问题)
         * [其它](#其它)
 * [数据结构相关](#数据结构相关)
-    * [栈和队列](#栈和队列)
-    * [哈希表](#哈希表)
-    * [字符串](#字符串)
-    * [数组与矩阵](#数组与矩阵)
     * [链表](#链表)
     * [树](#树)
         * [递归](#递归)
@@ -44,6 +40,10 @@
         * [前中后序遍历](#前中后序遍历)
         * [BST](#bst)
         * [Trie](#trie)
+    * [栈和队列](#栈和队列)
+    * [哈希表](#哈希表)
+    * [字符串](#字符串)
+    * [数组与矩阵](#数组与矩阵)
     * [图](#图)
         * [二分图](#二分图)
         * [拓扑排序](#拓扑排序)
@@ -55,9 +55,449 @@
 
 # 算法思想
 
+## 双指针
+
+双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
+
+**有序数组的 Two Sum** 
+
+[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
+
+```html
+Input: numbers={2, 7, 11, 15}, target=9
+Output: index1=1, index2=2
+```
+
+题目描述：在有序数组中找出两个数，使它们的和为 target。
+
+使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
+
+- 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
+- 如果 sum > target，移动较大的元素，使 sum 变小一些；
+- 如果 sum < target，移动较小的元素，使 sum 变大一些。
+
+```java
+public int[] twoSum(int[] numbers, int target) {
+    int i = 0, j = numbers.length - 1;
+    while (i < j) {
+        int sum = numbers[i] + numbers[j];
+        if (sum == target) {
+            return new int[]{i + 1, j + 1};
+        } else if (sum < target) {
+            i++;
+        } else {
+            j--;
+        }
+    }
+    return null;
+}
+```
+
+**两数平方和** 
+
+[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
+
+```html
+Input: 5
+Output: True
+Explanation: 1 * 1 + 2 * 2 = 5
+```
+
+题目描述：判断一个数是否为两个数的平方和，例如 5 = 1<sup>2</sup> + 2<sup>2</sup>。
+
+```java
+public boolean judgeSquareSum(int c) {
+    int i = 0, j = (int) Math.sqrt(c);
+    while (i <= j) {
+        int powSum = i * i + j * j;
+        if (powSum == c) {
+            return true;
+        } else if (powSum > c) {
+            j--;
+        } else {
+            i++;
+        }
+    }
+    return false;
+}
+```
+
+**反转字符串中的元音字符** 
+
+[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
+
+```html
+Given s = "leetcode", return "leotcede".
+```
+
+使用双指针，指向待反转的两个元音字符，一个指针从头向尾遍历，一个指针从尾到头遍历。
+
+```java
+private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
+
+public String reverseVowels(String s) {
+    int i = 0, j = s.length() - 1;
+    char[] result = new char[s.length()];
+    while (i <= j) {
+        char ci = s.charAt(i);
+        char cj = s.charAt(j);
+        if (!vowels.contains(ci)) {
+            result[i++] = ci;
+        } else if (!vowels.contains(cj)) {
+            result[j--] = cj;
+        } else {
+            result[i++] = cj;
+            result[j--] = ci;
+        }
+    }
+    return new String(result);
+}
+```
+
+**回文字符串** 
+
+[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
+
+```html
+Input: "abca"
+Output: True
+Explanation: You could delete the character 'c'.
+```
+
+题目描述：可以删除一个字符，判断是否能构成回文字符串。
+
+```java
+public boolean validPalindrome(String s) {
+    int i = -1, j = s.length();
+    while (++i < --j) {
+        if (s.charAt(i) != s.charAt(j)) {
+            return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
+        }
+    }
+    return true;
+}
+
+private boolean isPalindrome(String s, int i, int j) {
+    while (i < j) {
+        if (s.charAt(i++) != s.charAt(j--)) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+**归并两个有序数组** 
+
+[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
+
+```html
+Input:
+nums1 = [1,2,3,0,0,0], m = 3
+nums2 = [2,5,6],       n = 3
+
+Output: [1,2,2,3,5,6]
+```
+
+题目描述：把归并结果存到第一个数组上。
+
+需要从尾开始遍历，否则在 nums1 上归并得到的值会覆盖还未进行归并比较的值。
+
+```java
+public void merge(int[] nums1, int m, int[] nums2, int n) {
+    int index1 = m - 1, index2 = n - 1;
+    int indexMerge = m + n - 1;
+    while (index1 >= 0 || index2 >= 0) {
+        if (index1 < 0) {
+            nums1[indexMerge--] = nums2[index2--];
+        } else if (index2 < 0) {
+            nums1[indexMerge--] = nums1[index1--];
+        } else if (nums1[index1] > nums2[index2]) {
+            nums1[indexMerge--] = nums1[index1--];
+        } else {
+            nums1[indexMerge--] = nums2[index2--];
+        }
+    }
+}
+```
+
+**判断链表是否存在环** 
+
+[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
+
+使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
+
+```java
+public boolean hasCycle(ListNode head) {
+    if (head == null) {
+        return false;
+    }
+    ListNode l1 = head, l2 = head.next;
+    while (l1 != null && l2 != null && l2.next != null) {
+        if (l1 == l2) {
+            return true;
+        }
+        l1 = l1.next;
+        l2 = l2.next.next;
+    }
+    return false;
+}
+```
+
+**最长子序列** 
+
+[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
+
+```
+Input:
+s = "abpcplea", d = ["ale","apple","monkey","plea"]
+
+Output:
+"apple"
+```
+
+题目描述：删除 s 中的一些字符，使得它构成字符串列表 d 中的一个字符串，找出能构成的最长字符串。如果有多个相同长度的结果，返回字典序的最大字符串。
+
+```java
+public String findLongestWord(String s, List<String> d) {
+    String longestWord = "";
+    for (String target : d) {
+        int l1 = longestWord.length(), l2 = target.length();
+        if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
+            continue;
+        }
+        if (isValid(s, target)) {
+            longestWord = target;
+        }
+    }
+    return longestWord;
+}
+
+private boolean isValid(String s, String target) {
+    int i = 0, j = 0;
+    while (i < s.length() && j < target.length()) {
+        if (s.charAt(i) == target.charAt(j)) {
+            j++;
+        }
+        i++;
+    }
+    return j == target.length();
+}
+```
+
+## 排序
+
+### 快速选择
+
+用于求解  **Kth Element**  问题，使用快速排序的 partition() 进行实现。
+
+需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
+
+### 堆排序
+
+用于求解  **TopK Elements**  问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。
+
+堆排序也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。
+
+快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。
+
+可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
+
+**Kth Element** 
+
+[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
+
+**排序** ：时间复杂度 O(NlogN)，空间复杂度 O(1)
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    Arrays.sort(nums);
+    return nums[nums.length - k];
+}
+```
+
+**堆排序** ：时间复杂度 O(NlogK)，空间复杂度 O(K)。
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
+    for (int val : nums) {
+        pq.add(val);
+        if (pq.size() > k) // 维护堆的大小为 K
+            pq.poll();
+    }
+    return pq.peek();
+}
+```
+
+**快速选择** ：时间复杂度 O(N)，空间复杂度 O(1)
+
+```java
+public int findKthLargest(int[] nums, int k) {
+    k = nums.length - k;
+    int l = 0, h = nums.length - 1;
+    while (l < h) {
+        int j = partition(nums, l, h);
+        if (j == k) {
+            break;
+        } else if (j < k) {
+            l = j + 1;
+        } else {
+            h = j - 1;
+        }
+    }
+    return nums[k];
+}
+
+private int partition(int[] a, int l, int h) {
+    int i = l, j = h + 1;
+    while (true) {
+        while (a[++i] < a[l] && i < h) ;
+        while (a[--j] > a[l] && j > l) ;
+        if (i >= j) {
+            break;
+        }
+        swap(a, i, j);
+    }
+    swap(a, l, j);
+    return j;
+}
+
+private void swap(int[] a, int i, int j) {
+    int t = a[i];
+    a[i] = a[j];
+    a[j] = t;
+}
+```
+
+### 桶排序
+
+**出现频率最多的 k 个数** 
+
+[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
+
+```html
+Given [1,1,1,2,2,3] and k = 2, return [1,2].
+```
+
+设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。
+
+把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
+
+```java
+public List<Integer> topKFrequent(int[] nums, int k) {
+    Map<Integer, Integer> frequencyForNum = new HashMap<>();
+    for (int num : nums) {
+        frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
+    }
+    List<Integer>[] buckets = new ArrayList[nums.length + 1];
+    for (int key : frequencyForNum.keySet()) {
+        int frequency = frequencyForNum.get(key);
+        if (buckets[frequency] == null) {
+            buckets[frequency] = new ArrayList<>();
+        }
+        buckets[frequency].add(key);
+    }
+    List<Integer> topK = new ArrayList<>();
+    for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
+        if (buckets[i] != null) {
+            topK.addAll(buckets[i]);
+        }
+    }
+    return topK;
+}
+```
+
+**按照字符出现次数对字符串排序** 
+
+[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
+
+```html
+Input:
+"tree"
+
+Output:
+"eert"
+
+Explanation:
+'e' appears twice while 'r' and 't' both appear once.
+So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
+```
+
+```java
+public String frequencySort(String s) {
+    Map<Character, Integer> frequencyForNum = new HashMap<>();
+    for (char c : s.toCharArray())
+        frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
+
+    List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
+    for (char c : frequencyForNum.keySet()) {
+        int f = frequencyForNum.get(c);
+        if (frequencyBucket[f] == null) {
+            frequencyBucket[f] = new ArrayList<>();
+        }
+        frequencyBucket[f].add(c);
+    }
+    StringBuilder str = new StringBuilder();
+    for (int i = frequencyBucket.length - 1; i >= 0; i--) {
+        if (frequencyBucket[i] == null) {
+            continue;
+        }
+        for (char c : frequencyBucket[i]) {
+            for (int j = 0; j < i; j++) {
+                str.append(c);
+            }
+        }
+    }
+    return str.toString();
+}
+```
+
+### 荷兰国旗问题
+
+荷兰国旗包含三种颜色：红、白、蓝。
+
+有三种颜色的球，算法的目标是将这三种球按颜色顺序正确地排列。
+
+它其实是三向切分快速排序的一种变种，在三向切分快速排序中，每次切分都将数组分成三个区间：小于切分元素、等于切分元素、大于切分元素，而该算法是将数组分成三个区间：等于红色、等于白色、等于蓝色。
+
+<div align="center"> <img src="../pics//3b49dd67-2c40-4b81-8ad2-7bbb1fe2fcbd.png"/> </div><br>
+
+**按颜色进行排序** 
+
+[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
+
+```html
+Input: [2,0,2,1,1,0]
+Output: [0,0,1,1,2,2]
+```
+
+题目描述：只有 0/1/2 三种颜色。
+
+```java
+public void sortColors(int[] nums) {
+    int zero = -1, one = 0, two = nums.length;
+    while (one < two) {
+        if (nums[one] == 0) {
+            swap(nums, ++zero, one++);
+        } else if (nums[one] == 2) {
+            swap(nums, --two, one);
+        } else {
+            ++one;
+        }
+    }
+}
+
+private void swap(int[] nums, int i, int j) {
+    int t = nums[i];
+    nums[i] = nums[j];
+    nums[j] = t;
+}
+```
+
 ## 贪心思想
 
-贪心思想保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
+保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。
 
 **分配饼干** 
 
@@ -74,7 +514,7 @@ You need to output 2.
 
 题目描述：每个孩子都有一个满足度，每个饼干都有一个大小，只有饼干的大小大于等于一个孩子的满足度，该孩子才会获得满足。求解最多可以获得满足的孩子数量。
 
-因为最小的孩子最容易得到满足，因此先满足最小孩子。给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因此贪心策略
+给一个孩子的饼干应当尽量小又能满足该孩子，这样大饼干就能拿来给满足度比较大的孩子。因为最小的孩子最容易得到满足，所以先满足最小的孩子。
 
 证明：假设在某次选择中，贪心策略选择给当前满足度最小的孩子分配第 m 个饼干，第 m 个饼干为可以满足该孩子的最小饼干。假设存在一种最优策略，给该孩子分配第 n 个饼干，并且 m < n。我们可以发现，经过这一轮分配，贪心策略分配后剩下的饼干一定有一个比最优策略来得大。因此在后续的分配中，贪心策略一定能满足更多的孩子。也就是说不存在比贪心策略更优的策略，即贪心策略就是最优策略。
 
@@ -362,434 +802,6 @@ public int maxProfit(int[] prices) {
 }
 ```
 
-## 双指针
-
-双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
-
-**有序数组的 Two Sum** 
-
-[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
-
-```html
-Input: numbers={2, 7, 11, 15}, target=9
-Output: index1=1, index2=2
-```
-
-题目描述：在有序数组中找出两个数，使它们的和为 target。
-
-使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
-
-如果两个指针指向元素的和 sum == target，那么得到要求的结果；如果 sum > target，移动较大的元素，使 sum 变小一些；如果 sum < target，移动较小的元素，使 sum 变大一些。
-
-```java
-public int[] twoSum(int[] numbers, int target) {
-    int i = 0, j = numbers.length - 1;
-    while (i < j) {
-        int sum = numbers[i] + numbers[j];
-        if (sum == target) {
-            return new int[]{i + 1, j + 1};
-        } else if (sum < target) {
-            i++;
-        } else {
-            j--;
-        }
-    }
-    return null;
-}
-```
-
-**两数平方和** 
-
-[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
-
-```html
-Input: 5
-Output: True
-Explanation: 1 * 1 + 2 * 2 = 5
-```
-
-题目描述：判断一个数是否为两个数的平方和，例如 5 = 1<sup>2</sup> + 2<sup>2</sup>。
-
-```java
-public boolean judgeSquareSum(int c) {
-    int i = 0, j = (int) Math.sqrt(c);
-    while (i <= j) {
-        int powSum = i * i + j * j;
-        if (powSum == c) {
-            return true;
-        } else if (powSum > c) {
-            j--;
-        } else {
-            i++;
-        }
-    }
-    return false;
-}
-```
-
-**反转字符串中的元音字符** 
-
-[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
-
-```html
-Given s = "leetcode", return "leotcede".
-```
-
-使用双指针，指向待反转的两个元音字符，一个指针从头向尾遍历，一个指针从尾到头遍历。
-
-```java
-private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
-
-public String reverseVowels(String s) {
-    int i = 0, j = s.length() - 1;
-    char[] result = new char[s.length()];
-    while (i <= j) {
-        char ci = s.charAt(i);
-        char cj = s.charAt(j);
-        if (!vowels.contains(ci)) {
-            result[i++] = ci;
-        } else if (!vowels.contains(cj)) {
-            result[j--] = cj;
-        } else {
-            result[i++] = cj;
-            result[j--] = ci;
-        }
-    }
-    return new String(result);
-}
-```
-
-**回文字符串** 
-
-[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
-
-```html
-Input: "abca"
-Output: True
-Explanation: You could delete the character 'c'.
-```
-
-题目描述：可以删除一个字符，判断是否能构成回文字符串。
-
-```java
-public boolean validPalindrome(String s) {
-    int i = -1, j = s.length();
-    while (++i < --j) {
-        if (s.charAt(i) != s.charAt(j)) {
-            return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
-        }
-    }
-    return true;
-}
-
-private boolean isPalindrome(String s, int i, int j) {
-    while (i < j) {
-        if (s.charAt(i++) != s.charAt(j--)) {
-            return false;
-        }
-    }
-    return true;
-}
-```
-
-**归并两个有序数组** 
-
-[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
-
-```html
-Input:
-nums1 = [1,2,3,0,0,0], m = 3
-nums2 = [2,5,6],       n = 3
-
-Output: [1,2,2,3,5,6]
-```
-
-题目描述：把归并结果存到第一个数组上。
-
-需要从尾开始遍历，否则在 nums1 上归并得到的值会覆盖还未进行归并比较的值
-
-```java
-public void merge(int[] nums1, int m, int[] nums2, int n) {
-    int index1 = m - 1, index2 = n - 1;
-    int indexMerge = m + n - 1;
-    while (index1 >= 0 || index2 >= 0) {
-        if (index1 < 0) {
-            nums1[indexMerge--] = nums2[index2--];
-        } else if (index2 < 0) {
-            nums1[indexMerge--] = nums1[index1--];
-        } else if (nums1[index1] > nums2[index2]) {
-            nums1[indexMerge--] = nums1[index1--];
-        } else {
-            nums1[indexMerge--] = nums2[index2--];
-        }
-    }
-}
-```
-
-**判断链表是否存在环** 
-
-[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
-
-使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
-
-```java
-public boolean hasCycle(ListNode head) {
-    if (head == null) {
-        return false;
-    }
-    ListNode l1 = head, l2 = head.next;
-    while (l1 != null && l2 != null && l2.next != null) {
-        if (l1 == l2) {
-            return true;
-        }
-        l1 = l1.next;
-        l2 = l2.next.next;
-    }
-    return false;
-}
-```
-
-**最长子序列** 
-
-[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
-
-```
-Input:
-s = "abpcplea", d = ["ale","apple","monkey","plea"]
-
-Output:
-"apple"
-```
-
-题目描述：删除 s 中的一些字符，使得它构成字符串列表 d 中的一个字符串，找出能构成的最长字符串。如果有多个相同长度的结果，返回按字典序排序的最大字符串。
-
-```java
-public String findLongestWord(String s, List<String> d) {
-    String longestWord = "";
-    for (String target : d) {
-        int l1 = longestWord.length(), l2 = target.length();
-        if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
-            continue;
-        }
-        if (isValid(s, target)) {
-            longestWord = target;
-        }
-    }
-    return longestWord;
-}
-
-private boolean isValid(String s, String target) {
-    int i = 0, j = 0;
-    while (i < s.length() && j < target.length()) {
-        if (s.charAt(i) == target.charAt(j)) {
-            j++;
-        }
-        i++;
-    }
-    return j == target.length();
-}
-```
-
-## 排序
-
-### 快速选择
-
-一般用于求解  **Kth Element**  问题，可以在 O(N) 时间复杂度，O(1) 空间复杂度完成求解工作。
-
-与快速排序一样，快速选择一般需要先打乱数组，否则最坏情况下时间复杂度为 O(N<sup>2</sup>)。
-
-### 堆排序
-
-堆排序用于求解  **TopK Elements**  问题，通过维护一个大小为 K 的堆，堆中的元素就是 TopK Elements。当然它也可以用于求解 Kth Element 问题，堆顶元素就是 Kth Element。快速选择也可以求解 TopK Elements 问题，因为找到 Kth Element 之后，再遍历一次数组，所有小于等于 Kth Element 的元素都是 TopK Elements。可以看到，快速选择和堆排序都可以求解 Kth Element 和 TopK Elements 问题。
-
-**Kth Element** 
-
-[215. Kth Largest Element in an Array (Medium)](https://leetcode.com/problems/kth-largest-element-in-an-array/description/)
-
-**排序** ：时间复杂度 O(NlogN)，空间复杂度 O(1)
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    Arrays.sort(nums);
-    return nums[nums.length - k];
-}
-```
-
-**堆排序** ：时间复杂度 O(NlogK)，空间复杂度 O(K)。
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    PriorityQueue<Integer> pq = new PriorityQueue<>(); // 小顶堆
-    for (int val : nums) {
-        pq.add(val);
-        if (pq.size() > k) // 维护堆的大小为 K
-            pq.poll();
-    }
-    return pq.peek();
-}
-```
-
-**快速选择** ：时间复杂度 O(N)，空间复杂度 O(1)
-
-```java
-public int findKthLargest(int[] nums, int k) {
-    k = nums.length - k;
-    int l = 0, h = nums.length - 1;
-    while (l < h) {
-        int j = partition(nums, l, h);
-        if (j == k) {
-            break;
-        } else if (j < k) {
-            l = j + 1;
-        } else {
-            h = j - 1;
-        }
-    }
-    return nums[k];
-}
-
-private int partition(int[] a, int l, int h) {
-    int i = l, j = h + 1;
-    while (true) {
-        while (a[++i] < a[l] && i < h) ;
-        while (a[--j] > a[l] && j > l) ;
-        if (i >= j) {
-            break;
-        }
-        swap(a, i, j);
-    }
-    swap(a, l, j);
-    return j;
-}
-
-private void swap(int[] a, int i, int j) {
-    int t = a[i];
-    a[i] = a[j];
-    a[j] = t;
-}
-```
-
-### 桶排序
-
-**出现频率最多的 k 个数** 
-
-[347. Top K Frequent Elements (Medium)](https://leetcode.com/problems/top-k-frequent-elements/description/)
-
-```html
-Given [1,1,1,2,2,3] and k = 2, return [1,2].
-```
-
-设置若干个桶，每个桶存储出现频率相同的数，并且桶的下标代表桶中数出现的频率，即第 i 个桶中存储的数出现的频率为 i。把数都放到桶之后，从后向前遍历桶，最先得到的 k 个数就是出现频率最多的的 k 个数。
-
-```java
-public List<Integer> topKFrequent(int[] nums, int k) {
-    Map<Integer, Integer> frequencyForNum = new HashMap<>();
-    for (int num : nums) {
-        frequencyForNum.put(num, frequencyForNum.getOrDefault(num, 0) + 1);
-    }
-    List<Integer>[] buckets = new ArrayList[nums.length + 1];
-    for (int key : frequencyForNum.keySet()) {
-        int frequency = frequencyForNum.get(key);
-        if (buckets[frequency] == null) {
-            buckets[frequency] = new ArrayList<>();
-        }
-        buckets[frequency].add(key);
-    }
-    List<Integer> topK = new ArrayList<>();
-    for (int i = buckets.length - 1; i >= 0 && topK.size() < k; i--) {
-        if (buckets[i] != null) {
-            topK.addAll(buckets[i]);
-        }
-    }
-    return topK;
-}
-```
-
-**按照字符出现次数对字符串排序** 
-
-[451. Sort Characters By Frequency (Medium)](https://leetcode.com/problems/sort-characters-by-frequency/description/)
-
-```html
-Input:
-"tree"
-
-Output:
-"eert"
-
-Explanation:
-'e' appears twice while 'r' and 't' both appear once.
-So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
-```
-
-```java
-public String frequencySort(String s) {
-    Map<Character, Integer> frequencyForNum = new HashMap<>();
-    for (char c : s.toCharArray())
-        frequencyForNum.put(c, frequencyForNum.getOrDefault(c, 0) + 1);
-
-    List<Character>[] frequencyBucket = new ArrayList[s.length() + 1];
-    for (char c : frequencyForNum.keySet()) {
-        int f = frequencyForNum.get(c);
-        if (frequencyBucket[f] == null) {
-            frequencyBucket[f] = new ArrayList<>();
-        }
-        frequencyBucket[f].add(c);
-    }
-    StringBuilder str = new StringBuilder();
-    for (int i = frequencyBucket.length - 1; i >= 0; i--) {
-        if (frequencyBucket[i] == null) {
-            continue;
-        }
-        for (char c : frequencyBucket[i]) {
-            for (int j = 0; j < i; j++) {
-                str.append(c);
-            }
-        }
-    }
-    return str.toString();
-}
-```
-
-### 荷兰国旗问题
-
-荷兰国旗包含三种颜色：红、白、蓝。有这三种颜色的球，算法的目标是将这三种球按颜色顺序正确地排列。
-
-它其实是三向切分快速排序的一种变种，在三向切分快速排序中，每次切分都将数组分成三个区间：小于切分元素、等于切分元素、大于切分元素，而该算法是将数组分成三个区间：等于红色、等于白色、等于蓝色。
-
-<div align="center"> <img src="../pics//3b49dd67-2c40-4b81-8ad2-7bbb1fe2fcbd.png"/> </div><br>
-
-**按颜色进行排序** 
-
-[75. Sort Colors (Medium)](https://leetcode.com/problems/sort-colors/description/)
-
-```html
-Input: [2,0,2,1,1,0]
-Output: [0,0,1,1,2,2]
-```
-
-题目描述：只有 0/1/2 三种颜色。
-
-```java
-public void sortColors(int[] nums) {
-    int zero = -1, one = 0, two = nums.length;
-    while (one < two) {
-        if (nums[one] == 0) {
-            swap(nums, ++zero, one++);
-        } else if (nums[one] == 2) {
-            swap(nums, --two, one);
-        } else {
-            ++one;
-        }
-    }
-}
-
-private void swap(int[] nums, int i, int j) {
-    int t = nums[i];
-    nums[i] = nums[j];
-    nums[j] = t;
-}
-```
-
 ## 二分查找
 
 **正常实现** 
@@ -1065,6 +1077,53 @@ private int binarySearch(int[] nums, int target) {
 }
 ```
 
+## 分治
+
+**给表达式加括号** 
+
+[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
+
+```html
+Input: "2-1-1".
+
+((2-1)-1) = 0
+(2-(1-1)) = 2
+
+Output : [0, 2]
+```
+
+```java
+public List<Integer> diffWaysToCompute(String input) {
+    List<Integer> ways = new ArrayList<>();
+    for (int i = 0; i < input.length(); i++) {
+        char c = input.charAt(i);
+        if (c == '+' || c == '-' || c == '*') {
+            List<Integer> left = diffWaysToCompute(input.substring(0, i));
+            List<Integer> right = diffWaysToCompute(input.substring(i + 1));
+            for (int l : left) {
+                for (int r : right) {
+                    switch (c) {
+                        case '+':
+                            ways.add(l + r);
+                            break;
+                        case '-':
+                            ways.add(l - r);
+                            break;
+                        case '*':
+                            ways.add(l * r);
+                            break;
+                    }
+                }
+            }
+        }
+    }
+    if (ways.size() == 0) {
+        ways.add(Integer.valueOf(input));
+    }
+    return ways;
+}
+```
+
 ## 搜索
 
 深度优先搜索和广度优先搜索广泛运用于树和图中，但是它们的应用远远不止如此。
@@ -2312,53 +2371,6 @@ private void backtracking(int row) {
 }
 ```
 
-## 分治
-
-**给表达式加括号** 
-
-[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
-
-```html
-Input: "2-1-1".
-
-((2-1)-1) = 0
-(2-(1-1)) = 2
-
-Output : [0, 2]
-```
-
-```java
-public List<Integer> diffWaysToCompute(String input) {
-    List<Integer> ways = new ArrayList<>();
-    for (int i = 0; i < input.length(); i++) {
-        char c = input.charAt(i);
-        if (c == '+' || c == '-' || c == '*') {
-            List<Integer> left = diffWaysToCompute(input.substring(0, i));
-            List<Integer> right = diffWaysToCompute(input.substring(i + 1));
-            for (int l : left) {
-                for (int r : right) {
-                    switch (c) {
-                        case '+':
-                            ways.add(l + r);
-                            break;
-                        case '-':
-                            ways.add(l - r);
-                            break;
-                        case '*':
-                            ways.add(l * r);
-                            break;
-                    }
-                }
-            }
-        }
-    }
-    if (ways.size() == 0) {
-        ways.add(Integer.valueOf(input));
-    }
-    return ways;
-}
-```
-
 ## 动态规划
 
 递归和动态规划都是将原问题拆成多个子问题然后求解，他们之间最本质的区别是，动态规划保存了子问题的解，避免重复计算。
@@ -4036,973 +4048,6 @@ public int maximumProduct(int[] nums) {
 
 # 数据结构相关
 
-## 栈和队列
-
-**用栈实现队列** 
-
-[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
-
-```java
-class MyQueue {
-
-    private Stack<Integer> in = new Stack<>();
-    private Stack<Integer> out = new Stack<>();
-
-    public void push(int x) {
-        in.push(x);
-    }
-
-    public int pop() {
-        in2out();
-        return out.pop();
-    }
-
-    public int peek() {
-        in2out();
-        return out.peek();
-    }
-
-    private void in2out() {
-        if (out.isEmpty()) {
-            while (!in.isEmpty()) {
-                out.push(in.pop());
-            }
-        }
-    }
-
-    public boolean empty() {
-        return in.isEmpty() && out.isEmpty();
-    }
-}
-```
-
-**用队列实现栈** 
-
-[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
-
-在将一个元素 x 插入队列时，需要让除了 x 之外的所有元素出队列，再入队列。此时 x 在队首，第一个出队列，实现了后进先出顺序。
-
-```java
-class MyStack {
-
-    private Queue<Integer> queue;
-
-    public MyStack() {
-        queue = new LinkedList<>();
-    }
-
-    public void push(int x) {
-        queue.add(x);
-        int cnt = queue.size();
-        while (cnt-- > 1) {
-            queue.add(queue.poll());
-        }
-    }
-
-    public int pop() {
-        return queue.remove();
-    }
-
-    public int top() {
-        return queue.peek();
-    }
-
-    public boolean empty() {
-        return queue.isEmpty();
-    }
-}
-```
-
-**最小值栈** 
-
-[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
-
-```java
-class MinStack {
-
-    private Stack<Integer> dataStack;
-    private Stack<Integer> minStack;
-    private int min;
-
-    public MinStack() {
-        dataStack = new Stack<>();
-        minStack = new Stack<>();
-        min = Integer.MAX_VALUE;
-    }
-
-    public void push(int x) {
-        dataStack.add(x);
-        min = Math.min(min, x);
-        minStack.add(min);
-    }
-
-    public void pop() {
-        dataStack.pop();
-        minStack.pop();
-        min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
-    }
-
-    public int top() {
-        return dataStack.peek();
-    }
-
-    public int getMin() {
-        return minStack.peek();
-    }
-}
-```
-
-对于实现最小值队列问题，可以先将队列使用栈来实现，然后就将问题转换为最小值栈，这个问题出现在 编程之美：3.7。
-
-**用栈实现括号匹配** 
-
-[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
-
-```html
-"()[]{}"
-
-Output : true
-```
-
-```java
-public boolean isValid(String s) {
-    Stack<Character> stack = new Stack<>();
-    for (char c : s.toCharArray()) {
-        if (c == '(' || c == '{' || c == '[') {
-            stack.push(c);
-        } else {
-            if (stack.isEmpty()) {
-                return false;
-            }
-            char cStack = stack.pop();
-            boolean b1 = c == ')' && cStack != '(';
-            boolean b2 = c == ']' && cStack != '[';
-            boolean b3 = c == '}' && cStack != '{';
-            if (b1 || b2 || b3) {
-                return false;
-            }
-        }
-    }
-    return stack.isEmpty();
-}
-```
-
-**数组中元素与下一个比它大的元素之间的距离** 
-
-[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
-
-```html
-Input: [73, 74, 75, 71, 69, 72, 76, 73]
-Output: [1, 1, 4, 2, 1, 1, 0, 0]
-```
-
-在遍历数组时用 Stack 把数组中的数存起来，如果当前遍历的数比栈顶元素来的大，说明栈顶元素的下一个比它大的数就是当前元素。
-
-```java
-public int[] dailyTemperatures(int[] temperatures) {
-    int n = temperatures.length;
-    int[] dist = new int[n];
-    Stack<Integer> indexs = new Stack<>();
-    for (int curIndex = 0; curIndex < n; curIndex++) {
-        while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
-            int preIndex = indexs.pop();
-            dist[preIndex] = curIndex - preIndex;
-        }
-        indexs.add(curIndex);
-    }
-    return dist;
-}
-```
-
-**循环数组中比当前元素大的下一个元素** 
-
-[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
-
-```text
-Input: [1,2,1]
-Output: [2,-1,2]
-Explanation: The first 1's next greater number is 2; 
-The number 2 can't find next greater number; 
-The second 1's next greater number needs to search circularly, which is also 2.
-```
-
-与 739. Daily Temperatures (Medium) 不同的是，数组是循环数组，并且最后要求的不是距离而是下一个元素。
-
-```java
-public int[] nextGreaterElements(int[] nums) {
-    int n = nums.length;
-    int[] next = new int[n];
-    Arrays.fill(next, -1);
-    Stack<Integer> pre = new Stack<>();
-    for (int i = 0; i < n * 2; i++) {
-        int num = nums[i % n];
-        while (!pre.isEmpty() && nums[pre.peek()] < num) {
-            next[pre.pop()] = num;
-        }
-        if (i < n){
-            pre.push(i);
-        }
-    }
-    return next;
-}
-```
-
-## 哈希表
-
-哈希表使用 O(N) 空间复杂度存储数据，从而能够以 O(1) 时间复杂度求解问题。
-
-Java 中的  **HashSet**  用于存储一个集合，可以查找元素是否在集合中。
-
-如果元素有穷，并且范围不大，那么可以用一个布尔数组来存储一个元素是否存在。例如对于只有小写字符的元素，就可以用一个长度为 26 的布尔数组来存储一个字符集合，使得空间复杂度降低为 O(1)。
-
-Java 中的  **HashMap**  主要用于映射关系，从而把两个元素联系起来。
-
-在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中[Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)，利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
-
-HashMap 也可以用来对元素进行计数统计，此时键为元素，值为计数。和 HashSet 类似，如果元素有穷并且范围不大，可以用整型数组来进行统计。
-
-**数组中的两个数和为给定值** 
-
-[1. Two Sum (Easy)](https://leetcode.com/problems/two-sum/description/)
-
-可以先对数组进行排序，然后使用双指针方法或者二分查找方法。这样做的时间复杂度为 O(NlogN)，空间复杂度为 O(1)。
-
-用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i]，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
-
-```java
-public int[] twoSum(int[] nums, int target) {
-    HashMap<Integer, Integer> indexForNum = new HashMap<>();
-    for (int i = 0; i < nums.length; i++) {
-        if (indexForNum.containsKey(target - nums[i])) {
-            return new int[]{indexForNum.get(target - nums[i]), i};
-        } else {
-            indexForNum.put(nums[i], i);
-        }
-    }
-    return null;
-}
-```
-
-**判断数组是否含有重复元素** 
-
-[217. Contains Duplicate (Easy)](https://leetcode.com/problems/contains-duplicate/description/)
-
-```java
-public boolean containsDuplicate(int[] nums) {
-    Set<Integer> set = new HashSet<>();
-    for (int num : nums) {
-        set.add(num);
-    }
-    return set.size() < nums.length;
-}
-```
-
-**最长和谐序列** 
-
-[594. Longest Harmonious Subsequence (Easy)](https://leetcode.com/problems/longest-harmonious-subsequence/description/)
-
-```html
-Input: [1,3,2,2,5,2,3,7]
-Output: 5
-Explanation: The longest harmonious subsequence is [3,2,2,2,3].
-```
-
-和谐序列中最大数和最小数只差正好为 1，应该注意的是序列的元素不一定是数组的连续元素。
-
-```java
-public int findLHS(int[] nums) {
-    Map<Integer, Integer> countForNum = new HashMap<>();
-    for (int num : nums) {
-        countForNum.put(num, countForNum.getOrDefault(num, 0) + 1);
-    }
-    int longest = 0;
-    for (int num : countForNum.keySet()) {
-        if (countForNum.containsKey(num + 1)) {
-            longest = Math.max(longest, countForNum.get(num + 1) + countForNum.get(num));
-        }
-    }
-    return longest;
-}
-```
-
-**最长连续序列** 
-
-[128. Longest Consecutive Sequence (Hard)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
-
-```html
-Given [100, 4, 200, 1, 3, 2],
-The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
-```
-
-要求以 O(N) 的时间复杂度求解。
-
-```java
-public int longestConsecutive(int[] nums) {
-    Map<Integer, Integer> countForNum = new HashMap<>();
-    for (int num : nums) {
-        countForNum.put(num, 1);
-    }
-    for (int num : nums) {
-        forward(countForNum, num);
-    }
-    return maxCount(countForNum);
-}
-
-private int forward(Map<Integer, Integer> countForNum, int num) {
-    if (!countForNum.containsKey(num)) {
-        return 0;
-    }
-    int cnt = countForNum.get(num);
-    if (cnt > 1) {
-        return cnt;
-    }
-    cnt = forward(countForNum, num + 1) + 1;
-    countForNum.put(num, cnt);
-    return cnt;
-}
-
-private int maxCount(Map<Integer, Integer> countForNum) {
-    int max = 0;
-    for (int num : countForNum.keySet()) {
-        max = Math.max(max, countForNum.get(num));
-    }
-    return max;
-}
-```
-
-## 字符串
-
-**两个字符串包含的字符是否完全相同** 
-
-[242. Valid Anagram (Easy)](https://leetcode.com/problems/valid-anagram/description/)
-
-```html
-s = "anagram", t = "nagaram", return true.
-s = "rat", t = "car", return false.
-```
-
-字符串只包含小写字符，总共有 26 个小写字符。可以用 HashMap 来映射字符与出现次数。因为键的范围很小，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，然后比较两个字符串出现的字符数量是否相同。
-
-```java
-public boolean isAnagram(String s, String t) {
-    int[] cnts = new int[26];
-    for (char c : s.toCharArray()) {
-        cnts[c - 'a']++;
-    }
-    for (char c : t.toCharArray()) {
-        cnts[c - 'a']--;
-    }
-    for (int cnt : cnts) {
-        if (cnt != 0) {
-            return false;
-        }
-    }
-    return true;
-}
-```
-
-**计算一组字符集合可以组成的回文字符串的最大长度** 
-
-[409. Longest Palindrome (Easy)](https://leetcode.com/problems/longest-palindrome/description/)
-
-```html
-Input : "abccccdd"
-Output : 7
-Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
-```
-
-使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。
-
-因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
-
-```java
-public int longestPalindrome(String s) {
-    int[] cnts = new int[256];
-    for (char c : s.toCharArray()) {
-        cnts[c]++;
-    }
-    int palindrome = 0;
-    for (int cnt : cnts) {
-        palindrome += (cnt / 2) * 2;
-    }
-    if (palindrome < s.length()) {
-        palindrome++;   // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
-    }
-    return palindrome;
-}
-```
-
-**字符串同构** 
-
-[205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
-
-```html
-Given "egg", "add", return true.
-Given "foo", "bar", return false.
-Given "paper", "title", return true.
-```
-
-记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
-
-```java
-public boolean isIsomorphic(String s, String t) {
-    int[] preIndexOfS = new int[256];
-    int[] preIndexOfT = new int[256];
-    for (int i = 0; i < s.length(); i++) {
-        char sc = s.charAt(i), tc = t.charAt(i);
-        if (preIndexOfS[sc] != preIndexOfT[tc]) {
-            return false;
-        }
-        preIndexOfS[sc] = i + 1;
-        preIndexOfT[tc] = i + 1;
-    }
-    return true;
-}
-```
-
-**回文子字符串** 
-
-[647. Palindromic Substrings (Medium)](https://leetcode.com/problems/palindromic-substrings/description/)
-
-```html
-Input: "aaa"
-Output: 6
-Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
-```
-
-从字符串的某一位开始，尝试着去扩展子字符串。
-
-```java
-private int cnt = 0;
-
-public int countSubstrings(String s) {
-    for (int i = 0; i < s.length(); i++) {
-        extendSubstrings(s, i, i);     // 奇数长度
-        extendSubstrings(s, i, i + 1); // 偶数长度
-    }
-    return cnt;
-}
-
-private void extendSubstrings(String s, int start, int end) {
-    while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
-        start--;
-        end++;
-        cnt++;
-    }
-}
-```
-
-**判断一个整数是否是回文数** 
-
-[9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
-
-要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
-
-将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
-
-```java
-public boolean isPalindrome(int x) {
-    if (x == 0) {
-        return true;
-    }
-    if (x < 0 || x % 10 == 0) {
-        return false;
-    }
-    int right = 0;
-    while (x > right) {
-        right = right * 10 + x % 10;
-        x /= 10;
-    }
-    return x == right || x == right / 10;
-}
-```
-
-**统计二进制字符串中连续 1 和连续 0 数量相同的子字符串个数** 
-
-[696. Count Binary Substrings (Easy)](https://leetcode.com/problems/count-binary-substrings/description/)
-
-```html
-Input: "00110011"
-Output: 6
-Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
-```
-
-```java
-public int countBinarySubstrings(String s) {
-    int preLen = 0, curLen = 1, count = 0;
-    for (int i = 1; i < s.length(); i++) {
-        if (s.charAt(i) == s.charAt(i - 1)) {
-            curLen++;
-        } else {
-            preLen = curLen;
-            curLen = 1;
-        }
-
-        if (preLen >= curLen) {
-            count++;
-        }
-    }
-    return count;
-}
-```
-
-**字符串循环移位包含** 
-
-[编程之美：3.1](#)
-
-```html
-s1 = AABCD, s2 = CDAA
-Return : true
-```
-
-给定两个字符串 s1 和 s2，要求判定 s2 是否能够被 s1 做循环移位得到的字符串包含。
-
-s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2 是否是 s1s1 的子字符串即可。
-
-**字符串循环移位** 
-
-[编程之美：2.17](#)
-
-```html
-s = "abcd123" k = 3
-Return "123abcd"
-```
-
-将字符串向右循环移动 k 位。
-
-将 abcd123 中的 abcd 和 123 单独逆序，得到 dcba321，然后对整个字符串进行逆序，得到 123abcd。
-
-**字符串中单词的翻转** 
-
-[程序员代码面试指南](#)
-
-```html
-s = "I am a student"
-return "student a am I"
-```
-
-将每个单词逆序，然后将整个字符串逆序。
-
-## 数组与矩阵
-
-**把数组中的 0 移到末尾** 
-
-[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
-
-```html
-For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
-```
-
-```java
-public void moveZeroes(int[] nums) {
-    int idx = 0;
-    for (int num : nums) {
-        if (num != 0) {
-            nums[idx++] = num;
-        }
-    }
-    while (idx < nums.length) {
-        nums[idx++] = 0;
-    }
-}
-```
-
-**改变矩阵维度** 
-
-[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
-
-```html
-Input:
-nums =
-[[1,2],
- [3,4]]
-r = 1, c = 4
-
-Output:
-[[1,2,3,4]]
-
-Explanation:
-The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
-```
-
-```java
-public int[][] matrixReshape(int[][] nums, int r, int c) {
-    int m = nums.length, n = nums[0].length;
-    if (m * n != r * c) {
-        return nums;
-    }
-    int[][] reshapedNums = new int[r][c];
-    int index = 0;
-    for (int i = 0; i < r; i++) {
-        for (int j = 0; j < c; j++) {
-            reshapedNums[i][j] = nums[index / n][index % n];
-            index++;
-        }
-    }
-    return reshapedNums;
-}
-```
-
-**找出数组中最长的连续 1** 
-
-[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
-
-```java
-public int findMaxConsecutiveOnes(int[] nums) {
-    int max = 0, cur = 0;
-    for (int x : nums) {
-        cur = x == 0 ? 0 : cur + 1;
-        max = Math.max(max, cur);
-    }
-    return max;
-}
-```
-
-**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数** 
-
-[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
-
-```html
-Input: nums = [1,2,2,4]
-Output: [2,3]
-```
-
-```html
-Input: nums = [1,2,2,4]
-Output: [2,3]
-```
-
-最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
-
-主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。遍历数组，如果第 i 位上的元素不是 i + 1，那么一直交换第 i 位和 nums[i] - 1 位置上的元素。
-
-```java
-public int[] findErrorNums(int[] nums) {
-    for (int i = 0; i < nums.length; i++) {
-        while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
-            swap(nums, i, nums[i] - 1);
-        }
-    }
-    for (int i = 0; i < nums.length; i++) {
-        if (nums[i] != i + 1) {
-            return new int[]{nums[i], i + 1};
-        }
-    }
-    return null;
-}
-
-private void swap(int[] nums, int i, int j) {
-    int tmp = nums[i];
-    nums[i] = nums[j];
-    nums[j] = tmp;
-}
-```
-
-类似题目：
-
-- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
-- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
-
-**找出数组中重复的数，数组值在 [1, n] 之间** 
-
-[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
-
-要求不能修改数组，也不能使用额外的空间。
-
-二分查找解法：
-
-```java
-public int findDuplicate(int[] nums) {
-     int l = 1, h = nums.length - 1;
-     while (l <= h) {
-         int mid = l + (h - l) / 2;
-         int cnt = 0;
-         for (int i = 0; i < nums.length; i++) {
-             if (nums[i] <= mid) cnt++;
-         }
-         if (cnt > mid) h = mid - 1;
-         else l = mid + 1;
-     }
-     return l;
-}
-```
-
-双指针解法，类似于有环链表中找出环的入口：
-
-```java
-public int findDuplicate(int[] nums) {
-    int slow = nums[0], fast = nums[nums[0]];
-    while (slow != fast) {
-        slow = nums[slow];
-        fast = nums[nums[fast]];
-    }
-    fast = 0;
-    while (slow != fast) {
-        slow = nums[slow];
-        fast = nums[fast];
-    }
-    return slow;
-}
-```
-
-**有序矩阵查找** 
-
-[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
-
-```html
-[
-   [ 1,  5,  9],
-   [10, 11, 13],
-   [12, 13, 15]
-]
-```
-
-```java
-public boolean searchMatrix(int[][] matrix, int target) {
-    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
-    int m = matrix.length, n = matrix[0].length;
-    int row = 0, col = n - 1;
-    while (row < m && col >= 0) {
-        if (target == matrix[row][col]) return true;
-        else if (target < matrix[row][col]) col--;
-        else row++;
-    }
-    return false;
-}
-```
-
-**有序矩阵的 Kth Element** 
-
-[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
-
-```html
-matrix = [
-  [ 1,  5,  9],
-  [10, 11, 13],
-  [12, 13, 15]
-],
-k = 8,
-
-return 13.
-```
-
-解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
-
-二分查找解法：
-
-```java
-public int kthSmallest(int[][] matrix, int k) {
-    int m = matrix.length, n = matrix[0].length;
-    int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
-    while(lo <= hi) {
-        int mid = lo + (hi - lo) / 2;
-        int cnt = 0;
-        for(int i = 0; i < m; i++) {
-            for(int j = 0; j < n && matrix[i][j] <= mid; j++) {
-                cnt++;
-            }
-        }
-        if(cnt < k) lo = mid + 1;
-        else hi = mid - 1;
-    }
-    return lo;
-}
-```
-
-堆解法：
-
-```java
-public int kthSmallest(int[][] matrix, int k) {
-    int m = matrix.length, n = matrix[0].length;
-    PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
-    for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
-    for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
-        Tuple t = pq.poll();
-        if(t.x == m - 1) continue;
-        pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
-    }
-    return pq.poll().val;
-}
-
-class Tuple implements Comparable<Tuple> {
-    int x, y, val;
-    public Tuple(int x, int y, int val) {
-        this.x = x; this.y = y; this.val = val;
-    }
-
-    @Override
-    public int compareTo(Tuple that) {
-        return this.val - that.val;
-    }
-}
-```
-
-**数组相邻差值的个数** 
-
-[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
-
-```html
-Input: n = 3, k = 2
-Output: [1, 3, 2]
-Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
-```
-
-题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
-
-让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
-
-```java
-public int[] constructArray(int n, int k) {
-    int[] ret = new int[n];
-    ret[0] = 1;
-    for (int i = 1, interval = k; i <= k; i++, interval--) {
-        ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
-    }
-    for (int i = k + 1; i < n; i++) {
-        ret[i] = i + 1;
-    }
-    return ret;
-}
-```
-
-**数组的度** 
-
-[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
-
-```html
-Input: [1,2,2,3,1,4,2]
-Output: 6
-```
-
-题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
-
-```java
-public int findShortestSubArray(int[] nums) {
-    Map<Integer, Integer> numsCnt = new HashMap<>();
-    Map<Integer, Integer> numsLastIndex = new HashMap<>();
-    Map<Integer, Integer> numsFirstIndex = new HashMap<>();
-    for (int i = 0; i < nums.length; i++) {
-        int num = nums[i];
-        numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
-        numsLastIndex.put(num, i);
-        if (!numsFirstIndex.containsKey(num)) {
-            numsFirstIndex.put(num, i);
-        }
-    }
-    int maxCnt = 0;
-    for (int num : nums) {
-        maxCnt = Math.max(maxCnt, numsCnt.get(num));
-    }
-    int ret = nums.length;
-    for (int i = 0; i < nums.length; i++) {
-        int num = nums[i];
-        int cnt = numsCnt.get(num);
-        if (cnt != maxCnt) continue;
-        ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
-    }
-    return ret;
-}
-```
-
-**对角元素相等的矩阵** 
-
-[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
-
-```html
-1234
-5123
-9512
-
-In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
-```
-
-```java
-public boolean isToeplitzMatrix(int[][] matrix) {
-    for (int i = 0; i < matrix[0].length; i++) {
-        if (!check(matrix, matrix[0][i], 0, i)) {
-            return false;
-        }
-    }
-    for (int i = 0; i < matrix.length; i++) {
-        if (!check(matrix, matrix[i][0], i, 0)) {
-            return false;
-        }
-    }
-    return true;
-}
-
-private boolean check(int[][] matrix, int expectValue, int row, int col) {
-    if (row >= matrix.length || col >= matrix[0].length) {
-        return true;
-    }
-    if (matrix[row][col] != expectValue) {
-        return false;
-    }
-    return check(matrix, expectValue, row + 1, col + 1);
-}
-```
-
-**嵌套数组** 
-
-[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
-
-```html
-Input: A = [5,4,0,3,1,6,2]
-Output: 4
-Explanation:
-A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
-
-One of the longest S[K]:
-S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
-```
-
-题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
-
-```java
-public int arrayNesting(int[] nums) {
-    int max = 0;
-    for (int i = 0; i < nums.length; i++) {
-        int cnt = 0;
-        for (int j = i; nums[j] != -1; ) {
-            cnt++;
-            int t = nums[j];
-            nums[j] = -1; // 标记该位置已经被访问
-            j = t;
-
-        }
-        max = Math.max(max, cnt);
-    }
-    return max;
-}
-```
-
-**分隔数组** 
-
-[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
-
-```html
-Input: arr = [1,0,2,3,4]
-Output: 4
-Explanation:
-We can split into two chunks, such as [1, 0], [2, 3, 4].
-However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
-```
-
-题目描述：分隔数组，使得对每部分排序后数组就为有序。
-
-```java
-public int maxChunksToSorted(int[] arr) {
-    if (arr == null) return 0;
-    int ret = 0;
-    int right = arr[0];
-    for (int i = 0; i < arr.length; i++) {
-        right = Math.max(right, arr[i]);
-        if (right == i) ret++;
-    }
-    return ret;
-}
-```
-
 ## 链表
 
 链表是空节点，或者有一个值和一个指向下一个链表的指针，因此很多链表问题可以用递归来处理。
@@ -5036,7 +4081,10 @@ public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 }
 ```
 
-如果只是判断是否存在交点，那么就是另一个问题，即 [编程之美：3.6]() 的问题。有两种解法：把第一个链表的结尾连接到第二个链表的开头，看第二个链表是否存在环；或者直接比较两个链表的最后一个节点是否相同。
+如果只是判断是否存在交点，那么就是另一个问题，即 [编程之美 3.6]() 的问题。有两种解法：
+
+- 把第一个链表的结尾连接到第二个链表的开头，看第二个链表是否存在环；
+- 或者直接比较两个链表的最后一个节点是否相同。
 
 **链表反转** 
 
@@ -5101,7 +4149,7 @@ Given 1->1->2->3->3, return 1->2->3.
 
 ```java
 public ListNode deleteDuplicates(ListNode head) {
-    if(head == null || head.next == null) return head;
+    if (head == null || head.next == null) return head;
     head.next = deleteDuplicates(head.next);
     return head.val == head.next.val ? head.next : head;
 }
@@ -5423,8 +4471,8 @@ Input:
         3   2                     1   3
        /                           \   \
       5                             4   7
+
 Output:
-Merged tree:
          3
         / \
        4   5
@@ -5450,6 +4498,7 @@ public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
 
 ```html
 Given the below binary tree and sum = 22,
+
               5
              / \
             4   8
@@ -5457,10 +4506,11 @@ Given the below binary tree and sum = 22,
           11  13  4
          /  \      \
         7    2      1
+
 return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
 ```
 
-路径和定义为从 root 到 leaf 的所有节点的和
+路径和定义为从 root 到 leaf 的所有节点的和。
 
 ```java
 public boolean hasPathSum(TreeNode root, int sum) {
@@ -5852,7 +4902,7 @@ public List<Integer> preorderTraversal(TreeNode root) {
 
 [145. Binary Tree Postorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-postorder-traversal/description/)
 
-前序遍历为 root -> left -> right，后序遍历为 left -> right -> root，可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
+前序遍历为 root -> left -> right，后序遍历为 left -> right -> root。可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
 
 ```java
 public List<Integer> postorderTraversal(TreeNode root) {
@@ -6110,7 +5160,7 @@ public TreeNode sortedListToBST(ListNode head) {
     if (head == null) return null;
     if (head.next == null) return new TreeNode(head.val);
     ListNode preMid = preMid(head);
-    ListNode mid = preMid.next; 
+    ListNode mid = preMid.next;
     preMid.next = null;  // 断开链表
     TreeNode t = new TreeNode(mid.val);
     t.left = sortedListToBST(head);
@@ -6226,7 +5276,7 @@ private void inOrder(TreeNode node) {
 return [2].
 ```
 
-答案可能不止一个，也就是有多个值出现的次数一样多，并且是最大的。
+答案可能不止一个，也就是有多个值出现的次数一样多。
 
 ```java
 private int curCnt = 1;
@@ -6396,6 +5446,974 @@ class MapSum {
 }
 ```
 
+
+## 栈和队列
+
+**用栈实现队列** 
+
+[232. Implement Queue using Stacks (Easy)](https://leetcode.com/problems/implement-queue-using-stacks/description/)
+
+栈的顺序为后进先出，而队列的顺序为先进先出。使用两个栈实现队列，一个元素需要经过两个栈才能出队列，在经过第一个栈时元素顺序被反转，经过第二个栈时再次被反转，此时就是先进先出顺序。
+
+```java
+class MyQueue {
+
+    private Stack<Integer> in = new Stack<>();
+    private Stack<Integer> out = new Stack<>();
+
+    public void push(int x) {
+        in.push(x);
+    }
+
+    public int pop() {
+        in2out();
+        return out.pop();
+    }
+
+    public int peek() {
+        in2out();
+        return out.peek();
+    }
+
+    private void in2out() {
+        if (out.isEmpty()) {
+            while (!in.isEmpty()) {
+                out.push(in.pop());
+            }
+        }
+    }
+
+    public boolean empty() {
+        return in.isEmpty() && out.isEmpty();
+    }
+}
+```
+
+**用队列实现栈** 
+
+[225. Implement Stack using Queues (Easy)](https://leetcode.com/problems/implement-stack-using-queues/description/)
+
+在将一个元素 x 插入队列时，为了维护原来的后进先出顺序，需要让 x 插入队列首部。而队列的默认插入顺序是队列尾部，因此在将 x 插入队列尾部之后，需要让除了 x 之外的所有元素出队列，再入队列。
+
+```java
+class MyStack {
+
+    private Queue<Integer> queue;
+
+    public MyStack() {
+        queue = new LinkedList<>();
+    }
+
+    public void push(int x) {
+        queue.add(x);
+        int cnt = queue.size();
+        while (cnt-- > 1) {
+            queue.add(queue.poll());
+        }
+    }
+
+    public int pop() {
+        return queue.remove();
+    }
+
+    public int top() {
+        return queue.peek();
+    }
+
+    public boolean empty() {
+        return queue.isEmpty();
+    }
+}
+```
+
+**最小值栈** 
+
+[155. Min Stack (Easy)](https://leetcode.com/problems/min-stack/description/)
+
+```java
+class MinStack {
+
+    private Stack<Integer> dataStack;
+    private Stack<Integer> minStack;
+    private int min;
+
+    public MinStack() {
+        dataStack = new Stack<>();
+        minStack = new Stack<>();
+        min = Integer.MAX_VALUE;
+    }
+
+    public void push(int x) {
+        dataStack.add(x);
+        min = Math.min(min, x);
+        minStack.add(min);
+    }
+
+    public void pop() {
+        dataStack.pop();
+        minStack.pop();
+        min = minStack.isEmpty() ? Integer.MAX_VALUE : minStack.peek();
+    }
+
+    public int top() {
+        return dataStack.peek();
+    }
+
+    public int getMin() {
+        return minStack.peek();
+    }
+}
+```
+
+对于实现最小值队列问题，可以先将队列使用栈来实现，然后就将问题转换为最小值栈，这个问题出现在 编程之美：3.7。
+
+**用栈实现括号匹配** 
+
+[20. Valid Parentheses (Easy)](https://leetcode.com/problems/valid-parentheses/description/)
+
+```html
+"()[]{}"
+
+Output : true
+```
+
+```java
+public boolean isValid(String s) {
+    Stack<Character> stack = new Stack<>();
+    for (char c : s.toCharArray()) {
+        if (c == '(' || c == '{' || c == '[') {
+            stack.push(c);
+        } else {
+            if (stack.isEmpty()) {
+                return false;
+            }
+            char cStack = stack.pop();
+            boolean b1 = c == ')' && cStack != '(';
+            boolean b2 = c == ']' && cStack != '[';
+            boolean b3 = c == '}' && cStack != '{';
+            if (b1 || b2 || b3) {
+                return false;
+            }
+        }
+    }
+    return stack.isEmpty();
+}
+```
+
+**数组中元素与下一个比它大的元素之间的距离** 
+
+[739. Daily Temperatures (Medium)](https://leetcode.com/problems/daily-temperatures/description/)
+
+```html
+Input: [73, 74, 75, 71, 69, 72, 76, 73]
+Output: [1, 1, 4, 2, 1, 1, 0, 0]
+```
+
+在遍历数组时用栈把数组中的数存起来，如果当前遍历的数比栈顶元素来的大，说明栈顶元素的下一个比它大的数就是当前元素。
+
+```java
+public int[] dailyTemperatures(int[] temperatures) {
+    int n = temperatures.length;
+    int[] dist = new int[n];
+    Stack<Integer> indexs = new Stack<>();
+    for (int curIndex = 0; curIndex < n; curIndex++) {
+        while (!indexs.isEmpty() && temperatures[curIndex] > temperatures[indexs.peek()]) {
+            int preIndex = indexs.pop();
+            dist[preIndex] = curIndex - preIndex;
+        }
+        indexs.add(curIndex);
+    }
+    return dist;
+}
+```
+
+**循环数组中比当前元素大的下一个元素** 
+
+[503. Next Greater Element II (Medium)](https://leetcode.com/problems/next-greater-element-ii/description/)
+
+```text
+Input: [1,2,1]
+Output: [2,-1,2]
+Explanation: The first 1's next greater number is 2;
+The number 2 can't find next greater number;
+The second 1's next greater number needs to search circularly, which is also 2.
+```
+
+与 739. Daily Temperatures (Medium) 不同的是，数组是循环数组，并且最后要求的不是距离而是下一个元素。
+
+```java
+public int[] nextGreaterElements(int[] nums) {
+    int n = nums.length;
+    int[] next = new int[n];
+    Arrays.fill(next, -1);
+    Stack<Integer> pre = new Stack<>();
+    for (int i = 0; i < n * 2; i++) {
+        int num = nums[i % n];
+        while (!pre.isEmpty() && nums[pre.peek()] < num) {
+            next[pre.pop()] = num;
+        }
+        if (i < n){
+            pre.push(i);
+        }
+    }
+    return next;
+}
+```
+
+## 哈希表
+
+哈希表使用 O(N) 空间复杂度存储数据，并且以 O(1) 时间复杂度求解问题。
+
+- Java 中的  **HashSet**  用于存储一个集合，可以查找元素是否在集合中。如果元素有穷，并且范围不大，那么可以用一个布尔数组来存储一个元素是否存在。例如对于只有小写字符的元素，就可以用一个长度为 26 的布尔数组来存储一个字符集合，使得空间复杂度降低为 O(1)。
+
+- Java 中的  **HashMap**  主要用于映射关系，从而把两个元素联系起来。HashMap 也可以用来对元素进行计数统计，此时键为元素，值为计数。和 HashSet 类似，如果元素有穷并且范围不大，可以用整型数组来进行统计。在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中 [Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)，利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
+
+
+**数组中两个数的和为给定值** 
+
+[1. Two Sum (Easy)](https://leetcode.com/problems/two-sum/description/)
+
+可以先对数组进行排序，然后使用双指针方法或者二分查找方法。这样做的时间复杂度为 O(NlogN)，空间复杂度为 O(1)。
+
+用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i]，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
+
+```java
+public int[] twoSum(int[] nums, int target) {
+    HashMap<Integer, Integer> indexForNum = new HashMap<>();
+    for (int i = 0; i < nums.length; i++) {
+        if (indexForNum.containsKey(target - nums[i])) {
+            return new int[]{indexForNum.get(target - nums[i]), i};
+        } else {
+            indexForNum.put(nums[i], i);
+        }
+    }
+    return null;
+}
+```
+
+**判断数组是否含有重复元素** 
+
+[217. Contains Duplicate (Easy)](https://leetcode.com/problems/contains-duplicate/description/)
+
+```java
+public boolean containsDuplicate(int[] nums) {
+    Set<Integer> set = new HashSet<>();
+    for (int num : nums) {
+        set.add(num);
+    }
+    return set.size() < nums.length;
+}
+```
+
+**最长和谐序列** 
+
+[594. Longest Harmonious Subsequence (Easy)](https://leetcode.com/problems/longest-harmonious-subsequence/description/)
+
+```html
+Input: [1,3,2,2,5,2,3,7]
+Output: 5
+Explanation: The longest harmonious subsequence is [3,2,2,2,3].
+```
+
+和谐序列中最大数和最小数只差正好为 1，应该注意的是序列的元素不一定是数组的连续元素。
+
+```java
+public int findLHS(int[] nums) {
+    Map<Integer, Integer> countForNum = new HashMap<>();
+    for (int num : nums) {
+        countForNum.put(num, countForNum.getOrDefault(num, 0) + 1);
+    }
+    int longest = 0;
+    for (int num : countForNum.keySet()) {
+        if (countForNum.containsKey(num + 1)) {
+            longest = Math.max(longest, countForNum.get(num + 1) + countForNum.get(num));
+        }
+    }
+    return longest;
+}
+```
+
+**最长连续序列** 
+
+[128. Longest Consecutive Sequence (Hard)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
+
+```html
+Given [100, 4, 200, 1, 3, 2],
+The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
+```
+
+要求以 O(N) 的时间复杂度求解。
+
+```java
+public int longestConsecutive(int[] nums) {
+    Map<Integer, Integer> countForNum = new HashMap<>();
+    for (int num : nums) {
+        countForNum.put(num, 1);
+    }
+    for (int num : nums) {
+        forward(countForNum, num);
+    }
+    return maxCount(countForNum);
+}
+
+private int forward(Map<Integer, Integer> countForNum, int num) {
+    if (!countForNum.containsKey(num)) {
+        return 0;
+    }
+    int cnt = countForNum.get(num);
+    if (cnt > 1) {
+        return cnt;
+    }
+    cnt = forward(countForNum, num + 1) + 1;
+    countForNum.put(num, cnt);
+    return cnt;
+}
+
+private int maxCount(Map<Integer, Integer> countForNum) {
+    int max = 0;
+    for (int num : countForNum.keySet()) {
+        max = Math.max(max, countForNum.get(num));
+    }
+    return max;
+}
+```
+
+## 字符串
+
+**字符串循环移位包含** 
+
+[编程之美 3.1](#)
+
+```html
+s1 = AABCD, s2 = CDAA
+Return : true
+```
+
+给定两个字符串 s1 和 s2，要求判定 s2 是否能够被 s1 做循环移位得到的字符串包含。
+
+s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2 是否是 s1s1 的子字符串即可。
+
+**字符串循环移位** 
+
+[编程之美 2.17](#)
+
+```html
+s = "abcd123" k = 3
+Return "123abcd"
+```
+
+将字符串向右循环移动 k 位。
+
+将 abcd123 中的 abcd 和 123 单独翻转，得到 dcba321，然后对整个字符串进行翻转，得到 123abcd。
+
+**字符串中单词的翻转** 
+
+[程序员代码面试指南](#)
+
+```html
+s = "I am a student"
+Return "student a am I"
+```
+
+将每个单词翻转，然后将整个字符串翻转。
+
+**两个字符串包含的字符是否完全相同** 
+
+[242. Valid Anagram (Easy)](https://leetcode.com/problems/valid-anagram/description/)
+
+```html
+s = "anagram", t = "nagaram", return true.
+s = "rat", t = "car", return false.
+```
+
+可以用 HashMap 来映射字符与出现次数，然后比较两个字符串出现的字符数量是否相同。
+
+由于本题的字符串只包含 26 个小写字符，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，不再使用 HashMap。
+
+```java
+public boolean isAnagram(String s, String t) {
+    int[] cnts = new int[26];
+    for (char c : s.toCharArray()) {
+        cnts[c - 'a']++;
+    }
+    for (char c : t.toCharArray()) {
+        cnts[c - 'a']--;
+    }
+    for (int cnt : cnts) {
+        if (cnt != 0) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+**计算一组字符集合可以组成的回文字符串的最大长度** 
+
+[409. Longest Palindrome (Easy)](https://leetcode.com/problems/longest-palindrome/description/)
+
+```html
+Input : "abccccdd"
+Output : 7
+Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
+```
+
+使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。
+
+因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
+
+```java
+public int longestPalindrome(String s) {
+    int[] cnts = new int[256];
+    for (char c : s.toCharArray()) {
+        cnts[c]++;
+    }
+    int palindrome = 0;
+    for (int cnt : cnts) {
+        palindrome += (cnt / 2) * 2;
+    }
+    if (palindrome < s.length()) {
+        palindrome++;   // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
+    }
+    return palindrome;
+}
+```
+
+**字符串同构** 
+
+[205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
+
+```html
+Given "egg", "add", return true.
+Given "foo", "bar", return false.
+Given "paper", "title", return true.
+```
+
+记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
+
+```java
+public boolean isIsomorphic(String s, String t) {
+    int[] preIndexOfS = new int[256];
+    int[] preIndexOfT = new int[256];
+    for (int i = 0; i < s.length(); i++) {
+        char sc = s.charAt(i), tc = t.charAt(i);
+        if (preIndexOfS[sc] != preIndexOfT[tc]) {
+            return false;
+        }
+        preIndexOfS[sc] = i + 1;
+        preIndexOfT[tc] = i + 1;
+    }
+    return true;
+}
+```
+
+**回文子字符串个数** 
+
+[647. Palindromic Substrings (Medium)](https://leetcode.com/problems/palindromic-substrings/description/)
+
+```html
+Input: "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+```
+
+从字符串的某一位开始，尝试着去扩展子字符串。
+
+```java
+private int cnt = 0;
+
+public int countSubstrings(String s) {
+    for (int i = 0; i < s.length(); i++) {
+        extendSubstrings(s, i, i);     // 奇数长度
+        extendSubstrings(s, i, i + 1); // 偶数长度
+    }
+    return cnt;
+}
+
+private void extendSubstrings(String s, int start, int end) {
+    while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
+        start--;
+        end++;
+        cnt++;
+    }
+}
+```
+
+**判断一个整数是否是回文数** 
+
+[9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
+
+要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
+
+将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
+
+```java
+public boolean isPalindrome(int x) {
+    if (x == 0) {
+        return true;
+    }
+    if (x < 0 || x % 10 == 0) {
+        return false;
+    }
+    int right = 0;
+    while (x > right) {
+        right = right * 10 + x % 10;
+        x /= 10;
+    }
+    return x == right || x == right / 10;
+}
+```
+
+**统计二进制字符串中连续 1 和连续 0 数量相同的子字符串个数** 
+
+[696. Count Binary Substrings (Easy)](https://leetcode.com/problems/count-binary-substrings/description/)
+
+```html
+Input: "00110011"
+Output: 6
+Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
+```
+
+```java
+public int countBinarySubstrings(String s) {
+    int preLen = 0, curLen = 1, count = 0;
+    for (int i = 1; i < s.length(); i++) {
+        if (s.charAt(i) == s.charAt(i - 1)) {
+            curLen++;
+        } else {
+            preLen = curLen;
+            curLen = 1;
+        }
+
+        if (preLen >= curLen) {
+            count++;
+        }
+    }
+    return count;
+}
+```
+
+## 数组与矩阵
+
+**把数组中的 0 移到末尾** 
+
+[283. Move Zeroes (Easy)](https://leetcode.com/problems/move-zeroes/description/)
+
+```html
+For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
+```
+
+```java
+public void moveZeroes(int[] nums) {
+    int idx = 0;
+    for (int num : nums) {
+        if (num != 0) {
+            nums[idx++] = num;
+        }
+    }
+    while (idx < nums.length) {
+        nums[idx++] = 0;
+    }
+}
+```
+
+**改变矩阵维度** 
+
+[566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
+
+```html
+Input:
+nums =
+[[1,2],
+ [3,4]]
+r = 1, c = 4
+
+Output:
+[[1,2,3,4]]
+
+Explanation:
+The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
+```
+
+```java
+public int[][] matrixReshape(int[][] nums, int r, int c) {
+    int m = nums.length, n = nums[0].length;
+    if (m * n != r * c) {
+        return nums;
+    }
+    int[][] reshapedNums = new int[r][c];
+    int index = 0;
+    for (int i = 0; i < r; i++) {
+        for (int j = 0; j < c; j++) {
+            reshapedNums[i][j] = nums[index / n][index % n];
+            index++;
+        }
+    }
+    return reshapedNums;
+}
+```
+
+**找出数组中最长的连续 1** 
+
+[485. Max Consecutive Ones (Easy)](https://leetcode.com/problems/max-consecutive-ones/description/)
+
+```java
+public int findMaxConsecutiveOnes(int[] nums) {
+    int max = 0, cur = 0;
+    for (int x : nums) {
+        cur = x == 0 ? 0 : cur + 1;
+        max = Math.max(max, cur);
+    }
+    return max;
+}
+```
+
+**有序矩阵查找** 
+
+[240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
+
+```html
+[
+   [ 1,  5,  9],
+   [10, 11, 13],
+   [12, 13, 15]
+]
+```
+
+```java
+public boolean searchMatrix(int[][] matrix, int target) {
+    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
+    int m = matrix.length, n = matrix[0].length;
+    int row = 0, col = n - 1;
+    while (row < m && col >= 0) {
+        if (target == matrix[row][col]) return true;
+        else if (target < matrix[row][col]) col--;
+        else row++;
+    }
+    return false;
+}
+```
+
+**有序矩阵的 Kth Element** 
+
+[378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
+
+```html
+matrix = [
+  [ 1,  5,  9],
+  [10, 11, 13],
+  [12, 13, 15]
+],
+k = 8,
+
+return 13.
+```
+
+解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
+
+二分查找解法：
+
+```java
+public int kthSmallest(int[][] matrix, int k) {
+    int m = matrix.length, n = matrix[0].length;
+    int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
+    while (lo <= hi) {
+        int mid = lo + (hi - lo) / 2;
+        int cnt = 0;
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
+                cnt++;
+            }
+        }
+        if (cnt < k) lo = mid + 1;
+        else hi = mid - 1;
+    }
+    return lo;
+}
+```
+
+堆解法：
+
+```java
+public int kthSmallest(int[][] matrix, int k) {
+    int m = matrix.length, n = matrix[0].length;
+    PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
+    for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
+    for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
+        Tuple t = pq.poll();
+        if(t.x == m - 1) continue;
+        pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
+    }
+    return pq.poll().val;
+}
+
+class Tuple implements Comparable<Tuple> {
+    int x, y, val;
+    public Tuple(int x, int y, int val) {
+        this.x = x; this.y = y; this.val = val;
+    }
+
+    @Override
+    public int compareTo(Tuple that) {
+        return this.val - that.val;
+    }
+}
+```
+
+**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出重复的数和丢失的数** 
+
+[645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
+
+```html
+Input: nums = [1,2,2,4]
+Output: [2,3]
+```
+
+```html
+Input: nums = [1,2,2,4]
+Output: [2,3]
+```
+
+最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
+
+主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
+
+```java
+public int[] findErrorNums(int[] nums) {
+    for (int i = 0; i < nums.length; i++) {
+        while (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i]) {
+            swap(nums, i, nums[i] - 1);
+        }
+    }
+    for (int i = 0; i < nums.length; i++) {
+        if (nums[i] != i + 1) {
+            return new int[]{nums[i], i + 1};
+        }
+    }
+    return null;
+}
+
+private void swap(int[] nums, int i, int j) {
+    int tmp = nums[i];
+    nums[i] = nums[j];
+    nums[j] = tmp;
+}
+```
+
+类似题目：
+
+- [448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
+- [442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
+
+**找出数组中重复的数，数组值在 [1, n] 之间** 
+
+[287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
+
+要求不能修改数组，也不能使用额外的空间。
+
+二分查找解法：
+
+```java
+public int findDuplicate(int[] nums) {
+     int l = 1, h = nums.length - 1;
+     while (l <= h) {
+         int mid = l + (h - l) / 2;
+         int cnt = 0;
+         for (int i = 0; i < nums.length; i++) {
+             if (nums[i] <= mid) cnt++;
+         }
+         if (cnt > mid) h = mid - 1;
+         else l = mid + 1;
+     }
+     return l;
+}
+```
+
+双指针解法，类似于有环链表中找出环的入口：
+
+```java
+public int findDuplicate(int[] nums) {
+    int slow = nums[0], fast = nums[nums[0]];
+    while (slow != fast) {
+        slow = nums[slow];
+        fast = nums[nums[fast]];
+    }
+    fast = 0;
+    while (slow != fast) {
+        slow = nums[slow];
+        fast = nums[fast];
+    }
+    return slow;
+}
+```
+
+**数组相邻差值的个数** 
+
+[667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
+
+```html
+Input: n = 3, k = 2
+Output: [1, 3, 2]
+Explanation: The [1, 3, 2] has three different positive integers ranging from 1 to 3, and the [2, 1] has exactly 2 distinct integers: 1 and 2.
+```
+
+题目描述：数组元素为 1\~n 的整数，要求构建数组，使得相邻元素的差值不相同的个数为 k。
+
+让前 k+1 个元素构建出 k 个不相同的差值，序列为：1 k+1 2 k 3 k-1 ... k/2 k/2+1.
+
+```java
+public int[] constructArray(int n, int k) {
+    int[] ret = new int[n];
+    ret[0] = 1;
+    for (int i = 1, interval = k; i <= k; i++, interval--) {
+        ret[i] = i % 2 == 1 ? ret[i - 1] + interval : ret[i - 1] - interval;
+    }
+    for (int i = k + 1; i < n; i++) {
+        ret[i] = i + 1;
+    }
+    return ret;
+}
+```
+
+**数组的度** 
+
+[697. Degree of an Array (Easy)](https://leetcode.com/problems/degree-of-an-array/description/)
+
+```html
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+题目描述：数组的度定义为元素出现的最高频率，例如上面的数组度为 3。要求找到一个最小的子数组，这个子数组的度和原数组一样。
+
+```java
+public int findShortestSubArray(int[] nums) {
+    Map<Integer, Integer> numsCnt = new HashMap<>();
+    Map<Integer, Integer> numsLastIndex = new HashMap<>();
+    Map<Integer, Integer> numsFirstIndex = new HashMap<>();
+    for (int i = 0; i < nums.length; i++) {
+        int num = nums[i];
+        numsCnt.put(num, numsCnt.getOrDefault(num, 0) + 1);
+        numsLastIndex.put(num, i);
+        if (!numsFirstIndex.containsKey(num)) {
+            numsFirstIndex.put(num, i);
+        }
+    }
+    int maxCnt = 0;
+    for (int num : nums) {
+        maxCnt = Math.max(maxCnt, numsCnt.get(num));
+    }
+    int ret = nums.length;
+    for (int i = 0; i < nums.length; i++) {
+        int num = nums[i];
+        int cnt = numsCnt.get(num);
+        if (cnt != maxCnt) continue;
+        ret = Math.min(ret, numsLastIndex.get(num) - numsFirstIndex.get(num) + 1);
+    }
+    return ret;
+}
+```
+
+**对角元素相等的矩阵** 
+
+[766. Toeplitz Matrix (Easy)](https://leetcode.com/problems/toeplitz-matrix/description/)
+
+```html
+1234
+5123
+9512
+
+In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.
+```
+
+```java
+public boolean isToeplitzMatrix(int[][] matrix) {
+    for (int i = 0; i < matrix[0].length; i++) {
+        if (!check(matrix, matrix[0][i], 0, i)) {
+            return false;
+        }
+    }
+    for (int i = 0; i < matrix.length; i++) {
+        if (!check(matrix, matrix[i][0], i, 0)) {
+            return false;
+        }
+    }
+    return true;
+}
+
+private boolean check(int[][] matrix, int expectValue, int row, int col) {
+    if (row >= matrix.length || col >= matrix[0].length) {
+        return true;
+    }
+    if (matrix[row][col] != expectValue) {
+        return false;
+    }
+    return check(matrix, expectValue, row + 1, col + 1);
+}
+```
+
+**嵌套数组** 
+
+[565. Array Nesting (Medium)](https://leetcode.com/problems/array-nesting/description/)
+
+```html
+Input: A = [5,4,0,3,1,6,2]
+Output: 4
+Explanation:
+A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
+
+One of the longest S[K]:
+S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
+```
+
+题目描述：S[i] 表示一个集合，集合的第一个元素是 A[i]，第二个元素是 A[A[i]]，如此嵌套下去。求最大的 S[i]。
+
+```java
+public int arrayNesting(int[] nums) {
+    int max = 0;
+    for (int i = 0; i < nums.length; i++) {
+        int cnt = 0;
+        for (int j = i; nums[j] != -1; ) {
+            cnt++;
+            int t = nums[j];
+            nums[j] = -1; // 标记该位置已经被访问
+            j = t;
+
+        }
+        max = Math.max(max, cnt);
+    }
+    return max;
+}
+```
+
+**分隔数组** 
+
+[769. Max Chunks To Make Sorted (Medium)](https://leetcode.com/problems/max-chunks-to-make-sorted/description/)
+
+```html
+Input: arr = [1,0,2,3,4]
+Output: 4
+Explanation:
+We can split into two chunks, such as [1, 0], [2, 3, 4].
+However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
+```
+
+题目描述：分隔数组，使得对每部分排序后数组就为有序。
+
+```java
+public int maxChunksToSorted(int[] arr) {
+    if (arr == null) return 0;
+    int ret = 0;
+    int right = arr[0];
+    for (int i = 0; i < arr.length; i++) {
+        right = Math.max(right, arr[i]);
+        if (right == i) ret++;
+    }
+    return ret;
+}
+```
+
+
 ## 图
 
 ### 二分图
@@ -7056,3 +7074,4 @@ public int[] countBits(int num) {
 - 何海涛, 软件工程师. 剑指 Offer: 名企面试官精讲典型编程题[M]. 电子工业出版社, 2014.
 - 《编程之美》小组. 编程之美[M]. 电子工业出版社, 2008.
 - 左程云. 程序员代码面试指南[M]. 电子工业出版社, 2015.
+
diff --git a/notes/Leetcode-Database 题解.md b/notes/Leetcode-Database 题解.md
index 5ab726c5..5b4a3d00 100644
--- a/notes/Leetcode-Database 题解.md	
+++ b/notes/Leetcode-Database 题解.md	
@@ -461,7 +461,7 @@ Employee 表：
 +----+-------+--------+-----------+
 ```
 
-查找所有员工，他们的薪资大于其经理薪资。
+查找薪资大于其经理薪资的员工信息。
 
 ## SQL Schema
 
@@ -924,27 +924,27 @@ VALUES
 ```sql
 SELECT
     s1.id - 1 AS id,
-    s1.student 
+    s1.student
 FROM
-    seat s1 
+    seat s1
 WHERE
     s1.id MOD 2 = 0 UNION
 SELECT
     s2.id + 1 AS id,
-    s2.student 
+    s2.student
 FROM
-    seat s2 
+    seat s2
 WHERE
-    s2.id MOD 2 = 1 
+    s2.id MOD 2 = 1
     AND s2.id != ( SELECT max( s3.id ) FROM seat s3 ) UNION
 SELECT
     s4.id AS id,
-    s4.student 
+    s4.student
 FROM
-    seat s4 
+    seat s4
 WHERE
-    s4.id MOD 2 = 1 
-    AND s4.id = ( SELECT max( s5.id ) FROM seat s5 ) 
+    s4.id MOD 2 = 1
+    AND s4.id = ( SELECT max( s5.id ) FROM seat s5 )
 ORDER BY
     id;
 ```
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 47d5d94e..0c9bd1f9 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -633,11 +633,12 @@ public int RectCover(int n) {
 
 ## 解题思路
 
-当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；否则解在 [m + 1, h] 之间，令 l = m + 1。
+- 当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；
+- 否则解在 [m + 1, h] 之间，令 l = m + 1。
 
-因为 h 的赋值表达式为 h = m，因此循环体的循环条件应该为 l < h，详细解释请见 [Leetcode 题解](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md#%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE) 二分查找部分。
+因为 h 的赋值表达式为 h = m，因此循环体的循环条件应该为 l < h，详细解释请见 [Leetcode 题解](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md) 二分查找部分。
 
-但是如果出现 nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，因此需要切换到顺序查找。
+但是如果出现 nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，需要切换到顺序查找。
 
 复杂度：O(logN) + O(1)
 
@@ -1217,8 +1218,7 @@ public ListNode ReverseList(ListNode head) {
 ### 递归
 
 ```java
-public ListNode Merge(ListNode list1, ListNode list2)
-{
+public ListNode Merge(ListNode list1, ListNode list2) {
     if (list1 == null)
         return list2;
     if (list2 == null)
@@ -1236,8 +1236,7 @@ public ListNode Merge(ListNode list1, ListNode list2)
 ### 迭代
 
 ```java
-public ListNode Merge(ListNode list1, ListNode list2)
-{
+public ListNode Merge(ListNode list1, ListNode list2) {
     ListNode head = new ListNode(-1);
     ListNode cur = head;
     while (list1 != null && list2 != null) {
@@ -1269,15 +1268,13 @@ public ListNode Merge(ListNode list1, ListNode list2)
 ## 解题思路
 
 ```java
-public boolean HasSubtree(TreeNode root1, TreeNode root2)
-{
+public boolean HasSubtree(TreeNode root1, TreeNode root2) {
     if (root1 == null || root2 == null)
         return false;
     return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
 }
 
-private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
-{
+private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
     if (root2 == null)
         return true;
     if (root1 == null)
@@ -1298,9 +1295,10 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
 
 ## 解题思路
 
+### 递归
+
 ```java
-public void Mirror(TreeNode root)
-{
+public void Mirror(TreeNode root) {
     if (root == null)
         return;
     swap(root);
@@ -1308,14 +1306,36 @@ public void Mirror(TreeNode root)
     Mirror(root.right);
 }
 
-private void swap(TreeNode root)
-{
+private void swap(TreeNode root) {
     TreeNode t = root.left;
     root.left = root.right;
     root.right = t;
 }
 ```
 
+### 迭代
+
+```java
+public void Mirror(TreeNode root) {
+    Stack<TreeNode> stack = new Stack<>();
+    stack.push(root);
+    while (!stack.isEmpty()) {
+        TreeNode node = stack.pop();
+        if (node == null)
+            continue;
+        swap(node);
+        stack.push(node.left);
+        stack.push(node.right);
+    }
+}
+
+private void swap(TreeNode node) {
+    TreeNode t = node.left;
+    node.left = node.right;
+    node.right = t;
+}
+```
+
 # 28 对称的二叉树
 
 [NowCder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
@@ -1327,15 +1347,13 @@ private void swap(TreeNode root)
 ## 解题思路
 
 ```java
-boolean isSymmetrical(TreeNode pRoot)
-{
+boolean isSymmetrical(TreeNode pRoot) {
     if (pRoot == null)
         return true;
     return isSymmetrical(pRoot.left, pRoot.right);
 }
 
-boolean isSymmetrical(TreeNode t1, TreeNode t2)
-{
+boolean isSymmetrical(TreeNode t1, TreeNode t2) {
     if (t1 == null && t2 == null)
         return true;
     if (t1 == null || t2 == null)
@@ -1359,8 +1377,7 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2)
 ## 解题思路
 
 ```java
-public ArrayList<Integer> printMatrix(int[][] matrix)
-{
+public ArrayList<Integer> printMatrix(int[][] matrix) {
     ArrayList<Integer> ret = new ArrayList<>();
     int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
     while (r1 <= r2 && c1 <= c2) {
@@ -1394,25 +1411,21 @@ public ArrayList<Integer> printMatrix(int[][] matrix)
 private Stack<Integer> dataStack = new Stack<>();
 private Stack<Integer> minStack = new Stack<>();
 
-public void push(int node)
-{
+public void push(int node) {
     dataStack.push(node);
     minStack.push(minStack.isEmpty() ? node : Math.min(minStack.peek(), node));
 }
 
-public void pop()
-{
+public void pop() {
     dataStack.pop();
     minStack.pop();
 }
 
-public int top()
-{
+public int top() {
     return dataStack.peek();
 }
 
-public int min()
-{
+public int min() {
     return minStack.peek();
 }
 ```
@@ -1430,8 +1443,7 @@ public int min()
 使用一个栈来模拟压入弹出操作。
 
 ```java
-public boolean IsPopOrder(int[] pushSequence, int[] popSequence)
-{
+public boolean IsPopOrder(int[] pushSequence, int[] popSequence) {
     int n = pushSequence.length;
     Stack<Integer> stack = new Stack<>();
     for (int pushIndex = 0, popIndex = 0; pushIndex < n; pushIndex++) {
@@ -1464,8 +1476,7 @@ public boolean IsPopOrder(int[] pushSequence, int[] popSequence)
 不需要使用两个队列分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
 
 ```java
-public ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
-{
+public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
     Queue<TreeNode> queue = new LinkedList<>();
     ArrayList<Integer> ret = new ArrayList<>();
     queue.add(root);
@@ -1495,8 +1506,7 @@ public ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
 ## 解题思路
 
 ```java
-ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
-{
+ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     Queue<TreeNode> queue = new LinkedList<>();
     queue.add(pRoot);
@@ -1529,8 +1539,7 @@ ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
 ## 解题思路
 
 ```java
-public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
-{
+public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     Queue<TreeNode> queue = new LinkedList<>();
     queue.add(pRoot);
@@ -1571,15 +1580,13 @@ public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
 ## 解题思路
 
 ```java
-public boolean VerifySquenceOfBST(int[] sequence)
-{
+public boolean VerifySquenceOfBST(int[] sequence) {
     if (sequence == null || sequence.length == 0)
         return false;
     return verify(sequence, 0, sequence.length - 1);
 }
 
-private boolean verify(int[] sequence, int first, int last)
-{
+private boolean verify(int[] sequence, int first, int last) {
     if (last - first <= 1)
         return true;
     int rootVal = sequence[last];
@@ -1610,14 +1617,12 @@ private boolean verify(int[] sequence, int first, int last)
 ```java
 private ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
 
-public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target)
-{
+public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target) {
     backtracking(root, target, new ArrayList<>());
     return ret;
 }
 
-private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
-{
+private void backtracking(TreeNode node, int target, ArrayList<Integer> path) {
     if (node == null)
         return;
     path.add(node.val);
@@ -1641,8 +1646,7 @@ private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
 输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的 head。
 
 ```java
-public class RandomListNode
-{
+public class RandomListNode {
     int label;
     RandomListNode next = null;
     RandomListNode random = null;
@@ -1670,8 +1674,7 @@ public class RandomListNode
 <div align="center"> <img src="../pics//8f3b9519-d705-48fe-87ad-2e4052fc81d2.png" width="600"/> </div><br>
 
 ```java
-public RandomListNode Clone(RandomListNode pHead)
-{
+public RandomListNode Clone(RandomListNode pHead) {
     if (pHead == null)
         return null;
     // 插入新节点
@@ -1718,14 +1721,12 @@ public RandomListNode Clone(RandomListNode pHead)
 private TreeNode pre = null;
 private TreeNode head = null;
 
-public TreeNode Convert(TreeNode root)
-{
+public TreeNode Convert(TreeNode root) {
     inOrder(root);
     return head;
 }
 
-private void inOrder(TreeNode node)
-{
+private void inOrder(TreeNode node) {
     if (node == null)
         return;
     inOrder(node.left);
@@ -1752,21 +1753,18 @@ private void inOrder(TreeNode node)
 ```java
 private String deserializeStr;
 
-public String Serialize(TreeNode root)
-{
+public String Serialize(TreeNode root) {
     if (root == null)
         return "#";
     return root.val + " " + Serialize(root.left) + " " + Serialize(root.right);
 }
 
-public TreeNode Deserialize(String str)
-{
+public TreeNode Deserialize(String str) {
     deserializeStr = str;
     return Deserialize();
 }
 
-private TreeNode Deserialize()
-{
+private TreeNode Deserialize() {
     if (deserializeStr.length() == 0)
         return null;
     int index = deserializeStr.indexOf(" ");
@@ -1795,8 +1793,7 @@ private TreeNode Deserialize()
 ```java
 private ArrayList<String> ret = new ArrayList<>();
 
-public ArrayList<String> Permutation(String str)
-{
+public ArrayList<String> Permutation(String str) {
     if (str.length() == 0)
         return ret;
     char[] chars = str.toCharArray();
@@ -1805,8 +1802,7 @@ public ArrayList<String> Permutation(String str)
     return ret;
 }
 
-private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
-{
+private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s) {
     if (s.length() == chars.length) {
         ret.add(s.toString());
         return;
@@ -1836,8 +1832,7 @@ private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s)
 使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
 
 ```java
-public int MoreThanHalfNum_Solution(int[] nums)
-{
+public int MoreThanHalfNum_Solution(int[] nums) {
     int majority = nums[0];
     for (int i = 1, cnt = 1; i < nums.length; i++) {
         cnt = nums[i] == majority ? cnt + 1 : cnt - 1;
@@ -1868,8 +1863,7 @@ public int MoreThanHalfNum_Solution(int[] nums)
 快速排序的 partition() 方法，会返回一个整数 j 使得 a[l..j-1] 小于等于 a[j]，且 a[j+1..h] 大于等于 a[j]，此时 a[j] 就是数组的第 j 大元素。可以利用这个特性找出数组的第 K 个元素，这种找第 K 个元素的算法称为快速选择算法。
 
 ```java
-public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
-{
+public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
     ArrayList<Integer> ret = new ArrayList<>();
     if (k > nums.length || k <= 0)
         return ret;
@@ -1880,8 +1874,7 @@ public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
     return ret;
 }
 
-public void findKthSmallest(int[] nums, int k)
-{
+public void findKthSmallest(int[] nums, int k) {
     int l = 0, h = nums.length - 1;
     while (l < h) {
         int j = partition(nums, l, h);
@@ -1894,8 +1887,7 @@ public void findKthSmallest(int[] nums, int k)
     }
 }
 
-private int partition(int[] nums, int l, int h)
-{
+private int partition(int[] nums, int l, int h) {
     int p = nums[l];     /* 切分元素 */
     int i = l, j = h + 1;
     while (true) {
@@ -1909,8 +1901,7 @@ private int partition(int[] nums, int l, int h)
     return j;
 }
 
-private void swap(int[] nums, int i, int j)
-{
+private void swap(int[] nums, int i, int j) {
     int t = nums[i];
     nums[i] = nums[j];
     nums[j] = t;
@@ -1927,8 +1918,7 @@ private void swap(int[] nums, int i, int j)
 维护一个大小为 K 的最小堆过程如下：在添加一个元素之后，如果大顶堆的大小大于 K，那么需要将大顶堆的堆顶元素去除。
 
 ```java
-public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k)
-{
+public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
     if (k > nums.length || k <= 0)
         return new ArrayList<>();
     PriorityQueue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> o2 - o1);
@@ -1959,8 +1949,7 @@ private PriorityQueue<Integer> right = new PriorityQueue<>();
 /* 当前数据流读入的元素个数 */
 private int N = 0;
 
-public void Insert(Integer val)
-{
+public void Insert(Integer val) {
     /* 插入要保证两个堆存于平衡状态 */
     if (N % 2 == 0) {
         /* N 为偶数的情况下插入到右半边。
@@ -1975,8 +1964,7 @@ public void Insert(Integer val)
     N++;
 }
 
-public Double GetMedian()
-{
+public Double GetMedian() {
     if (N % 2 == 0)
         return (left.peek() + right.peek()) / 2.0;
     else
@@ -1998,16 +1986,14 @@ public Double GetMedian()
 private int[] cnts = new int[256];
 private Queue<Character> queue = new LinkedList<>();
 
-public void Insert(char ch)
-{
+public void Insert(char ch) {
     cnts[ch]++;
     queue.add(ch);
     while (!queue.isEmpty() && cnts[queue.peek()] > 1)
         queue.poll();
 }
 
-public char FirstAppearingOnce()
-{
+public char FirstAppearingOnce() {
     return queue.isEmpty() ? '#' : queue.peek();
 }
 ```
@@ -2023,8 +2009,7 @@ public char FirstAppearingOnce()
 ## 解题思路
 
 ```java
-public int FindGreatestSumOfSubArray(int[] nums)
-{
+public int FindGreatestSumOfSubArray(int[] nums) {
     if (nums == null || nums.length == 0)
         return 0;
     int greatestSum = Integer.MIN_VALUE;
@@ -2044,8 +2029,7 @@ public int FindGreatestSumOfSubArray(int[] nums)
 ## 解题思路
 
 ```java
-public int NumberOf1Between1AndN_Solution(int n)
-{
+public int NumberOf1Between1AndN_Solution(int n) {
     int cnt = 0;
     for (int m = 1; m <= n; m *= 10) {
         int a = n / m, b = n % m;
@@ -2066,8 +2050,7 @@ public int NumberOf1Between1AndN_Solution(int n)
 ## 解题思路
 
 ```java
-public int getDigitAtIndex(int index)
-{
+public int getDigitAtIndex(int index) {
     if (index < 0)
         return -1;
     int place = 1;  // 1 表示个位，2 表示 十位...
@@ -2085,8 +2068,7 @@ public int getDigitAtIndex(int index)
  * place 位数的数字组成的字符串长度
  * 10, 90, 900, ...
  */
-private int getAmountOfPlace(int place)
-{
+private int getAmountOfPlace(int place) {
     if (place == 1)
         return 10;
     return (int) Math.pow(10, place - 1) * 9;
@@ -2096,8 +2078,7 @@ private int getAmountOfPlace(int place)
  * place 位数的起始数字
  * 0, 10, 100, ...
  */
-private int getBeginNumberOfPlace(int place)
-{
+private int getBeginNumberOfPlace(int place) {
     if (place == 1)
         return 0;
     return (int) Math.pow(10, place - 1);
@@ -2106,8 +2087,7 @@ private int getBeginNumberOfPlace(int place)
 /**
  * 在 place 位数组成的字符串中，第 index 个数
  */
-private int getDigitAtIndex(int index, int place)
-{
+private int getDigitAtIndex(int index, int place) {
     int beginNumber = getBeginNumberOfPlace(place);
     int shiftNumber = index / place;
     String number = (beginNumber + shiftNumber) + "";
@@ -2129,8 +2109,7 @@ private int getDigitAtIndex(int index, int place)
 可以看成是一个排序问题，在比较两个字符串 S1 和 S2 的大小时，应该比较的是 S1+S2 和 S2+S1 的大小，如果 S1+S2 < S2+S1，那么应该把 S1 排在前面，否则应该把 S2 排在前面。
 
 ```java
-public String PrintMinNumber(int[] numbers)
-{
+public String PrintMinNumber(int[] numbers) {
     if (numbers == null || numbers.length == 0)
         return "";
     int n = numbers.length;
@@ -2156,8 +2135,7 @@ public String PrintMinNumber(int[] numbers)
 ## 解题思路
 
 ```java
-public int numDecodings(String s)
-{
+public int numDecodings(String s) {
     if (s == null || s.length() == 0)
         return 0;
     int n = s.length();
@@ -2200,8 +2178,7 @@ public int numDecodings(String s)
 应该用动态规划求解，而不是深度优先搜索，深度优先搜索过于复杂，不是最优解。
 
 ```java
-public int getMost(int[][] values)
-{
+public int getMost(int[][] values) {
     if (values == null || values.length == 0 || values[0].length == 0)
         return 0;
     int n = values[0].length;
@@ -2224,8 +2201,7 @@ public int getMost(int[][] values)
 ## 解题思路
 
 ```java
-public int longestSubStringWithoutDuplication(String str)
-{
+public int longestSubStringWithoutDuplication(String str) {
     int curLen = 0;
     int maxLen = 0;
     int[] preIndexs = new int[26];
@@ -2257,8 +2233,7 @@ public int longestSubStringWithoutDuplication(String str)
 ## 解题思路
 
 ```java
-public int GetUglyNumber_Solution(int N)
-{
+public int GetUglyNumber_Solution(int N) {
     if (N <= 6)
         return N;
     int i2 = 0, i3 = 0, i5 = 0;
@@ -2291,8 +2266,7 @@ public int GetUglyNumber_Solution(int N)
 最直观的解法是使用 HashMap 对出现次数进行统计，但是考虑到要统计的字符范围有限，因此可以使用整型数组代替 HashMap。
 
 ```java
-public int FirstNotRepeatingChar(String str)
-{
+public int FirstNotRepeatingChar(String str) {
     int[] cnts = new int[256];
     for (int i = 0; i < str.length(); i++)
         cnts[str.charAt(i)]++;
@@ -2306,8 +2280,7 @@ public int FirstNotRepeatingChar(String str)
 以上实现的空间复杂度还不是最优的。考虑到只需要找到只出现一次的字符，那么我们只需要统计的次数信息只有 0,1,更大，使用两个比特位就能存储这些信息。
 
 ```java
-public int FirstNotRepeatingChar2(String str)
-{
+public int FirstNotRepeatingChar2(String str) {
     BitSet bs1 = new BitSet(256);
     BitSet bs2 = new BitSet(256);
     for (char c : str.toCharArray()) {
@@ -2339,15 +2312,13 @@ public int FirstNotRepeatingChar2(String str)
 private long cnt = 0;
 private int[] tmp;  // 在这里创建辅助数组，而不是在 merge() 递归函数中创建
 
-public int InversePairs(int[] nums)
-{
+public int InversePairs(int[] nums) {
     tmp = new int[nums.length];
     mergeSort(nums, 0, nums.length - 1);
     return (int) (cnt % 1000000007);
 }
 
-private void mergeSort(int[] nums, int l, int h)
-{
+private void mergeSort(int[] nums, int l, int h) {
     if (h - l < 1)
         return;
     int m = l + (h - l) / 2;
@@ -2356,8 +2327,7 @@ private void mergeSort(int[] nums, int l, int h)
     merge(nums, l, m, h);
 }
 
-private void merge(int[] nums, int l, int m, int h)
-{
+private void merge(int[] nums, int l, int m, int h) {
     int i = l, j = m + 1, k = l;
     while (i <= m || j <= h) {
         if (i > m)
@@ -2392,8 +2362,7 @@ private void merge(int[] nums, int l, int m, int h)
 当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部重新开始访问链表 B；同样地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部重新开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。
 
 ```java
-public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2)
-{
+public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
     ListNode l1 = pHead1, l2 = pHead2;
     while (l1 != l2) {
         l1 = (l1 == null) ? pHead2 : l1.next;
@@ -2420,15 +2389,13 @@ Output:
 ## 解题思路
 
 ```java
-public int GetNumberOfK(int[] nums, int K)
-{
+public int GetNumberOfK(int[] nums, int K) {
     int first = binarySearch(nums, K);
     int last = binarySearch(nums, K + 1);
     return (first == nums.length || nums[first] != K) ? 0 : last - first;
 }
 
-private int binarySearch(int[] nums, int K)
-{
+private int binarySearch(int[] nums, int K) {
     int l = 0, h = nums.length;
     while (l < h) {
         int m = l + (h - l) / 2;
@@ -2453,14 +2420,12 @@ private int binarySearch(int[] nums, int K)
 private TreeNode ret;
 private int cnt = 0;
 
-public TreeNode KthNode(TreeNode pRoot, int k)
-{
+public TreeNode KthNode(TreeNode pRoot, int k) {
     inOrder(pRoot, k);
     return ret;
 }
 
-private void inOrder(TreeNode root, int k)
-{
+private void inOrder(TreeNode root, int k) {
     if (root == null || cnt >= k)
         return;
     inOrder(root.left, k);
@@ -2484,8 +2449,7 @@ private void inOrder(TreeNode root, int k)
 ## 解题思路
 
 ```java
-public int TreeDepth(TreeNode root)
-{
+public int TreeDepth(TreeNode root) {
     return root == null ? 0 : 1 + Math.max(TreeDepth(root.left), TreeDepth(root.right));
 }
 ```
@@ -2505,14 +2469,12 @@ public int TreeDepth(TreeNode root)
 ```java
 private boolean isBalanced = true;
 
-public boolean IsBalanced_Solution(TreeNode root)
-{
+public boolean IsBalanced_Solution(TreeNode root) {
     height(root);
     return isBalanced;
 }
 
-private int height(TreeNode root)
-{
+private int height(TreeNode root) {
     if (root == null || !isBalanced)
         return 0;
     int left = height(root.left);
@@ -2538,8 +2500,7 @@ private int height(TreeNode root)
 diff &= -diff 得到出 diff 最右侧不为 0 的位，也就是不存在重复的两个元素在位级表示上最右侧不同的那一位，利用这一位就可以将两个元素区分开来。
 
 ```java
-public void FindNumsAppearOnce(int[] nums, int num1[], int num2[])
-{
+public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
     int diff = 0;
     for (int num : nums)
         diff ^= num;
@@ -2570,8 +2531,7 @@ public void FindNumsAppearOnce(int[] nums, int num1[], int num2[])
 - 如果 sum < target，移动较小的元素，使 sum 变大一些。
 
 ```java
-public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum)
-{
+public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum) {
     int i = 0, j = array.length - 1;
     while (i < j) {
         int cur = array[i] + array[j];
@@ -2604,8 +2564,7 @@ public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum)
 ## 解题思路
 
 ```java
-public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum)
-{
+public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
     ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
     int start = 1, end = 2;
     int curSum = 3;
@@ -2648,8 +2607,7 @@ public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum)
 正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
 
 ```java
-public String ReverseSentence(String str)
-{
+public String ReverseSentence(String str) {
     int n = str.length();
     char[] chars = str.toCharArray();
     int i = 0, j = 0;
@@ -2664,14 +2622,12 @@ public String ReverseSentence(String str)
     return new String(chars);
 }
 
-private void reverse(char[] c, int i, int j)
-{
+private void reverse(char[] c, int i, int j) {
     while (i < j)
         swap(c, i++, j--);
 }
 
-private void swap(char[] c, int i, int j)
-{
+private void swap(char[] c, int i, int j) {
     char t = c[i];
     c[i] = c[j];
     c[j] = t;
@@ -2691,8 +2647,7 @@ private void swap(char[] c, int i, int j)
 先将 "abc" 和 "XYZdef" 分别翻转，得到 "cbafedZYX"，然后再把整个字符串翻转得到 "XYZdefabc"。
 
 ```java
-public String LeftRotateString(String str, int n)
-{
+public String LeftRotateString(String str, int n) {
     if (n >= str.length())
         return str;
     char[] chars = str.toCharArray();
@@ -2702,14 +2657,12 @@ public String LeftRotateString(String str, int n)
     return new String(chars);
 }
 
-private void reverse(char[] chars, int i, int j)
-{
+private void reverse(char[] chars, int i, int j) {
     while (i < j)
         swap(chars, i++, j--);
 }
 
-private void swap(char[] chars, int i, int j)
-{
+private void swap(char[] chars, int i, int j) {
     char t = chars[i];
     chars[i] = chars[j];
     chars[j] = t;
@@ -2727,8 +2680,7 @@ private void swap(char[] chars, int i, int j)
 ## 解题思路
 
 ```java
-public ArrayList<Integer> maxInWindows(int[] num, int size)
-{
+public ArrayList<Integer> maxInWindows(int[] num, int size) {
     ArrayList<Integer> ret = new ArrayList<>();
     if (size > num.length || size < 1)
         return ret;
@@ -2762,8 +2714,7 @@ public ArrayList<Integer> maxInWindows(int[] num, int size)
 空间复杂度：O(N<sup>2</sup>)
 
 ```java
-public List<Map.Entry<Integer, Double>> dicesSum(int n)
-{
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
     final int face = 6;
     final int pointNum = face * n;
     long[][] dp = new long[n + 1][pointNum + 1];
@@ -2790,8 +2741,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n)
 空间复杂度：O(N)
 
 ```java
-public List<Map.Entry<Integer, Double>> dicesSum(int n)
-{
+public List<Map.Entry<Integer, Double>> dicesSum(int n) {
     final int face = 6;
     final int pointNum = face * n;
     long[][] dp = new long[2][pointNum + 1];
@@ -2829,8 +2779,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n)
 ## 解题思路
 
 ```java
-public boolean isContinuous(int[] nums)
-{
+public boolean isContinuous(int[] nums) {
     if (nums.length < 5)
         return false;
     Arrays.sort(nums);
@@ -2861,8 +2810,7 @@ public boolean isContinuous(int[] nums)
 约瑟夫环，圆圈长度为 n 的解可以看成长度为 n-1 的解再加上报数的长度 m。因为是圆圈，所以最后需要对 n 取余。
 
 ```java
-public int LastRemaining_Solution(int n, int m)
-{
+public int LastRemaining_Solution(int n, int m) {
     if (n == 0)     /* 特殊输入的处理 */
         return -1;
     if (n == 1)     /* 返回条件 */
@@ -2884,8 +2832,7 @@ public int LastRemaining_Solution(int n, int m)
 使用贪心策略，假设第 i 轮进行卖出操作，买入操作价格应该在 i 之前并且价格最低。
 
 ```java
-public int maxProfit(int[] prices)
-{
+public int maxProfit(int[] prices) {
     if (prices == null || prices.length == 0)
         return 0;
     int soFarMin = prices[0];
@@ -2915,8 +2862,7 @@ public int maxProfit(int[] prices)
 以下实现中，递归的返回条件为 n <= 0，取非后就是 n > 0，递归的主体部分为 sum += Sum_Solution(n - 1)，转换为条件语句后就是 (sum += Sum_Solution(n - 1)) > 0。
 
 ```java
-public int Sum_Solution(int n)
-{
+public int Sum_Solution(int n) {
     int sum = n;
     boolean b = (n > 0) && ((sum += Sum_Solution(n - 1)) > 0);
     return sum;
@@ -2938,8 +2884,7 @@ a ^ b 表示没有考虑进位的情况下两数的和，(a & b) << 1 就是进
 递归会终止的原因是 (a & b) << 1 最右边会多一个 0，那么继续递归，进位最右边的 0 会慢慢增多，最后进位会变为 0，递归终止。
 
 ```java
-public int Add(int a, int b)
-{
+public int Add(int a, int b) {
     return b == 0 ? a : Add(a ^ b, (a & b) << 1);
 }
 ```
@@ -2955,8 +2900,7 @@ public int Add(int a, int b)
 ## 解题思路
 
 ```java
-public int[] multiply(int[] A)
-{
+public int[] multiply(int[] A) {
     int n = A.length;
     int[] B = new int[n];
     for (int i = 0, product = 1; i < n; product *= A[i], i++)       /* 从左往右累乘 */
@@ -2988,8 +2932,7 @@ Output:
 ## 解题思路
 
 ```java
-public int StrToInt(String str)
-{
+public int StrToInt(String str) {
     if (str == null || str.length() == 0)
         return 0;
     boolean isNegative = str.charAt(0) == '-';
diff --git a/notes/攻击技术.md b/notes/攻击技术.md
index c99071a0..d5a735ba 100644
--- a/notes/攻击技术.md
+++ b/notes/攻击技术.md
@@ -29,7 +29,7 @@
 
 ## 危害
 
-- 窃取用户的 Cookie 值
+- 窃取用户的 Cookie
 - 伪造虚假的输入表单骗取个人信息
 - 显示伪造的文章或者图片
 
@@ -47,7 +47,7 @@
 
 富文本编辑器允许用户输入 HTML 代码，就不能简单地将 `<` 等字符进行过滤了，极大地提高了 XSS 攻击的可能性。
 
-富文本编辑器通常采用 XSS filter 来防范 XSS 攻击，可以定义一些标签白名单或者黑名单，从而不允许有攻击性的 HTML 代码的输入。
+富文本编辑器通常采用 XSS filter 来防范 XSS 攻击，通过定义一些标签白名单或者黑名单，从而不允许有攻击性的 HTML 代码的输入。
 
 以下例子中，form 和 script 等标签都被转义，而 h 和 p 等标签将会保留。
 
@@ -131,7 +131,7 @@ http://www.examplebank.com/withdraw?account=AccoutName&amount=1000&for=PayeeName
 <img src="http://www.examplebank.com/withdraw?account=Alice&amount=1000&for=Badman">。
 ```
 
-如果有账户名为 Alice 的用户访问了恶意站点，而她之前刚访问过银行不久，登录信息尚未过期，那么她就会损失 1000 资金。
+如果有账户名为 Alice 的用户访问了恶意站点，而她之前刚访问过银行不久，登录信息尚未过期，那么她就会损失 1000 美元。
 
 这种恶意的网址可以有很多种形式，藏身于网页中的许多地方。此外，攻击者也不需要控制放置恶意网址的网站。例如他可以将这种地址藏在论坛，博客等任何用户生成内容的网站中。这意味着如果服务器端没有合适的防御措施的话，用户即使访问熟悉的可信网站也有受攻击的危险。
 
@@ -153,8 +153,6 @@ Referer 首部字段位于 HTTP 报文中，用于标识请求来源的地址。
 
 因为 CSRF 攻击是在用户无意识的情况下发生的，所以要求用户输入验证码可以让用户知道自己正在做的操作。
 
-也可以要求用户输入验证码来进行校验。
-
 # 三、SQL 注入攻击
 
 ## 概念
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index 151ffc0f..e7ef804b 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -383,13 +383,13 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 ## Record Locks
 
-锁定的对象是记录的索引，而不是记录本身。
+锁定一个记录上的索引，而不是记录本身。
 
 如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
 
 ## Gap Locks
 
-锁定一个范围内的索引，例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
+锁定索引之间的间隙，但是不包含索引本身。例如当一个事务执行以下语句，其它事务就不能在 t.c 中插入 15。
 
 ```sql
 SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
@@ -397,28 +397,14 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 ## Next-Key Locks
 
-它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录，也锁定范围内的索引。在 user 中有以下记录：
+它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定范围内的索引。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
 
 ```sql
-|   id | last_name   | first_name   |   age |
-|------|-------------|--------------|-------|
-|    4 | stark       | tony         |    21 |
-|    1 | tom         | hiddleston   |    30 |
-|    3 | morgan      | freeman      |    40 |
-|    5 | jeff        | dean         |    50 |
-|    2 | donald      | trump        |    80 |
-+------|-------------|--------------|-------+
-```
-
-那么就需要锁定以下范围：
-
-```sql
-(-∞, 21]
-(21, 30]
-(30, 40]
-(40, 50]
-(50, 80]
-(80, ∞)
+(negative infinity, 10]
+(10, 11]
+(11, 13]
+(13, 20]
+(20, positive infinity)
 ```
 
 # 七、关系数据库设计理论
diff --git a/notes/算法.md b/notes/算法.md
index fb878742..cb5d4dbe 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -1623,10 +1623,10 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ## 2-3 查找树
 
-<div align="center"> <img src="../pics//ff396233-1bb1-4e74-8bc2-d7c90146f5dd.png" width="250"/> </div><br>
-
 2-3 查找树引入了 2- 节点和 3- 节点，目的是为了让树平衡。一颗完美平衡的 2-3 查找树的所有空链接到根节点的距离应该是相同的。
 
+<div align="center"> <img src="../pics//ff396233-1bb1-4e74-8bc2-d7c90146f5dd.png" width="250"/> </div><br>
+
 ### 1. 插入操作
 
 插入操作和 BST 的插入操作有很大区别，BST 的插入操作是先进行一次未命中的查找，然后再将节点插入到对应的空链接上。但是 2-3 查找树如果也这么做的话，那么就会破坏了平衡性。它是将新节点插入到叶子节点上。
@@ -2042,24 +2042,26 @@ public class SparseVector {
 
 ## 汉诺塔
 
+<div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
+
 这是一个经典的递归问题，分为三步求解：
 
-1. 将 n-1 个圆盘从 from -> buffer
-2. 将 1 个圆盘从 from -> to
-3. 将 n-1 个圆盘从 buffer -> to
+- 将 n-1 个圆盘从 from -> buffer
+
+<div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
+
+- 将 1 个圆盘从 from -> to
+
+<div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
+
+- 将 n-1 个圆盘从 buffer -> to
+
+<div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
 
 如果只有一个圆盘，那么只需要进行一次移动操作。
 
 从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然 a<sub>n</sub> = 2<sup>n</sup> - 1，n 个圆盘需要移动 2<sup>n</sup> - 1 次。
 
-<div align="center"> <img src="../pics//54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
-
-<div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
-
 ```java
 public class Hanoi {
     public static void move(int n, String from, String buffer, String to) {
@@ -2105,7 +2107,7 @@ from H1 to H3
 
 生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到根节点，根节点代表的字符的编码就是这个路径编码。
 
-<div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png"/> </div><br>
+<div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png" width="400"/> </div><br>
 
 ```java
 public class Huffman {
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index 40f0ca9d..805764a9 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -79,7 +79,7 @@
 实现可扩展主要有两种方式：
 
 - 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
-- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以用过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
+- 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以通过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
 
 # 四、可用性
 
diff --git a/notes/设计模式.md b/notes/设计模式.md
index ae44213e..b8f4dc9b 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -98,7 +98,9 @@ public static synchronized Singleton getUniqueInstance() {
 
 （三）饿汉式-线程安全
 
-线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。但是直接实例化的方式也丢失了延迟实例化带来的节约资源的优势。
+线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。
+
+但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
 
 ```java
 private static Singleton uniqueInstance = new Singleton();
@@ -106,7 +108,7 @@ private static Singleton uniqueInstance = new Singleton();
 
 （四）双重校验锁-线程安全
 
-uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行。也就是说，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
+uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
 
 双重校验锁先判断 uniqueInstance 是否已经被实例化，如果没有被实例化，那么才对实例化语句进行加锁。
 
@@ -131,7 +133,7 @@ public class Singleton {
 }
 ```
 
-考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程同时执行 if 语句，那么两个线程就会同时进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，也就是说会进行两次实例化，从而产生了两个实例。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
+考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程同时执行 if 语句，那么两个线程就会同时进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，那么就会进行两次实例化，从而产生了两个实例。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
 
 ```java
 if (uniqueInstance == null) {
@@ -157,7 +159,7 @@ uniqueInstance 采用 volatile 关键字修饰也是很有必要的。`uniqueIns
 
 这种方式不仅具有延迟初始化的好处，而且由虚拟机提供了对线程安全的支持。
 
-```source-java
+```java
 public class Singleton {
 
     private Singleton() {
@@ -299,7 +301,7 @@ public class Client {
 
 ### 意图
 
-定义了一个创建对象的接口，但由子类决定要实例化哪个类。工厂方法把实例化推迟到子类。
+定义了一个创建对象的接口，但由子类决定要实例化哪个类。工厂方法把实例化操作推迟到子类。
 
 ### 类图
 
@@ -1851,11 +1853,7 @@ No gumball dispensed
 
 ### 与状态模式的比较
 
-状态模式的类图和策略模式类似，并且都是能够动态改变对象的行为。
-
-但是状态模式是通过状态转移来改变 Context 所组合的 State 对象，而策略模式是通过 Context 本身的决策来改变组合的 Strategy 对象。
-
-所谓的状态转移，是指 Context 在运行过程中由于一些条件发生改变而使得 State 对象发生改变，注意必须要是在运行过程中。
+状态模式的类图和策略模式类似，并且都是能够动态改变对象的行为。但是状态模式是通过状态转移来改变 Context 所组合的 State 对象，而策略模式是通过 Context 本身的决策来改变组合的 Strategy 对象。所谓的状态转移，是指 Context 在运行过程中由于一些条件发生改变而使得 State 对象发生改变，注意必须要是在运行过程中。
 
 状态模式主要是用来解决状态转移的问题，当状态发生转移了，那么 Context 对象就会改变它的行为；而策略模式主要是用来封装一组可以互相替代的算法族，并且可以根据需要动态地去替换 Context 使用的算法。
 
@@ -1969,7 +1967,7 @@ public abstract class CaffeineBeverage {
 ```
 
 ```java
-public class Coffee extends CaffeineBeverage{
+public class Coffee extends CaffeineBeverage {
     @Override
     void brew() {
         System.out.println("Coffee.brew");
@@ -1983,7 +1981,7 @@ public class Coffee extends CaffeineBeverage{
 ```
 
 ```java
-public class Tea extends CaffeineBeverage{
+public class Tea extends CaffeineBeverage {
     @Override
     void brew() {
         System.out.println("Tea.brew");
@@ -2238,7 +2236,7 @@ Number of items:     6
 
 ### 意图
 
-使用什么都不做的空对象来替代 NULL。
+使用什么都不做的空对象来代替 NULL。
 
 一个方法返回 NULL，意味着方法的调用端需要去检查返回值是否是 NULL，这么做会导致非常多的冗余的检查代码。并且如果某一个调用端忘记了做这个检查返回值，而直接使用返回的对象，那么就有可能抛出空指针异常。
 
@@ -2393,7 +2391,7 @@ public abstract class TV {
 ```
 
 ```java
-public class Sony extends TV{
+public class Sony extends TV {
     @Override
     public void on() {
         System.out.println("Sony.on()");
@@ -2412,7 +2410,7 @@ public class Sony extends TV{
 ```
 
 ```java
-public class RCA extends TV{
+public class RCA extends TV {
     @Override
     public void on() {
         System.out.println("RCA.on()");
@@ -2551,9 +2549,6 @@ public abstract class Component {
 ```
 
 ```java
-import java.util.ArrayList;
-import java.util.List;
-
 public class Composite extends Component {
 
     private List<Component> child;
@@ -2659,7 +2654,7 @@ Composite:root
 
 ### 类图
 
-装饰者（Decorator）和具体组件（ConcreteComponent）都继承自组件（Component），具体组件的方法实现不需要依赖于其它对象，而装饰者组合了一个组件，这样它可以装饰其它装饰者或者具体组件。所谓装饰，就是把这个装饰者套在被装饰上，从而动态扩展被装饰者的功能。装饰者的方法有一部分是自己的，这属于它的功能，然后调用被装饰者的方法实现，从而也保留了被装饰者的功能。可以看到，具体组件应当是装饰层次的最低层，因为只有具体组件的方法实现不需要依赖于其它对象。
+装饰者（Decorator）和具体组件（ConcreteComponent）都继承自组件（Component），具体组件的方法实现不需要依赖于其它对象，而装饰者组合了一个组件，这样它可以装饰其它装饰者或者具体组件。所谓装饰，就是把这个装饰者套在被装饰者之上，从而动态扩展被装饰者的功能。装饰者的方法有一部分是自己的，这属于它的功能，然后调用被装饰者的方法实现，从而也保留了被装饰者的功能。可以看到，具体组件应当是装饰层次的最低层，因为只有具体组件的方法实现不需要依赖于其它对象。
 
 <div align="center"> <img src="../pics//137c593d-0a9e-47b8-a9e6-b71f540b82dd.png"/> </div><br>
 
@@ -2770,7 +2765,7 @@ public class Client {
 
 ### 实现
 
-观看电影需要操作很多电器，使用外观模式可以实现一键看电影功能。
+观看电影需要操作很多电器，使用外观模式实现一键看电影功能。
 
 ```java
 public class SubSystem {
@@ -2811,7 +2806,7 @@ public class Client {
 
 ### 设计原则
 
-最少知识原则：只和你的密友谈话。也就是客户对象所需要交互的对象应当尽可能少。
+最少知识原则：只和你的密友谈话。也就是说客户对象所需要交互的对象应当尽可能少。
 
 ## 6. 享元（Flyweight）
 
@@ -2822,8 +2817,8 @@ public class Client {
 ### 类图
 
 - Flyweight：享元对象
-- IntrinsicState：内部状态，相同的项元对象共享
-- ExtrinsicState：外部状态
+- IntrinsicState：内部状态，享元对象共享内部状态
+- ExtrinsicState：外部状态，每个享元对象的外部状态不同
 
 <div align="center"> <img src="../pics//d52270b4-9097-4667-9f18-f405fc661c99.png"/> </div><br>
 
@@ -2854,8 +2849,6 @@ public class ConcreteFlyweight implements Flyweight {
 ```
 
 ```java
-import java.util.HashMap;
-
 public class FlyweightFactory {
 
     private HashMap<String, Flyweight> flyweights = new HashMap<>();
diff --git a/pics/VP4n3i8m34Ntd28NQ4_0KCJ2q044Oez.png b/pics/VP4n3i8m34Ntd28NQ4_0KCJ2q044Oez.png
new file mode 100644
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From 01a916cce7112ff95995413348e68a3baeb71e45 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 14:40:05 +0800
Subject: [PATCH 097/172] auto commit

---
 notes/缓存.md | 14 ++++++++------
 notes/集群.md | 22 ++++++++++++----------
 2 files changed, 20 insertions(+), 16 deletions(-)

diff --git a/notes/缓存.md b/notes/缓存.md
index 449598ff..9a9f2810 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -35,10 +35,10 @@
 
 # 二、LRU
 
-以下是一个基于 双向队列 + HashMap 的 LRU 算法实现，对算法的解释如下：
+以下是一个基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
 
 - 最基本的思路是当访问某个节点时，将其从原来的位置删除，并重新插入到链表头部，这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就删除链表尾部的节点。
-- 为了使删除操作时间复杂度为 O(1)，那么就不能采用遍历的方式找到某个节点。HashMap 存储这 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
+- 为了使删除操作时间复杂度为 O(1)，那么就不能采用遍历的方式找到某个节点。HashMap 存储着 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
 
 ```java
 public class LRU<K, V> implements Iterable<K> {
@@ -143,6 +143,10 @@ public class LRU<K, V> implements Iterable<K> {
 }
 ```
 
+源代码：
+
+- [CyC2018/Algorithm](https://github.com/CyC2018/Algorithm/tree/master/Caching)
+
 # 三、缓存位置
 
 ## 浏览器
@@ -165,9 +169,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。
 
-相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。
-
-不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
 
 ## 数据库缓存
 
@@ -263,7 +265,7 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
 
-数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真是节点好，从而使得数据分布也更加均匀。
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
 
 参考资料：
 
diff --git a/notes/集群.md b/notes/集群.md
index 717ae637..15657e43 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -1,6 +1,6 @@
 <!-- GFM-TOC -->
 * [一、负载均衡](#一负载均衡)
-    * [算法实现](#算法实现)
+    * [负载均衡算法](#负载均衡算法)
     * [转发实现](#转发实现)
 * [二、集群下的 Session 管理](#二集群下的-session-管理)
     * [Sticky Session](#sticky-session)
@@ -11,27 +11,27 @@
 
 # 一、负载均衡
 
-集群中的应用服务器通常被设计成无状态，用户可以请求任何一个节点（应用服务器）。
+集群中的应用服务器（节点）通常被设计成无状态，用户可以请求任何一个应用服务器。
 
 负载均衡器会根据集群中每个节点的负载情况，将用户请求转发到合适的节点上。
 
 负载均衡器可以用来实现高可用以及伸缩性：
 
-- 高可用：当某个节点故障时，负载均衡器不会将用户请求转发到该节点上，从而保证所有服务持续可用；
+- 高可用：当某个节点故障时，负载均衡器会将用户请求转发到另外的节点上，从而保证所有服务持续可用；
 - 伸缩性：可以很容易地添加移除节点。
 
 负载均衡运行过程包含两个部分：
 
 1. 根据负载均衡算法得到请求转发的节点；
-2. 将请求进行转发；
+2. 将请求进行转发。
 
-## 算法实现
+## 负载均衡算法
 
 ### 1. 轮询（Round Robin）
 
 轮询算法把每个请求轮流发送到每个服务器上。
 
-下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。最后，(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
+下图中，一共有 6 个客户端产生了 6 个请求，这 6 个请求按 (1, 2, 3, 4, 5, 6) 的顺序发送。(1, 3, 5) 的请求会被发送到服务器 1，(2, 4, 6) 的请求会被发送到服务器 2。
 
 <div align="center"> <img src="../pics//2766d04f-7dad-42e4-99d1-60682c9d5c61.jpg"/> </div><br>
 
@@ -51,7 +51,7 @@
 
 由于每个请求的连接时间不一样，使用轮询或者加权轮询算法的话，可能会让一台服务器当前连接数过大，而另一台服务器的连接过小，造成负载不均衡。
 
-例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开。该系统继续运行时，服务器 2 会承担过大的负载。
+例如下图中，(1, 3, 5) 请求会被发送到服务器 1，但是 (1, 3) 很快就断开连接，此时只有 (5) 请求连接服务器 1；(2, 4, 6) 请求被发送到服务器 2，只有 (2) 的连接断开，此时 (6, 4) 请求连接服务器 2。该系统继续运行时，服务器 2 会承担过大的负载。
 
 <div align="center"> <img src="../pics//3b0d1aa8-d0e0-46c2-8fd1-736bf08a11aa.jpg"/> </div><br>
 
@@ -100,7 +100,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 ### 2. DNS 域名解析
 
-在 DNS 解析域名的同时使用负载均衡算法计算服务器地址。
+在 DNS 解析域名的同时使用负载均衡算法计算服务器 IP 地址。
 
 优点：
 
@@ -151,7 +151,9 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 在链路层根据负载均衡算法计算源服务器的 MAC 地址，并修改请求数据包的目的 MAC 地址，并进行转发。
 
-通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，直接转发给客户端，避免了负载均衡服务器的成为瓶颈。这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
+通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，可以直接转发给客户端，避免了负载均衡服务器的成为瓶颈。
+
+这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
 
 这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
 
@@ -188,7 +190,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 ## Session Server
 
-使用一个单独的服务器存储 Session 数据，可以使用 MySQL，也使用 Redis 或者 Memcached 这种内存型数据库。
+使用一个单独的服务器存储 Session 数据，可以使用传统的 MySQL，也使用 Redis 或者 Memcached 这种内存型数据库。
 
 优点：
 

From 4168b1ccab7084aae8955e3756647e4a9f27dd6d Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 14:47:06 +0800
Subject: [PATCH 098/172] auto commit

---
 notes/消息队列.md                             |   4 ++--
 pics/685a692f-8f76-4cac-baac-b68e2df9a30f.jpg | Bin 0 -> 31744 bytes
 2 files changed, 2 insertions(+), 2 deletions(-)
 create mode 100644 pics/685a692f-8f76-4cac-baac-b68e2df9a30f.jpg

diff --git a/notes/消息队列.md b/notes/消息队列.md
index 74155802..c5687e8a 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -18,7 +18,7 @@
 
 消息生产者向消息队列中发送了一个消息之后，只能被一个消费者消费一次。
 
-<div align="center"> <img src="../pics//09b52bcb-88ba-4e36-8244-b375f16ad116.jpg"/> </div><br>
+<div align="center"> <img src="../pics//685a692f-8f76-4cac-baac-b68e2df9a30f.jpg"/> </div><br>
 
 ## 发布/订阅
 
@@ -72,7 +72,7 @@
 
 ## 接收端的可靠性
 
-接收端能够从消息中间件成功消费一次消息。
+接收端能够从消息队列成功消费一次消息。
 
 实现方法：
 
diff --git a/pics/685a692f-8f76-4cac-baac-b68e2df9a30f.jpg b/pics/685a692f-8f76-4cac-baac-b68e2df9a30f.jpg
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HcmV?d00001


From 4b569bd08c027fd4c0b15c48d320f9b54044b7c6 Mon Sep 17 00:00:00 2001
From: no one <1422537078@qq.com>
Date: Mon, 13 Aug 2018 16:45:09 +0800
Subject: [PATCH 099/172] =?UTF-8?q?Update=20Java=20=E5=AE=B9=E5=99=A8.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

更正错别字 LUR 为LRU
---
 notes/Java 容器.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index af118e4c..4ae57b25 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -1012,7 +1012,7 @@ protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
 以下是使用 LinkedHashMap 实现的一个 LRU 缓存：
 
 - 设定最大缓存空间 MAX_ENTRIES  为 3；
-- 使用 LinkedHashMap 的构造函数将 accessOrder 设置为 true，开启 LUR 顺序；
+- 使用 LinkedHashMap 的构造函数将 accessOrder 设置为 true，开启 LRU 顺序；
 - 覆盖 removeEldestEntry() 方法实现，在节点多于 MAX_ENTRIES 就会将最近最久未使用的数据移除。
 
 ```java

From 5438ab4b60bfab62efe4596fca20efc433054fe9 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 17:40:07 +0800
Subject: [PATCH 100/172] auto commit

---
 notes/Java 基础.md | 109 ++++++++++++++++++++++++++++-----------------
 1 file changed, 68 insertions(+), 41 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index bcb94acc..625fb35b 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -62,7 +62,10 @@ int y = x;         // 拆箱
 
 ## 缓存池
 
-new Integer(123) 与 Integer.valueOf(123) 的区别在于，new Integer(123) 每次都会新建一个对象，而 Integer.valueOf(123) 可能会使用缓存对象，因此多次使用 Integer.valueOf(123) 会取得同一个对象的引用。
+new Integer(123) 与 Integer.valueOf(123) 的区别在于：
+
+- new Integer(123) 每次都会新建一个对象
+- Integer.valueOf(123) 会使用缓存池中的对象，多次调用会取得同一个对象的引用。
 
 ```java
 Integer x = new Integer(123);
@@ -73,14 +76,6 @@ Integer k = Integer.valueOf(123);
 System.out.println(z == k);   // true
 ```
 
-编译器会在自动装箱过程调用 valueOf() 方法，因此多个 Integer 实例使用自动装箱来创建并且值相同，那么就会引用相同的对象。
-
-```java
-Integer m = 123;
-Integer n = 123;
-System.out.println(m == n); // true
-```
-
 valueOf() 方法的实现比较简单，就是先判断值是否在缓存池中，如果在的话就直接返回缓存池的内容。
 
 ```java
@@ -125,7 +120,15 @@ static {
 }
 ```
 
-Java 还将一些其它基本类型的值放在缓冲池中，包含以下这些：
+编译器会在自动装箱过程调用 valueOf() 方法，因此多个 Integer 实例使用自动装箱来创建并且值相同，那么就会引用相同的对象。
+
+```java
+Integer m = 123;
+Integer n = 123;
+System.out.println(m == n); // true
+```
+
+基本类型对应的缓冲池如下：
 
 - boolean values true and false
 - all byte values
@@ -133,7 +136,7 @@ Java 还将一些其它基本类型的值放在缓冲池中，包含以下这些
 - int values between -128 and 127
 - char in the range \u0000 to \u007F
 
-因此在使用这些基本类型对应的包装类型时，就可以直接使用缓冲池中的对象。
+在使用这些基本类型对应的包装类型时，就可以直接使用缓冲池中的对象。
 
 [StackOverflow : Differences between new Integer(123), Integer.valueOf(123) and just 123
 ](https://stackoverflow.com/questions/9030817/differences-between-new-integer123-integer-valueof123-and-just-123)
@@ -186,15 +189,15 @@ String 不可变性天生具备线程安全，可以在多个线程中安全地
 
 - String 不可变，因此是线程安全的
 - StringBuilder 不是线程安全的
-- StringBuffer 是线程安全的，内部使用 synchronized 来同步
+- StringBuffer 是线程安全的，内部使用 synchronized 进行同步
 
 [StackOverflow : String, StringBuffer, and StringBuilder](https://stackoverflow.com/questions/2971315/string-stringbuffer-and-stringbuilder)
 
 ## String.intern()
 
-使用 String.intern() 可以保证相同内容的字符串变量引用相同的内存对象。
+使用 String.intern() 可以保证相同内容的字符串变量引用同一的内存对象。
 
-下面示例中，s1 和 s2 采用 new String() 的方式新建了两个不同对象，而 s3 是通过 s1.intern() 方法取得一个对象引用，这个方法首先把 s1 引用的对象放到 String Pool（字符串常量池）中，然后返回这个对象引用。因此 s3 和 s1 引用的是同一个字符串常量池的对象。
+下面示例中，s1 和 s2 采用 new String() 的方式新建了两个不同对象，而 s3 是通过 s1.intern() 方法取得一个对象引用。intern() 首先把 s1 引用的对象放到 String Pool（字符串常量池）中，然后返回这个对象引用。因此 s3 和 s1 引用的是同一个字符串常量池的对象。
 
 ```java
 String s1 = new String("aaa");
@@ -214,7 +217,7 @@ System.out.println(s4 == s5);  // true
 
 在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到 Native Method 中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
 
-- [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning) 
+- [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning)
 - [深入解析 String#intern](https://tech.meituan.com/in_depth_understanding_string_intern.html)
 
 # 三、运算
@@ -223,7 +226,7 @@ System.out.println(s4 == s5);  // true
 
 Java 的参数是以值传递的形式传入方法中，而不是引用传递。
 
-以下代码中 Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。但是如果在方法中改变对象的字段值会改变原对象该字段值，因为改变的是同一个地址指向的内容。
+以下代码中 Dog dog 的 dog 是一个指针，存储的是对象的地址。在将一个参数传入一个方法时，本质上是将对象的地址以值的方式传递到形参中。因此在方法中改变指针引用的对象，那么这两个指针此时指向的是完全不同的对象，一方改变其所指向对象的内容对另一方没有影响。
 
 ```java
 public class Dog {
@@ -234,7 +237,11 @@ public class Dog {
     }
 
     String getName() {
-        return name;
+        return this.name;
+    }
+
+    void setName(String name) {
+        this.name = name;
     }
 
     String getObjectAddress() {
@@ -262,6 +269,22 @@ public class PassByValueExample {
 }
 ```
 
+但是如果在方法中改变对象的字段值会改变原对象该字段值，因为改变的是同一个地址指向的内容。
+
+```java
+class PassByValueExample {
+    public static void main(String[] args) {
+        Dog dog = new Dog("A");
+        func(dog);
+        System.out.println(dog.getName());          // B
+    }
+
+    private static void func(Dog dog) {
+        dog.setName("B");
+    }
+}
+```
+
 [StackOverflow: Is Java “pass-by-reference” or “pass-by-value”?](https://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value)
 
 ## float 与 double
@@ -317,7 +340,7 @@ switch (s) {
 }
 ```
 
-switch 不支持 long，是因为 switch 的设计初衷是为那些只需要对少数的几个值进行等值判断，如果值过于复杂，那么还是用 if 比较合适。
+switch 不支持 long，是因为 switch 的设计初衷是对那些只有少数的几个值进行等值判断，如果值过于复杂，那么还是用 if 比较合适。
 
 ```java
 // long x = 111;
@@ -341,33 +364,37 @@ Java 中有三个访问权限修饰符：private、protected 以及 public，如
 
 可以对类或类中的成员（字段以及方法）加上访问修饰符。
 
-- 成员可见表示其它类可以用这个类的实例对象访问到该成员；
 - 类可见表示其它类可以用这个类创建实例对象。
+- 成员可见表示其它类可以用这个类的实例对象访问到该成员；
 
 protected 用于修饰成员，表示在继承体系中成员对于子类可见，但是这个访问修饰符对于类没有意义。
 
 设计良好的模块会隐藏所有的实现细节，把它的 API 与它的实现清晰地隔离开来。模块之间只通过它们的 API 进行通信，一个模块不需要知道其他模块的内部工作情况，这个概念被称为信息隐藏或封装。因此访问权限应当尽可能地使每个类或者成员不被外界访问。
 
-如果子类的方法覆盖了父类的方法，那么子类中该方法的访问级别不允许低于父类的访问级别。这是为了确保可以使用父类实例的地方都可以使用子类实例，也就是确保满足里氏替换原则。
+如果子类的方法重写了父类的方法，那么子类中该方法的访问级别不允许低于父类的访问级别。这是为了确保可以使用父类实例的地方都可以使用子类实例，也就是确保满足里氏替换原则。
 
-字段决不能是公有的，因为这么做的话就失去了对这个字段修改行为的控制，客户端可以对其随意修改。可以使用公有的 getter 和 setter 方法来替换公有字段。
+字段决不能是公有的，因为这么做的话就失去了对这个字段修改行为的控制，客户端可以对其随意修改。例如下面的例子中，AccessExample 拥有 id 共有字段，如果在某个时刻，我们想要使用 int 去存储 id 字段，那么就需要去修改所有的客户端代码。
 
 ```java
 public class AccessExample {
-    public int x;
+    public String id;
 }
 ```
 
+可以使用公有的 getter 和 setter 方法来替换公有字段，这样的话就可以控制对字段的修改行为。
+
+
 ```java
 public class AccessExample {
-    private int x;
 
-    public int getX() {
-        return x;
+    private int id;
+
+    public String getId() {
+        return id + "";
     }
 
-    public void setX(int x) {
-        this.x = x;
+    public void setId(String id) {
+        this.id = Integer.valueOf(id);
     }
 }
 ```
@@ -387,7 +414,7 @@ public class AccessWithInnerClassExample {
     }
 
     public int getValue() {
-        return innerClass.x; // 直接访问
+        return innerClass.x;  // 直接访问
     }
 }
 ```
@@ -396,7 +423,7 @@ public class AccessWithInnerClassExample {
 
 **1. 抽象类** 
 
-抽象类和抽象方法都使用 abstract 进行声明。抽象类一般会包含抽象方法，抽象方法一定位于抽象类中。
+抽象类和抽象方法都使用 abstract 关键字进行声明。抽象类一般会包含抽象方法，抽象方法一定位于抽象类中。
 
 抽象类和普通类最大的区别是，抽象类不能被实例化，需要继承抽象类才能实例化其子类。
 
@@ -415,7 +442,7 @@ public abstract class AbstractClassExample {
 ```
 
 ```java
-public class AbstractExtendClassExample extends AbstractClassExample{
+public class AbstractExtendClassExample extends AbstractClassExample {
     @Override
     public void func1() {
         System.out.println("func1");
@@ -448,7 +475,7 @@ public interface InterfaceExample {
     }
 
     int x = 123;
-    // int y;                // Variable 'y' might not have been initialized
+    // int y;               // Variable 'y' might not have been initialized
     public int z = 0;       // Modifier 'public' is redundant for interface fields
     // private int k = 0;   // Modifier 'private' not allowed here
     // protected int l = 0; // Modifier 'protected' not allowed here
@@ -477,21 +504,21 @@ System.out.println(InterfaceExample.x);
 - 从设计层面上看，抽象类提供了一种 IS-A 关系，那么就必须满足里式替换原则，即子类对象必须能够替换掉所有父类对象。而接口更像是一种 LIKE-A 关系，它只是提供一种方法实现契约，并不要求接口和实现接口的类具有 IS-A 关系。
 - 从使用上来看，一个类可以实现多个接口，但是不能继承多个抽象类。
 - 接口的字段只能是 static 和 final 类型的，而抽象类的字段没有这种限制。
-- 接口的方法只能是 public 的，而抽象类的方法可以有多种访问权限。
+- 接口的成员只能是 public 的，而抽象类的成员可以有多种访问权限。
 
 **4. 使用选择** 
 
-使用抽象类：
-
-- 需要在几个相关的类中共享代码。
-- 需要能控制继承来的成员的访问权限，而不是都为 public。
-- 需要继承非静态（non-static）和非常量（non-final）字段。
-
 使用接口：
 
 - 需要让不相关的类都实现一个方法，例如不相关的类都可以实现 Compareable 接口中的 compareTo() 方法；
 - 需要使用多重继承。
 
+使用抽象类：
+
+- 需要在几个相关的类中共享代码。
+- 需要能控制继承来的成员的访问权限，而不是都为 public。
+- 需要继承非静态和非常量字段。
+
 在很多情况下，接口优先于抽象类，因为接口没有抽象类严格的类层次结构要求，可以灵活地为一个类添加行为。并且从 Java 8 开始，接口也可以有默认的方法实现，使得修改接口的成本也变的很低。
 
 - [深入理解 abstract class 和 interface](https://www.ibm.com/developerworks/cn/java/l-javainterface-abstract/)
@@ -499,8 +526,8 @@ System.out.println(InterfaceExample.x);
 
 ## super
 
-- 访问父类的构造函数：可以使用 super() 函数访问父类的构造函数，从而完成一些初始化的工作。
-- 访问父类的成员：如果子类覆盖了父类的中某个方法的实现，可以通过使用 super 关键字来引用父类的方法实现。
+- 访问父类的构造函数：可以使用 super() 函数访问父类的构造函数，从而委托父类完成一些初始化的工作。
+- 访问父类的成员：如果子类重写了父类的中某个方法的实现，可以通过使用 super 关键字来引用父类的方法实现。
 
 ```java
 public class SuperExample {
@@ -549,7 +576,7 @@ SuperExtendExample.func()
 
 ## 重写与重载
 
-- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法，子类的返回值类型要等于或者小于父类的返回值；
+- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法。子类的返回值类型要等于或者小于父类的返回值；
 
 - 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载。
 

From 7746f5d72f8fddfa9a7f6ab48aa64507ff845020 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 19:57:04 +0800
Subject: [PATCH 101/172] auto commit

---
 notes/Java 基础.md | 41 +++++++++++++++++++++++------------------
 1 file changed, 23 insertions(+), 18 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 625fb35b..adf899fa 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -650,7 +650,7 @@ x.equals(null); // false;
 **2. equals() 与 ==** 
 
 - 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
-- 对于引用类型，== 判断两个实例是否引用同一个对象，而 equals() 判断引用的对象是否等价，根据引用对象 equals() 方法的具体实现来进行比较。
+- 对于引用类型，== 判断两个变量是否引用同一个对象，而 equals() 判断引用的对象是否等价。
 
 ```java
 Integer x = new Integer(1);
@@ -663,7 +663,7 @@ System.out.println(x == y);      // false
 
 - 检查是否为同一个对象的引用，如果是直接返回 true；
 - 检查是否是同一个类型，如果不是，直接返回 false；
-- 将 Object 实例进行转型；
+- 将 Object 对象进行转型；
 - 判断每个关键域是否相等。
 
 ```java
@@ -694,11 +694,11 @@ public class EqualExample {
 
 ## hashCode()
 
-hasCode() 返回散列值，而 equals() 是用来判断两个实例是否等价。等价的两个实例散列值一定要相同，但是散列值相同的两个实例不一定等价。
+hasCode() 返回散列值，而 equals() 是用来判断两个对象是否等价。等价的两个对象散列值一定相同，但是散列值相同的两个对象不一定等价。
 
-在覆盖 equals() 方法时应当总是覆盖 hashCode() 方法，保证等价的两个实例散列值也相等。
+在覆盖 equals() 方法时应当总是覆盖 hashCode() 方法，保证等价的两个对象散列值也相等。
 
-下面的代码中，新建了两个等价的实例，并将它们添加到 HashSet 中。我们希望将这两个实例当成一样的，只在集合中添加一个实例，但是因为 EqualExample 没有实现 hasCode() 方法，因此这两个实例的散列值是不同的，最终导致集合添加了两个等价的实例。
+下面的代码中，新建了两个等价的对象，并将它们添加到 HashSet 中。我们希望将这两个对象当成一样的，只在集合中添加一个对象，但是因为 EqualExample 没有实现 hasCode() 方法，因此这两个对象的散列值是不同的，最终导致集合添加了两个等价的对象。
 
 ```java
 EqualExample e1 = new EqualExample(1, 1, 1);
@@ -710,7 +710,7 @@ set.add(e2);
 System.out.println(set.size());   // 2
 ```
 
-理想的散列函数应当具有均匀性，即不相等的实例应当均匀分布到所有可能的散列值上。这就要求了散列函数要把所有域的值都考虑进来，可以将每个域都当成 R 进制的某一位，然后组成一个 R 进制的整数。R 一般取 31，因为它是一个奇素数，如果是偶数的话，当出现乘法溢出，信息就会丢失，因为与 2 相乘相当于向左移一位。
+理想的散列函数应当具有均匀性，即不相等的对象应当均匀分布到所有可能的散列值上。这就要求了散列函数要把所有域的值都考虑进来，可以将每个域都当成 R 进制的某一位，然后组成一个 R 进制的整数。R 一般取 31，因为它是一个奇素数，如果是偶数的话，当出现乘法溢出，信息就会丢失，因为与 2 相乘相当于向左移一位。
 
 一个数与 31 相乘可以转换成移位和减法：`31*x == (x<<5)-x`，编译器会自动进行这个优化。
 
@@ -752,7 +752,7 @@ ToStringExample@4554617c
 
 **1. cloneable** 
 
-clone() 是 Object 的 protect 方法，它不是 public，一个类不显式去重写 clone()，其它类就不能直接去调用该类实例的 clone() 方法。
+clone() 是 Object 的 protected 方法，它不是 public，一个类不显式去重写 clone()，其它类就不能直接去调用该类实例的 clone() 方法。
 
 ```java
 public class CloneExample {
@@ -795,6 +795,8 @@ java.lang.CloneNotSupportedException: CloneExample
 
 以上抛出了 CloneNotSupportedException，这是因为 CloneExample 没有实现 Cloneable 接口。
 
+应该注意的是，clone() 方法并不是 Cloneable 接口的方法，而是 Object 的一个 protected 方法。Cloneable 接口只是规定，如果一个类没有实现 Cloneable 接口又调用了 clone() 方法，就会抛出 CloneNotSupportedException。
+
 ```java
 public class CloneExample implements Cloneable {
     private int a;
@@ -807,12 +809,9 @@ public class CloneExample implements Cloneable {
 }
 ```
 
-应该注意的是，clone() 方法并不是 Cloneable 接口的方法，而是 Object 的一个 protected 方法。Cloneable 接口只是规定，如果一个类没有实现 Cloneable 接口又调用了 clone() 方法，就会抛出 CloneNotSupportedException。
+**2. 浅拷贝** 
 
-**2. 深拷贝与浅拷贝** 
-
-- 浅拷贝：拷贝实例和原始实例的引用类型引用同一个对象；
-- 深拷贝：拷贝实例和原始实例的引用类型引用不同对象。
+拷贝对象和原始对象的引用类型引用同一个对象。
 
 ```java
 public class ShallowCloneExample implements Cloneable {
@@ -852,6 +851,10 @@ e1.set(2, 222);
 System.out.println(e2.get(2)); // 222
 ```
 
+**3. 深拷贝** 
+
+拷贝对象和原始对象的引用类型引用不同对象。
+
 ```java
 public class DeepCloneExample implements Cloneable {
     private int[] arr;
@@ -895,6 +898,8 @@ e1.set(2, 222);
 System.out.println(e2.get(2)); // 2
 ```
 
+**4. clone() 的替代方案** 
+
 使用 clone() 方法来拷贝一个对象即复杂又有风险，它会抛出异常，并且还需要类型转换。Effective Java 书上讲到，最好不要去使用 clone()，可以使用拷贝构造函数或者拷贝工厂来拷贝一个对象。
 
 ```java
@@ -964,7 +969,7 @@ private 方法隐式地被指定为 final，如果在子类中定义的方法和
 
 **1. 静态变量** 
 
-- 静态变量：类所有的实例都共享静态变量，可以直接通过类名来访问它；静态变量在内存中只存在一份。
+- 静态变量：又称为类变量，也就是说这个变量属于类的，类所有的实例都共享静态变量，可以直接通过类名来访问它；静态变量在内存中只存在一份。
 - 实例变量：每创建一个实例就会产生一个实例变量，它与该实例同生共死。
 
 ```java
@@ -983,7 +988,7 @@ public class A {
 
 **2. 静态方法** 
 
-静态方法在类加载的时候就存在了，它不依赖于任何实例，所以静态方法必须有实现，也就是说它不能是抽象方法（abstract）。
+静态方法在类加载的时候就存在了，它不依赖于任何实例。所以静态方法必须有实现，也就是说它不能是抽象方法（abstract）。
 
 ```java
 public abstract class A {
@@ -1023,7 +1028,6 @@ public class A {
         A a2 = new A();
     }
 }
-
 ```
 
 ```html
@@ -1055,12 +1059,12 @@ public class OuterClass {
 
 **5. 静态导包** 
 
+在使用静态变量和方法时不用再指明 ClassName，从而简化代码，但可读性大大降低。
+
 ```java
 import static com.xxx.ClassName.*
 ```
 
-在使用静态变量和方法时不用再指明 ClassName，从而简化代码，但可读性大大降低。
-
 **6. 初始化顺序** 
 
 静态变量和静态语句块优先于实例变量和普通语句块，静态变量和静态语句块的初始化顺序取决于它们在代码中的顺序。
@@ -1102,11 +1106,12 @@ public InitialOrderTest() {
 - 子类（实例变量、普通语句块）
 - 子类（构造函数）
 
+
 # 七、反射
 
 每个类都有一个  **Class**  对象，包含了与类有关的信息。当编译一个新类时，会产生一个同名的 .class 文件，该文件内容保存着 Class 对象。
 
-类加载相当于 Class 对象的加载。类在第一次使用时才动态加载到 JVM 中，可以使用 Class.forName("com.mysql.jdbc.Driver") 这种方式来控制类的加载，该方法会返回一个 Class 对象。
+类加载相当于 Class 对象的加载。类在第一次使用时才动态加载到 JVM 中，可以使用 `Class.forName("com.mysql.jdbc.Driver")` 这种方式来控制类的加载，该方法会返回一个 Class 对象。
 
 反射可以提供运行时的类信息，并且这个类可以在运行时才加载进来，甚至在编译时期该类的 .class 不存在也可以加载进来。
 

From ddb9e86224f11781f3befca6eb30aa5ef276abb5 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 20:44:07 +0800
Subject: [PATCH 102/172] auto commit

---
 notes/Java 容器.md | 48 +++++++++++++++++++++++++---------------------
 1 file changed, 26 insertions(+), 22 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 4ae57b25..9959e2ca 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -21,7 +21,7 @@
 
 # 一、概览
 
-容器主要包括 Collection 和 Map 两种，Collection 又包含了 List、Set 以及 Queue。
+容器主要包括 Collection 和 Map 两种，Collection 存储着对象的集合，而 Map 存储着键值对（两个对象）的映射表。
 
 ## Collection
 
@@ -29,9 +29,9 @@
 
 ### 1. Set
 
-- HashSet：基于哈希表实现，支持快速查找。但不支持有序性操作，例如根据一个范围查找元素的操作。并且失去了元素的插入顺序信息，也就是说使用 Iterator 遍历 HashSet 得到的结果是不确定的。
+- TreeSet：基于红黑树实现，支持有序性操作，例如根据一个范围查找元素的操作。但是查找效率不如 HashSet，HashSet 查找的时间复杂度为 O(1)，TreeSet 则为 O(logN)。
 
-- TreeSet：基于红黑树实现，支持有序性操作，但是查找效率不如 HashSet，HashSet 查找时间复杂度为 O(1)，TreeSet 则为 O(logN)。
+- HashSet：基于哈希表实现，支持快速查找，但不支持有序性操作。并且失去了元素的插入顺序信息，也就是说使用 Iterator 遍历 HashSet 得到的结果是不确定的。
 
 - LinkedHashSet：具有 HashSet 的查找效率，且内部使用双向链表维护元素的插入顺序。
 
@@ -53,13 +53,14 @@
 
 <div align="center"> <img src="../pics//SoWkIImgAStDuUBAp2j9BKfBJ4vLy4q.png"/> </div><br>
 
-- HashMap：基于哈希表实现；
+- TreeMap：基于红黑树实现。
+
+- HashMap：基于哈希表实现。
 
 - HashTable：和 HashMap 类似，但它是线程安全的，这意味着同一时刻多个线程可以同时写入 HashTable 并且不会导致数据不一致。它是遗留类，不应该去使用它。现在可以使用 ConcurrentHashMap 来支持线程安全，并且 ConcurrentHashMap 的效率会更高，因为 ConcurrentHashMap 引入了分段锁。
 
 - LinkedHashMap：使用双向链表来维护元素的顺序，顺序为插入顺序或者最近最少使用（LRU）顺序。
 
-- TreeMap：基于红黑树实现。
 
 # 二、容器中的设计模式
 
@@ -296,14 +297,14 @@ public synchronized E get(int index) {
 }
 ```
 
-### 2. 与 ArrayList 的区别
+### 2. 与 ArrayList 的比较
 
 - Vector 是同步的，因此开销就比 ArrayList 要大，访问速度更慢。最好使用 ArrayList 而不是 Vector，因为同步操作完全可以由程序员自己来控制；
 - Vector 每次扩容请求其大小的 2 倍空间，而 ArrayList 是 1.5 倍。
 
 ### 3. 替代方案
 
-为了获得线程安全的 ArrayList，可以使用 `Collections.synchronizedList();` 得到一个线程安全的 ArrayList。
+可以使用 `Collections.synchronizedList();` 得到一个线程安全的 ArrayList。
 
 ```java
 List<String> list = new ArrayList<>();
@@ -322,7 +323,7 @@ List<String> list = new CopyOnWriteArrayList<>();
 
 写操作在一个复制的数组上进行，读操作还是在原始数组中进行，读写分离，互不影响。
 
-写操作需要加锁，防止同时并发写入时导致的写入数据丢失。
+写操作需要加锁，防止并发写入时导致写入数据丢失。
 
 写操作结束之后需要把原始数组指向新的复制数组。
 
@@ -386,9 +387,9 @@ transient Node<E> first;
 transient Node<E> last;
 ```
 
-<div align="center"> <img src="../pics//49495c95-52e5-4c9a-b27b-92cf235ff5ec.png"/> </div><br>
+<div align="center"> <img src="../pics//49495c95-52e5-4c9a-b27b-92cf235ff5ec.png" width="500"/> </div><br>
 
-### 2. ArrayList 与 LinkedList
+### 2. 与 ArrayList 的比较
 
 - ArrayList 基于动态数组实现，LinkedList 基于双向链表实现；
 - ArrayList 支持随机访问，LinkedList 不支持；
@@ -406,7 +407,7 @@ transient Node<E> last;
 transient Entry[] table;
 ```
 
-其中，Entry 就是存储数据的键值对，它包含了四个字段。从 next 字段我们可以看出 Entry 是一个链表，即数组中的每个位置被当成一个桶，一个桶存放一个链表，链表中存放哈希值相同的 Entry。也就是说，HashMap 使用拉链法来解决冲突。
+Entry 存储着键值对。它包含了四个字段，从 next 字段我们可以看出 Entry 是一个链表。即数组中的每个位置被当成一个桶，一个桶存放一个链表。HashMap 使用拉链法来解决冲突，同一个链表中存放哈希值相同的 Entry。
 
 <div align="center"> <img src="../pics//8fe838e3-ef77-4f63-bf45-417b6bc5c6bb.png" width="600"/> </div><br>
 
@@ -634,8 +635,8 @@ y&(x-1) : 00000010
 这个性质和 y 对 x 取模效果是一样的：
 
 ```
-x   : 00010000
 y   : 10110010
+x   : 00010000
 y%x : 00000010
 ```
 
@@ -693,7 +694,7 @@ void addEntry(int hash, K key, V value, int bucketIndex) {
 }
 ```
 
-扩容使用 resize() 实现，需要注意的是，扩容操作同样需要把旧 table 的所有键值对重新插入新的 table 中，因此这一步是很费时的。
+扩容使用 resize() 实现，需要注意的是，扩容操作同样需要把 oldTable 的所有键值对重新插入 newTable 中，因此这一步是很费时的。
 
 ```java
 void resize(int newCapacity) {
@@ -739,7 +740,10 @@ capacity     : 00010000
 new capacity : 00100000
 ```
 
-对于一个 Key，它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；如果为 1，那么得到的结果为原来的结果 +16。
+对于一个 Key，
+
+- 它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；
+- 如果为 1，那么得到的结果为原来的结果 +16。
 
 ### 7. 扩容-计算数组容量
 
@@ -778,7 +782,7 @@ static final int tableSizeFor(int cap) {
 
 从 JDK 1.8 开始，一个桶存储的链表长度大于 8 时会将链表转换为红黑树。
 
-### 9. HashMap 与 HashTable
+### 9. 与 HashTable 的比较
 
 - HashTable 使用 synchronized 来进行同步。
 - HashMap 可以插入键为 null 的 Entry。
@@ -939,7 +943,7 @@ transient LinkedHashMap.Entry<K,V> head;
 transient LinkedHashMap.Entry<K,V> tail;
 ```
 
-accessOrder 决定了顺序，默认为 false，此时使用的是插入顺序。
+accessOrder 决定了顺序，默认为 false，此时维护的是插入顺序。
 
 ```java
 final boolean accessOrder;
@@ -954,7 +958,7 @@ void afterNodeInsertion(boolean evict) { }
 
 ### afterNodeAccess()
 
-当一个节点被访问时，如果 accessOrder 为 true，则会将 该节点移到链表尾部。也就是说指定为 LRU 顺序之后，在每次访问一个节点时，会将这个节点移到链表尾部，保证链表尾部是最近访问的节点，那么链表首部就是最近最久未使用的节点。
+当一个节点被访问时，如果 accessOrder 为 true，则会将该节点移到链表尾部。也就是说指定为 LRU 顺序之后，在每次访问一个节点时，会将这个节点移到链表尾部，保证链表尾部是最近访问的节点，那么链表首部就是最近最久未使用的节点。
 
 ```java
 void afterNodeAccess(Node<K,V> e) { // move node to last
@@ -1003,8 +1007,8 @@ removeEldestEntry() 默认为 false，如果需要让它为 true，需要继承
 
 ```java
 protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
-        return false;
-    }
+    return false;
+}
 ```
 
 ### LRU 缓存
@@ -1059,14 +1063,14 @@ private static class Entry<K,V> extends WeakReference<Object> implements Map.Ent
 
 ### ConcurrentCache
 
-Tomcat 中的 ConcurrentCache 就使用了 WeakHashMap 来实现缓存功能。
+Tomcat 中的 ConcurrentCache 使用了 WeakHashMap 来实现缓存功能。
 
 ConcurrentCache 采取的是分代缓存：
 
 - 经常使用的对象放入 eden 中，eden 使用 ConcurrentHashMap 实现，不用担心会被回收（伊甸园）；
 - 不常用的对象放入 longterm，longterm 使用 WeakHashMap 实现，这些老对象会被垃圾收集器回收。
-- 当调用  get() 方法时，会先从 eden 区获取，如果没有找到的话再到 longterm 获取，当从 longterm 获取到就把对象放入 eden 中，保证频繁被访问的节点不容易被回收。
-- 当调用 put() 方法时，如果缓存当前容量大小超过了 size，那么就将 eden 中的所有对象都放入 longterm 中，利用虚拟机回收掉一部分不经常使用的对象。
+- 当调用  get() 方法时，会先从 eden 区获取，如果没有找到的话再到 longterm 获取，当从 longterm 获取到就把对象放入 eden 中，从而保证经常被访问的节点不容易被回收。
+- 当调用 put() 方法时，如果 eden 的大小超过了 size，那么就将 eden 中的所有对象都放入 longterm 中，利用虚拟机回收掉一部分不经常使用的对象。
 
 ```java
 public final class ConcurrentCache<K, V> {

From 4a8835bf2e8f774285e6d0cc181bc65685301852 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 22:03:33 +0800
Subject: [PATCH 103/172] auto commit

---
 notes/Java 虚拟机.md | 406 +++++++++++++++++++++----------------------
 1 file changed, 203 insertions(+), 203 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 136ca5ad..891211da 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -8,15 +8,22 @@
     * [运行时常量池](#运行时常量池)
     * [直接内存](#直接内存)
 * [二、垃圾收集](#二垃圾收集)
-    * [判断一个对象是否存活](#判断一个对象是否存活)
+    * [判断一个对象是否可被回收](#判断一个对象是否可被回收)
+    * [引用类型](#引用类型)
     * [垃圾收集算法](#垃圾收集算法)
     * [垃圾收集器](#垃圾收集器)
-    * [内存分配与回收策略](#内存分配与回收策略)
-* [三、类加载机制](#三类加载机制)
+* [三、内存分配与回收策略](#三内存分配与回收策略)
+    * [Minor GC 和 Full GC](#minor-gc-和-full-gc)
+    * [内存分配策略](#内存分配策略)
+    * [Full GC 的触发条件](#full-gc-的触发条件)
+* [四、类加载机制](#四类加载机制)
     * [类的生命周期](#类的生命周期)
-    * [类初始化时机](#类初始化时机)
     * [类加载过程](#类加载过程)
-    * [类加载器](#类加载器)
+    * [类初始化时机](#类初始化时机)
+    * [类与类加载器](#类与类加载器)
+    * [类加载器分类](#类加载器分类)
+    * [双亲委派模型](#双亲委派模型)
+    * [自定义类加载器实现](#自定义类加载器实现)
 * [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
@@ -48,19 +55,17 @@ java -Xss=512M HackTheJava
 
 ## 本地方法栈
 
-本地方法不是用 Java 实现，对待这些方法需要特别处理。
+本地方法一般是用其它语言（C、C++ 或汇编语言等）编写的，并且被编译为基于本机硬件和操作系统的程序，对待这些方法需要特别处理。
 
-与 Java 虚拟机栈类似，它们之间的区别只不过是本地方法栈为本地方法服务。
-
-本地方法一般是用其它语言（C、C++ 或汇编语言等）编写的, 并且被编译为基于本机硬件和操作系统的程序。
+本地方法栈与 Java 虚拟机栈类似，它们之间的区别只不过是本地方法栈为本地方法服务。
 
 <div align="center"> <img src="../pics//JNI-Java-Native-Interface.jpg" width="350"/> </div><br>
 
 ## 堆
 
-所有对象实例都在这里分配内存，是垃圾收集的主要区域（"GC 堆"）。
+所有对象都在这里分配内存，是垃圾收集的主要区域（"GC 堆"）。
 
-现代的垃圾收集器基本都是采用分代收集算法，针对不同的对象采取不同的垃圾回收算法，可以将堆分成两块：
+现代的垃圾收集器基本都是采用分代收集算法，针对不同类型的对象采取不同的垃圾回收算法，可以将堆分成两块：
 
 - 新生代（Young Generation）
 - 老年代（Old Generation）
@@ -83,9 +88,7 @@ java -Xms=1M -Xmx=2M HackTheJava
 
 用于存放已被加载的类信息、常量、静态变量、即时编译器编译后的代码等数据。
 
-和堆一样不需要连续的内存，并且可以动态扩展。
-
-动态扩展失败一样会抛出 OutOfMemoryError 异常。
+和堆一样不需要连续的内存，并且可以动态扩展，动态扩展失败一样会抛出 OutOfMemoryError 异常。
 
 对这块区域进行垃圾回收的主要目标是对常量池的回收和对类的卸载，但是一般比较难实现。
 
@@ -107,15 +110,15 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 # 二、垃圾收集
 
-垃圾回收主要是针对堆和方法区进行。
+垃圾收集主要是针对堆和方法区进行。
 
 程序计数器、虚拟机栈和本地方法栈这三个区域属于线程私有的，只存在于线程的生命周期内，线程结束之后也会消失，因此不需要对这三个区域进行垃圾回收。
 
-## 判断一个对象是否存活
+## 判断一个对象是否可被回收
 
 ### 1. 引用计数算法
 
-给对象添加一个引用计数器，当对象增加一个引用时计数器加 1，引用失效时计数器减 1。引用计数不为 0 的对象仍然存活。
+给对象添加一个引用计数器，当对象增加一个引用时计数器加 1，引用失效时计数器减 1。引用计数为 0 的对象可被回收。
 
 两个对象出现循环引用的情况下，此时引用计数器永远不为 0，导致无法对它们进行回收。
 
@@ -123,6 +126,7 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ```java
 public class ReferenceCountingGC {
+
     public Object instance = null;
 
     public static void main(String[] args) {
@@ -147,61 +151,7 @@ Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC
 - 方法区中类静态属性引用的对象
 - 方法区中的常量引用的对象
 
-### 3. 引用类型
-
-无论是通过引用计算算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
-
-Java 具有四种强度不同的引用类型。
-
-**（一）强引用** 
-
-被强引用关联的对象不会被回收。
-
-使用 new 一个新对象的方式来创建强引用。
-
-```java
-Object obj = new Object();
-```
-
-**（二）软引用** 
-
-被软引用关联的对象只有在内存不够的情况下才会被回收。
-
-使用 SoftReference 类来创建软引用。
-
-```java
-Object obj = new Object();
-SoftReference<Object> sf = new SoftReference<Object>(obj);
-obj = null;  // 使对象只被软引用关联
-```
-
-**（三）弱引用** 
-
-被弱引用关联的对象一定会被回收，也就是说它只能存活到下一次垃圾收集。
-
-使用 WeakReference 类来实现弱引用。
-
-```java
-Object obj = new Object();
-WeakReference<Object> wf = new WeakReference<Object>(obj);
-obj = null;
-```
-
-**（四）虚引用** 
-
-又称为幽灵引用或者幻影引用。一个对象是否有虚引用的存在，完全不会对其生存时间构成影响，也无法通过虚引用取得一个对象实例。
-
-为一个对象设置虚引用关联的唯一目的就是能在这个对象被回收时收到一个系统通知。
-
-使用 PhantomReference 来实现虚引用。
-
-```java
-Object obj = new Object();
-PhantomReference<Object> pf = new PhantomReference<Object>(obj);
-obj = null;
-```
-
-### 4. 方法区的回收
+### 3. 方法区的回收
 
 因为方法区主要存放永久代对象，而永久代对象的回收率比新生代低很多，因此在方法区上进行回收性价比不高。
 
@@ -217,12 +167,66 @@ obj = null;
 
 可以通过 -Xnoclassgc 参数来控制是否对类进行卸载。
 
-### 5. finalize()
+### 4. finalize()
 
 finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。但是 try-finally 等方式可以做的更好，并且该方法运行代价高昂，不确定性大，无法保证各个对象的调用顺序，因此最好不要使用。
 
 当一个对象可被回收时，如果需要执行该对象的 finalize() 方法，那么就有可能通过在该方法中让对象重新被引用，从而实现自救。自救只能进行一次，如果回收的对象之前调用了 finalize() 方法自救，后面回收时不会调用 finalize() 方法。
 
+## 引用类型
+
+无论是通过引用计算算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
+
+Java 具有四种强度不同的引用类型。
+
+### 1. 强引用
+
+被强引用关联的对象不会被回收。
+
+使用 new 一个新对象的方式来创建强引用。
+
+```java
+Object obj = new Object();
+```
+
+### 2. 软引用
+
+被软引用关联的对象只有在内存不够的情况下才会被回收。
+
+使用 SoftReference 类来创建软引用。
+
+```java
+Object obj = new Object();
+SoftReference<Object> sf = new SoftReference<Object>(obj);
+obj = null;  // 使对象只被软引用关联
+```
+
+### 3. 弱引用
+
+被弱引用关联的对象一定会被回收，也就是说它只能存活到下一次垃圾回收发生之前。
+
+使用 WeakReference 类来实现弱引用。
+
+```java
+Object obj = new Object();
+WeakReference<Object> wf = new WeakReference<Object>(obj);
+obj = null;
+```
+
+### 4. 虚引用
+
+又称为幽灵引用或者幻影引用。一个对象是否有虚引用的存在，完全不会对其生存时间构成影响，也无法通过虚引用取得一个对象。
+
+为一个对象设置虚引用关联的唯一目的就是能在这个对象被回收时收到一个系统通知。
+
+使用 PhantomReference 来实现虚引用。
+
+```java
+Object obj = new Object();
+PhantomReference<Object> pf = new PhantomReference<Object>(obj);
+obj = null;
+```
+
 ## 垃圾收集算法
 
 ### 1. 标记 - 清除
@@ -376,20 +380,21 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 
 更详细内容请参考：[Getting Started with the G1 Garbage Collector](http://www.oracle.com/webfolder/technetwork/tutorials/obe/java/G1GettingStarted/index.html)
 
-## 内存分配与回收策略
+# 三、内存分配与回收策略
 
-### 1. Minor GC 和 Full GC
+## Minor GC 和 Full GC
 
 - Minor GC：发生在新生代上，因为新生代对象存活时间很短，因此 Minor GC 会频繁执行，执行的速度一般也会比较快。
+
 - Full GC：发生在老年代上，老年代对象其存活时间长，因此 Full GC 很少执行，执行速度会比 Minor GC 慢很多。
 
-### 2. 内存分配策略
+## 内存分配策略
 
-（一）对象优先在 Eden 分配
+### 1. 对象优先在 Eden 分配
 
 大多数情况下，对象在新生代 Eden 区分配，当 Eden 区空间不够时，发起 Minor GC。
 
-（二）大对象直接进入老年代
+### 2. 大对象直接进入老年代
 
 大对象是指需要连续内存空间的对象，最典型的大对象是那种很长的字符串以及数组。
 
@@ -397,41 +402,41 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 
 -XX:PretenureSizeThreshold，大于此值的对象直接在老年代分配，避免在 Eden 区和 Survivor 区之间的大量内存复制。
 
-（三）长期存活的对象进入老年代
+### 3. 长期存活的对象进入老年代
 
 为对象定义年龄计数器，对象在 Eden 出生并经过 Minor GC 依然存活，将移动到 Survivor 中，年龄就增加 1 岁，增加到一定年龄则移动到老年代中。
 
 -XX:MaxTenuringThreshold 用来定义年龄的阈值。
 
-（四）动态对象年龄判定
+### 4. 动态对象年龄判定
 
 虚拟机并不是永远地要求对象的年龄必须达到 MaxTenuringThreshold 才能晋升老年代，如果在 Survivor 中相同年龄所有对象大小的总和大于 Survivor 空间的一半，则年龄大于或等于该年龄的对象可以直接进入老年代，无需等到 MaxTenuringThreshold 中要求的年龄。
 
-（五）空间分配担保
+### 5. 空间分配担保
 
 在发生 Minor GC 之前，虚拟机先检查老年代最大可用的连续空间是否大于新生代所有对象总空间，如果条件成立的话，那么 Minor GC 可以确认是安全的。
 
 如果不成立的话虚拟机会查看 HandlePromotionFailure 设置值是否允许担保失败，如果允许那么就会继续检查老年代最大可用的连续空间是否大于历次晋升到老年代对象的平均大小，如果大于，将尝试着进行一次 Minor GC；如果小于，或者 HandlePromotionFailure 设置不允许冒险，那么就要进行一次 Full GC。
 
-### 3. Full GC 的触发条件
+## Full GC 的触发条件
 
 对于 Minor GC，其触发条件非常简单，当 Eden 空间满时，就将触发一次 Minor GC。而 Full GC 则相对复杂，有以下条件：
 
-（一）调用 System.gc()
+### 1. 调用 System.gc()
 
 只是建议虚拟机执行 Full GC，但是虚拟机不一定真正去执行。不建议使用这种方式，而是让虚拟机管理内存。
 
-（二）老年代空间不足
+### 2. 老年代空间不足
 
 老年代空间不足的常见场景为前文所讲的大对象直接进入老年代、长期存活的对象进入老年代等。
 
 为了避免以上原因引起的 Full GC，应当尽量不要创建过大的对象以及数组。除此之外，可以通过 -Xmn 虚拟机参数调大新生代的大小，让对象尽量在新生代被回收掉，不进入老年代。还可以通过 -XX:MaxTenuringThreshold 调大对象进入老年代的年龄，让对象在新生代多存活一段时间。
 
-（三）空间分配担保失败
+### 3. 空间分配担保失败
 
 使用复制算法的 Minor GC 需要老年代的内存空间作担保，如果担保失败会执行一次 Full GC。具体内容请参考上面的第五小节。
 
-（四）JDK 1.7 及以前的永久代空间不足
+### 4. JDK 1.7 及以前的永久代空间不足
 
 在 JDK 1.7 及以前，HotSpot 虚拟机中的方法区是用永久代实现的，永久代中存放的为一些 Class 的信息、常量、静态变量等数据。
 
@@ -439,13 +444,13 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 
 为避免以上原因引起的 Full GC，可采用的方法为增大永久代空间或转为使用 CMS GC。
 
-（五）Concurrent Mode Failure
+### 5. Concurrent Mode Failure
 
 执行 CMS GC 的过程中同时有对象要放入老年代，而此时老年代空间不足（可能是 GC 过程中浮动垃圾过多导致暂时性的空间不足），便会报 Concurrent Mode Failure 错误，并触发 Full GC。
 
-# 三、类加载机制
+# 四、类加载机制
 
-类是在运行期间动态加载的。
+类是在运行期间第一次使用时动态加载的，而不是编译时期一次性加载。因为如果在编译时期一次性加载，那么会占用很多的内存。
 
 ## 类的生命周期
 
@@ -461,10 +466,108 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 - 使用（Using）
 - 卸载（Unloading）
 
+## 类加载过程
+
+包含了加载、验证、准备、解析和初始化这 5 个阶段。
+
+### 1. 加载
+
+加载是类加载的一个阶段，注意不要混淆。
+
+加载过程完成以下三件事：
+
+- 通过一个类的全限定名来获取定义此类的二进制字节流。
+- 将这个字节流所代表的静态存储结构转化为方法区的运行时存储结构。
+- 在内存中生成一个代表这个类的 Class 对象，作为方法区这个类的各种数据的访问入口。
+
+其中二进制字节流可以从以下方式中获取：
+
+- 从 ZIP 包读取，成为 JAR、EAR、WAR 格式的基础。
+- 从网络中获取，最典型的应用是 Applet。
+- 运行时计算生成，例如动态代理技术，在 java.lang.reflect.Proxy 使用 ProxyGenerator.generateProxyClass 的代理类的二进制字节流。
+- 由其他文件生成，例如由 JSP 文件生成对应的 Class 类。
+
+### 2. 验证
+
+确保 Class 文件的字节流中包含的信息符合当前虚拟机的要求，并且不会危害虚拟机自身的安全。
+
+### 3. 准备
+
+类变量是被 static 修饰的变量，准备阶段为类变量分配内存并设置初始值，使用的是方法区的内存。
+
+实例变量不会在这阶段分配内存，它将会在对象实例化时随着对象一起分配在堆中。
+
+注意，实例化不是类加载的一个过程，类加载发生在所有实例化操作之前，并且类加载只进行一次，实例化可以进行多次。
+
+初始值一般为 0 值，例如下面的类变量 value 被初始化为 0 而不是 123。
+
+```java
+public static int value = 123;
+```
+
+如果类变量是常量，那么会按照表达式来进行初始化，而不是赋值为 0。
+
+```java
+public static final int value = 123;
+```
+
+### 4. 解析
+
+将常量池的符号引用替换为直接引用的过程。
+
 其中解析过程在某些情况下可以在初始化阶段之后再开始，这是为了支持 Java 的动态绑定。
 
+### 5. 初始化
+
+初始化阶段才真正开始执行类中定义的 Java 程序代码。初始化阶段即虚拟机执行类构造器 &lt;clinit>() 方法的过程。
+
+在准备阶段，类变量已经赋过一次系统要求的初始值，而在初始化阶段，根据程序员通过程序制定的主观计划去初始化类变量和其它资源。
+
+&lt;clinit>() 方法具有以下特点：
+
+- 是由编译器自动收集类中所有类变量的赋值动作和静态语句块中的语句合并产生的，编译器收集的顺序由语句在源文件中出现的顺序决定。特别注意的是，静态语句块只能访问到定义在它之前的类变量，定义在它之后的类变量只能赋值，不能访问。例如以下代码：
+
+```java
+public class Test {
+    static {
+        i = 0;                // 给变量赋值可以正常编译通过
+        System.out.print(i);  // 这句编译器会提示“非法向前引用”
+    }
+    static int i = 1;
+}
+```
+
+- 与类的构造函数（或者说实例构造器 &lt;init>()）不同，不需要显式的调用父类的构造器。虚拟机会自动保证在子类的 &lt;clinit>() 方法运行之前，父类的 &lt;clinit>() 方法已经执行结束。因此虚拟机中第一个执行 &lt;clinit>() 方法的类肯定为 java.lang.Object。
+
+- 由于父类的 &lt;clinit>() 方法先执行，也就意味着父类中定义的静态语句块要优先于子类的变量赋值操作。例如以下代码：
+
+```java
+static class Parent {
+    public static int A = 1;
+    static {
+        A = 2;
+    }
+}
+
+static class Sub extends Parent {
+    public static int B = A;
+}
+
+public static void main(String[] args) {
+     System.out.println(Sub.B);  // 2
+}
+```
+
+- &lt;clinit>() 方法对于类或接口不是必须的，如果一个类中不包含静态语句块，也没有对类变量的赋值操作，编译器可以不为该类生成 &lt;clinit>() 方法。
+
+- 接口中不可以使用静态语句块，但仍然有类变量初始化的赋值操作，因此接口与类一样都会生成 &lt;clinit>() 方法。但接口与类不同的是，执行接口的 &lt;clinit>() 方法不需要先执行父接口的 &lt;clinit>() 方法。只有当父接口中定义的变量使用时，父接口才会初始化。另外，接口的实现类在初始化时也一样不会执行接口的 &lt;clinit>() 方法。
+
+- 虚拟机会保证一个类的 &lt;clinit>() 方法在多线程环境下被正确的加锁和同步，如果多个线程同时初始化一个类，只会有一个线程执行这个类的 &lt;clinit>() 方法，其它线程都会阻塞等待，直到活动线程执行 &lt;clinit>() 方法完毕。如果在一个类的 &lt;clinit>() 方法中有耗时的操作，就可能造成多个线程阻塞，在实际过程中此种阻塞很隐蔽。
+
 ## 类初始化时机
 
+### 1. 主动引用
+
 虚拟机规范中并没有强制约束何时进行加载，但是规范严格规定了有且只有下列五种情况必须对类进行初始化（加载、验证、准备都会随之发生）：
 
 - 遇到 new、getstatic、putstatic、invokestatic 这四条字节码指令时，如果类没有进行过初始化，则必须先触发其初始化。最常见的生成这 4 条指令的场景是：使用 new 关键字实例化对象的时候；读取或设置一个类的静态字段（被 final 修饰、已在编译期把结果放入常量池的静态字段除外）的时候；以及调用一个类的静态方法的时候。
@@ -477,6 +580,8 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 
 - 当使用 JDK 1.7 的动态语言支持时，如果一个 java.lang.invoke.MethodHandle 实例最后的解析结果为 REF_getStatic, REF_putStatic, REF_invokeStatic 的方法句柄，并且这个方法句柄所对应的类没有进行过初始化，则需要先触发其初始化；
 
+### 2. 被动引用
+
 以上 5 种场景中的行为称为对一个类进行主动引用。除此之外，所有引用类的方式都不会触发初始化，称为被动引用。被动引用的常见例子包括：
 
 - 通过子类引用父类的静态字段，不会导致子类初始化。
@@ -497,118 +602,13 @@ SuperClass[] sca = new SuperClass[10];
 System.out.println(ConstClass.HELLOWORLD);
 ```
 
-## 类加载过程
+## 类与类加载器
 
-包含了加载、验证、准备、解析和初始化这 5 个阶段。
-
-### 1. 加载
-
-加载是类加载的一个阶段，注意不要混淆。
-
-加载过程完成以下三件事：
-
-- 通过一个类的全限定名来获取定义此类的二进制字节流。
-- 将这个字节流所代表的静态存储结构转化为方法区的运行时存储结构。
-- 在内存中生成一个代表这个类的 Class 对象，作为方法区这个类的各种数据的访问入口。
-
-其中二进制字节流可以从以下方式中获取：
-
-- 从 ZIP 包读取，这很常见，最终成为日后 JAR、EAR、WAR 格式的基础。
-- 从网络中获取，这种场景最典型的应用是 Applet。
-- 运行时计算生成，这种场景使用得最多得就是动态代理技术，在 java.lang.reflect.Proxy 中，就是用了 ProxyGenerator.generateProxyClass 的代理类的二进制字节流。
-- 由其他文件生成，典型场景是 JSP 应用，即由 JSP 文件生成对应的 Class 类。
-- 从数据库读取，这种场景相对少见，例如有些中间件服务器（如 SAP Netweaver）可以选择把程序安装到数据库中来完成程序代码在集群间的分发。
-...
-
-### 2. 验证
-
-确保 Class 文件的字节流中包含的信息符合当前虚拟机的要求，并且不会危害虚拟机自身的安全。
-
-- 文件格式验证：验证字节流是否符合 Class 文件格式的规范，并且能被当前版本的虚拟机处理。
-- 元数据验证：对字节码描述的信息进行语义分析，以保证其描述的信息符合 Java 语言规范的要求。
-- 字节码验证：通过数据流和控制流分析，确保程序语义是合法、符合逻辑的。
-- 符号引用验证：发生在虚拟机将符号引用转换为直接引用的时候，对类自身以外（常量池中的各种符号引用）的信息进行匹配性校验。
-
-### 3. 准备
-
-类变量是被 static 修饰的变量，准备阶段为类变量分配内存并设置初始值，使用的是方法区的内存。
-
-实例变量不会在这阶段分配内存，它将会在对象实例化时随着对象一起分配在 Java 堆中。（实例化不是类加载的一个过程，类加载发生在所有实例化操作之前，并且类加载只进行一次，实例化可以进行多次）
-
-初始值一般为 0 值，例如下面的类变量 value 被初始化为 0 而不是 123。
-
-```java
-public static int value = 123;
-```
-
-如果类变量是常量，那么会按照表达式来进行初始化，而不是赋值为 0。
-
-```java
-public static final int value = 123;
-```
-
-### 4. 解析
-
-将常量池的符号引用替换为直接引用的过程。
-
-### 5. 初始化
-
-初始化阶段才真正开始执行类中的定义的 Java 程序代码。初始化阶段即虚拟机执行类构造器 &lt;clinit>() 方法的过程。
-
-在准备阶段，类变量已经赋过一次系统要求的初始值，而在初始化阶段，根据程序员通过程序制定的主观计划去初始化类变量和其它资源。
-
-&lt;clinit>() 方法具有以下特点：
-
-- 是由编译器自动收集类中所有类变量的赋值动作和静态语句块（static{} 块）中的语句合并产生的，编译器收集的顺序由语句在源文件中出现的顺序决定。特别注意的是，静态语句块只能访问到定义在它之前的类变量，定义在它之后的类变量只能赋值，不能访问。例如以下代码：
-
-```java
-public class Test {
-    static {
-        i = 0;                // 给变量赋值可以正常编译通过
-        System.out.print(i);  // 这句编译器会提示“非法向前引用”
-    }
-    static int i = 1;
-}
-```
-
-- 与类的构造函数（或者说实例构造器 &lt;init>()）不同，不需要显式的调用父类的构造器。虚拟机会自动保证在子类的 &lt;clinit>() 方法运行之前，父类的 &lt;clinit>() 方法已经执行结束。因此虚拟机中第一个执行 &lt;clinit>() 方法的类肯定为 java.lang.Object。
-
-- 由于父类的 &lt;clinit>() 方法先执行，也就意味着父类中定义的静态语句块要优于子类的变量赋值操作。例如以下代码：
-
-```java
-static class Parent {
-    public static int A = 1;
-    static {
-        A = 2;
-    }
-}
-
-static class Sub extends Parent {
-    public static int B = A;
-}
-
-public static void main(String[] args) {
-     System.out.println(Sub.B);  // 输出结果是父类中的静态变量 A 的值，也就是 2。
-}
-```
-
-- &lt;clinit>() 方法对于类或接口不是必须的，如果一个类中不包含静态语句块，也没有对类变量的赋值操作，编译器可以不为该类生成 &lt;clinit>() 方法。
-
-- 接口中不可以使用静态语句块，但仍然有类变量初始化的赋值操作，因此接口与类一样都会生成 &lt;clinit>() 方法。但接口与类不同的是，执行接口的 &lt;clinit>() 方法不需要先执行父接口的 &lt;clinit>() 方法。只有当父接口中定义的变量使用时，父接口才会初始化。另外，接口的实现类在初始化时也一样不会执行接口的 &lt;clinit>() 方法。
-
-- 虚拟机会保证一个类的 &lt;clinit>() 方法在多线程环境下被正确的加锁和同步，如果多个线程同时初始化一个类，只会有一个线程执行这个类的 &lt;clinit>() 方法，其它线程都会阻塞等待，直到活动线程执行 &lt;clinit>() 方法完毕。如果在一个类的 &lt;clinit>() 方法中有耗时的操作，就可能造成多个线程阻塞，在实际过程中此种阻塞很隐蔽。
-
-## 类加载器
-
-在 Java 虚拟机外部实现，以便让应用程序自己决定如何去获取所需要的类。
-
-### 1. 类与类加载器
-
-两个类相等：类本身相等，并且使用同一个类加载器进行加载。这是因为每一个类加载器都拥有一个独立的类名称空间。
+两个类相等需要类本身相等，并且使用同一个类加载器进行加载。这是因为每一个类加载器都拥有一个独立的类名称空间。
 
 这里的相等，包括类的 Class 对象的 equals() 方法、isAssignableFrom() 方法、isInstance() 方法的返回结果为 true，也包括使用 instanceof 关键字做对象所属关系判定结果为 true。
 
-### 2. 类加载器分类
+## 类加载器分类
 
 从 Java 虚拟机的角度来讲，只存在以下两种不同的类加载器：
 
@@ -624,7 +624,7 @@ public static void main(String[] args) {
 
 - 应用程序类加载器（Application ClassLoader）这个类加载器是由 AppClassLoader（sun.misc.Launcher$AppClassLoader）实现的。由于这个类加载器是 ClassLoader 中的 getSystemClassLoader() 方法的返回值，因此一般称为系统类加载器。它负责加载用户类路径（ClassPath）上所指定的类库，开发者可以直接使用这个类加载器，如果应用程序中没有自定义过自己的类加载器，一般情况下这个就是程序中默认的类加载器。
 
-### 3. 双亲委派模型
+## 双亲委派模型
 
 应用程序都是由三种类加载器相互配合进行加载的，如果有必要，还可以加入自己定义的类加载器。
 
@@ -632,17 +632,17 @@ public static void main(String[] args) {
 
 <div align="center"> <img src="../pics//class_loader_hierarchy.png" width="600"/> </div><br>
 
-（一）工作过程
+### 1. 工作过程
 
 一个类加载器首先将类加载请求传送到父类加载器，只有当父类加载器无法完成类加载请求时才尝试加载。
 
-（二）好处
+### 2. 好处
 
 使得 Java 类随着它的类加载器一起具有一种带有优先级的层次关系，从而使得基础类得到统一。
 
-例如 java.lang.Object 存放在 rt.jar 中，如果编写另外一个 java.lang.Object 的类并放到 ClassPath 中，程序可以编译通过。因为双亲委派模型的存在，所以在 rt.jar 中的 Object 比在 ClassPath 中的 Object 优先级更高，因为 rt.jar 中的 Object 使用的是启动类加载器，而 ClassPath 中的 Object 使用的是应用程序类加载器。正因为 rt.jar 中的 Object 优先级更高，因为程序中所有的 Object 都是这个 Object。
+例如 java.lang.Object 存放在 rt.jar 中，如果编写另外一个 java.lang.Object 的类并放到 ClassPath 中，程序可以编译通过。由于双亲委派模型的存在，所以在 rt.jar 中的 Object 比在 ClassPath 中的 Object 优先级更高，这是因为 rt.jar 中的 Object 使用的是启动类加载器，而 ClassPath 中的 Object 使用的是应用程序类加载器。rt.jar 中的 Object 优先级更高，那么程序中所有的 Object 都是这个 Object。
 
-（三）实现
+### 3. 实现
 
 以下是抽象类 java.lang.ClassLoader 的代码片段，其中的 loadClass() 方法运行过程如下：先检查类是否已经加载过，如果没有则让父类加载器去加载。当父类加载器加载失败时抛出 ClassNotFoundException，此时尝试自己去加载。
 
@@ -690,7 +690,7 @@ public abstract class ClassLoader {
 }
 ```
 
-### 4. 自定义类加载器实现
+## 自定义类加载器实现
 
 FileSystemClassLoader 是自定义类加载器，继承自 java.lang.ClassLoader，用于加载文件系统上的类。它首先根据类的全名在文件系统上查找类的字节代码文件（.class 文件），然后读取该文件内容，最后通过 defineClass() 方法来把这些字节代码转换成 java.lang.Class 类的实例。
 

From 5bbc6c381f8d4e2259701e117deab22f8163f913 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 22:22:55 +0800
Subject: [PATCH 104/172] auto commit

---
 notes/Java IO.md | 82 +++++++++++++++++++++++++++++++-----------------
 1 file changed, 54 insertions(+), 28 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index 1e3aedcd..a14ea902 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -2,8 +2,17 @@
 * [一、概览](#一概览)
 * [二、磁盘操作](#二磁盘操作)
 * [三、字节操作](#三字节操作)
+    * [实现文件复制](#实现文件复制)
+    * [装饰者模式](#装饰者模式)
 * [四、字符操作](#四字符操作)
+    * [编码与解码](#编码与解码)
+    * [String](#string)
+    * [Reader 与 Writer](#reader-与-writer)
+    * [实现逐行输出文本文件的内容](#实现逐行输出文本文件的内容)
 * [五、对象操作](#五对象操作)
+    * [序列化](#序列化)
+    * [Serializable](#serializable)
+    * [transient](#transient)
 * [六、网络操作](#六网络操作)
     * [InetAddress](#inetaddress)
     * [URL](#url)
@@ -37,7 +46,7 @@ Java 的 I/O 大概可以分成以下几类：
 
 File 类可以用于表示文件和目录的信息，但是它不表示文件的内容。
 
-递归地输出一个目录下所有文件：
+递归地列出一个目录下所有文件：
 
 ```java
 public static void listAllFiles(File dir) {
@@ -56,7 +65,7 @@ public static void listAllFiles(File dir) {
 
 # 三、字节操作
 
-使用字节流操作进行文件复制：
+## 实现文件复制
 
 ```java
 public static void copyFile(String src, String dist) throws IOException {
@@ -77,13 +86,15 @@ public static void copyFile(String src, String dist) throws IOException {
 }
 ```
 
-<div align="center"> <img src="../pics//DP-Decorator-java.io.png" width="500"/> </div><br>
+## 装饰者模式
 
 Java I/O 使用了装饰者模式来实现。以 InputStream 为例，
 
 - InputStream 是抽象组件；
 - FileInputStream 是 InputStream 的子类，属于具体组件，提供了字节流的输入操作；
-- FilterInputStream 属于抽象装饰者，装饰者用于装饰组件，为组件提供额外的功能，例如 BufferedInputStream 为 FileInputStream 提供缓存的功能。
+- FilterInputStream 属于抽象装饰者，装饰者用于装饰组件，为组件提供额外的功能。例如 BufferedInputStream 为 FileInputStream 提供缓存的功能。
+
+<div align="center"> <img src="../pics//DP-Decorator-java.io.png" width="500"/> </div><br>
 
 实例化一个具有缓存功能的字节流对象时，只需要在 FileInputStream 对象上再套一层 BufferedInputStream 对象即可。
 
@@ -96,27 +107,7 @@ DataInputStream 装饰者提供了对更多数据类型进行输入的操作，
 
 # 四、字符操作
 
-不管是磁盘还是网络传输，最小的存储单元都是字节，而不是字符。但是在程序中操作的通常是字符形式的数据，因此需要提供对字符进行操作的方法。
-
-- InputStreamReader 实现从字节流解码成字符流；
-- OutputStreamWriter 实现字符流编码成为字节流。
-
-逐行输出文本文件的内容：
-
-```java
-public static void readFileContent(String filePath) throws IOException {
-    FileReader fileReader = new FileReader(filePath);
-    BufferedReader bufferedReader = new BufferedReader(fileReader);
-    String line;
-    while ((line = bufferedReader.readLine()) != null) {
-        System.out.println(line);
-    }
-    // 装饰者模式使得 BufferedReader 组合了一个 Reader 对象
-    // 在调用 BufferedReader 的 close() 方法时会去调用 fileReader 的 close() 方法
-    // 因此只要一个 close() 调用即可
-    bufferedReader.close();
-}
-```
+## 编码与解码
 
 编码就是把字符转换为字节，而解码是把字节重新组合成字符。
 
@@ -130,6 +121,8 @@ UTF-16be 中的 be 指的是 Big Endian，也就是大端。相应地也有 UTF-
 
 Java 使用双字节编码 UTF-16be，这不是指 Java 只支持这一种编码方式，而是说 char 这种类型使用 UTF-16be 进行编码。char 类型占 16 位，也就是两个字节，Java 使用这种双字节编码是为了让一个中文或者一个英文都能使用一个 char 来存储。
 
+## String
+
 String 可以看成一个字符序列，可以指定一个编码方式将它编码为字节序列，也可以指定一个编码方式将一个字节序列解码为 String。
 
 ```java
@@ -145,13 +138,46 @@ System.out.println(str2);
 byte[] bytes = str1.getBytes();
 ```
 
+## Reader 与 Writer
+
+不管是磁盘还是网络传输，最小的存储单元都是字节，而不是字符。但是在程序中操作的通常是字符形式的数据，因此需要提供对字符进行操作的方法。
+
+- InputStreamReader 实现从字节流解码成字符流；
+- OutputStreamWriter 实现字符流编码成为字节流。
+
+## 实现逐行输出文本文件的内容
+
+```java
+public static void readFileContent(String filePath) throws IOException {
+
+    FileReader fileReader = new FileReader(filePath);
+    BufferedReader bufferedReader = new BufferedReader(fileReader);
+
+    String line;
+    while ((line = bufferedReader.readLine()) != null) {
+        System.out.println(line);
+    }
+
+    // 装饰者模式使得 BufferedReader 组合了一个 Reader 对象
+    // 在调用 BufferedReader 的 close() 方法时会去调用 Reader 的 close() 方法
+    // 因此只要一个 close() 调用即可
+    bufferedReader.close();
+}
+```
+
 # 五、对象操作
 
+## 序列化
+
 序列化就是将一个对象转换成字节序列，方便存储和传输。
 
 - 序列化：ObjectOutputStream.writeObject()
 - 反序列化：ObjectInputStream.readObject()
 
+不会对静态变量进行序列化，因为序列化只是保存对象的状态，静态变量属于类的状态。
+
+## Serializable
+
 序列化的类需要实现 Serializable 接口，它只是一个标准，没有任何方法需要实现，但是如果不去实现它的话而进行序列化，会抛出异常。
 
 ```java
@@ -184,11 +210,11 @@ private static class A implements Serializable {
 }
 ```
 
-不会对静态变量进行序列化，因为序列化只是保存对象的状态，静态变量属于类的状态。
+## transient
 
 transient 关键字可以使一些属性不会被序列化。
 
-ArrayList 中存储数据的数组是用 transient 修饰的，因为这个数组是动态扩展的，并不是所有的空间都被使用，因此就不需要所有的内容都被序列化。通过重写序列化和反序列化方法，使得可以只序列化数组中有内容的那部分数据。
+ArrayList 中存储数据的数组 elementData 是用 transient 修饰的，因为这个数组是动态扩展的，并不是所有的空间都被使用，因此就不需要所有的内容都被序列化。通过重写序列化和反序列化方法，使得可以只序列化数组中有内容的那部分数据。
 
 ```java
 private transient Object[] elementData;
@@ -249,8 +275,8 @@ public static void main(String[] args) throws IOException {
 
 ## Datagram
 
-- DatagramPacket：数据包类
 - DatagramSocket：通信类
+- DatagramPacket：数据包类
 
 # 七、NIO
 

From eacd9d07ac00a5a3b0b5821b26f1cbdd7832f174 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 13 Aug 2018 22:31:36 +0800
Subject: [PATCH 105/172] auto commit

---
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 1 file changed, 0 insertions(+), 0 deletions(-)
 create mode 100644 pics/urlnuri.jpg

diff --git a/pics/urlnuri.jpg b/pics/urlnuri.jpg
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zm9yR2jLF$2gBsg1KpjB4C5uGVYno8)lP>cXmHHzqG;2fG(}Ab6JIho&o;WR=8x`1!
Z^)LBv<@9{(=>Na{|K~B!|J(NeKLBL<vkCwJ

literal 0
HcmV?d00001


From 2abb5464c8ce9062a60f60a8e59f7ff9f39fd6d6 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E4=BC=9F?= <18056981502@163.com>
Date: Tue, 14 Aug 2018 01:03:00 +0800
Subject: [PATCH 106/172] =?UTF-8?q?=E8=87=AA=E9=80=89=E9=94=81=20->=20?=
 =?UTF-8?q?=E8=87=AA=E6=97=8B=E9=94=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 89ab7cac..28e69e36 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1594,7 +1594,7 @@ ThreadLocal 从理论上讲并不是用来解决多线程并发问题的，因
 
 互斥同步的进入阻塞状态的开销都很大，应该尽量避免。在许多应用中，共享数据的锁定状态只会持续很短的一段时间。自旋锁的思想是让一个线程在请求一个共享数据的锁时执行忙循环（自旋）一段时间，如果在这段时间内能获得锁，就可以避免进入阻塞状态。
 
-自选锁虽然能避免进入阻塞状态从而减少开销，但是它需要进行忙循环操作占用 CPU 时间，它只适用于共享数据的锁定状态很短的场景。
+自旋锁虽然能避免进入阻塞状态从而减少开销，但是它需要进行忙循环操作占用 CPU 时间，它只适用于共享数据的锁定状态很短的场景。
 
 在 JDK 1.6 中引入了自适应的自旋锁。自适应意味着自旋的次数不再固定了，而是由前一次在同一个锁上的自旋次数及锁的拥有者的状态来决定。
 

From 7fc6ceac3c621239257bae76eb169593deca5292 Mon Sep 17 00:00:00 2001
From: zhangzhikai <zhangzhikai@corp.netease.com>
Date: Tue, 14 Aug 2018 14:45:21 +0800
Subject: [PATCH 107/172] =?UTF-8?q?=E4=BF=AE=E6=94=B9CyclicBarrier?=
 =?UTF-8?q?=E9=83=A8=E5=88=86=E9=94=99=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index ab3528f2..e7d09e1f 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -759,7 +759,7 @@ run..run..run..run..run..run..run..run..run..run..end
 
 用来控制多个线程互相等待，只有当多个线程都到达时，这些线程才会继续执行。
 
-和 CountdownLatch 相似，都是通过维护计数器来实现的。但是它的计数器是递增的，每次执行 await() 方法之后，计数器会加 1，直到计数器的值和设置的值相等，等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 的计数器可以循环使用，所以它才叫做循环屏障。
+和 CountdownLatch 相似，都是通过维护计数器来实现的。初始化的时候计数器count与构造函数传入的参数parties相同，每次有线程调用await方法时候，计数器count不断减1，知道count减为0时等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 可以通过调用reset方法循环使用，所以它才叫做循环屏障。
 
 下图应该从下往上看才正确。
 

From 3056dea7b21034cbd1017dcd7652b7576ddd0be7 Mon Sep 17 00:00:00 2001
From: zhangzhikai <zhangzhikai@corp.netease.com>
Date: Tue, 14 Aug 2018 14:49:12 +0800
Subject: [PATCH 108/172] =?UTF-8?q?=E4=BF=AE=E6=94=B9CyclicBarrier?=
 =?UTF-8?q?=E9=83=A8=E5=88=86=E9=94=99=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index e7d09e1f..e59cb9e7 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -759,7 +759,7 @@ run..run..run..run..run..run..run..run..run..run..end
 
 用来控制多个线程互相等待，只有当多个线程都到达时，这些线程才会继续执行。
 
-和 CountdownLatch 相似，都是通过维护计数器来实现的。初始化的时候计数器count与构造函数传入的参数parties相同，每次有线程调用await方法时候，计数器count不断减1，知道count减为0时等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 可以通过调用reset方法循环使用，所以它才叫做循环屏障。
+和 CountdownLatch 相似，都是通过维护计数器来实现的。初始化的时候计数器count与构造函数传入的参数parties相同，每次有线程调用await方法时候，计数器count不断减1，直到count减为0时等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 可以通过调用reset方法循环使用，所以它才叫做循环屏障。
 
 下图应该从下往上看才正确。
 

From 819060dc94049c77cec80eaed2d0f67156128353 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 14 Aug 2018 21:07:46 +0800
Subject: [PATCH 109/172] auto commit

---
 notes/Java 并发.md       | 108 ++++++++++++++++++++++-----------------
 notes/剑指 offer 题解.md |  41 ++++++++++-----
 2 files changed, 89 insertions(+), 60 deletions(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 89ab7cac..9cd5955b 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -733,7 +733,6 @@ java.util.concurrent（J.U.C）大大提高了并发性能，AQS 被认为是 J.
 
 ```java
 public class CountdownLatchExample {
-
     public static void main(String[] args) throws InterruptedException {
         final int totalThread = 10;
         CountDownLatch countDownLatch = new CountDownLatch(totalThread);
@@ -759,15 +758,30 @@ run..run..run..run..run..run..run..run..run..run..end
 
 用来控制多个线程互相等待，只有当多个线程都到达时，这些线程才会继续执行。
 
-和 CountdownLatch 相似，都是通过维护计数器来实现的。但是它的计数器是递增的，每次执行 await() 方法之后，计数器会加 1，直到计数器的值和设置的值相等，等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 的计数器可以循环使用，所以它才叫做循环屏障。
+和 CountdownLatch 相似，都是通过维护计数器来实现的。线程执行 await() 方法之后计数器会减 1，并进行等待，直到计数器为 0，所有调用 awati() 方法而在等待的线程才能继续执行。
 
-下图应该从下往上看才正确。
+CyclicBarrier 和 CountdownLatch 的一个区别是，CyclicBarrier 的计数器通过调用 reset() 方法可以循环使用，所以它才叫做循环屏障。
+
+CyclicBarrier 有两个构造函数，其中 parties 指示计数器的初始值，barrierAction 在所有线程都到达屏障的时候会执行一次。
+
+```java
+public CyclicBarrier(int parties, Runnable barrierAction) {
+    if (parties <= 0) throw new IllegalArgumentException();
+    this.parties = parties;
+    this.count = parties;
+    this.barrierCommand = barrierAction;
+}
+
+public CyclicBarrier(int parties) {
+    this(parties, null);
+}
+```
 
 <div align="center"> <img src="../pics//CyclicBarrier.png" width=""/> </div><br>
 
 ```java
 public class CyclicBarrierExample {
-    public static void main(String[] args) throws InterruptedException {
+    public static void main(String[] args) {
         final int totalThread = 10;
         CyclicBarrier cyclicBarrier = new CyclicBarrier(totalThread);
         ExecutorService executorService = Executors.newCachedThreadPool();
@@ -776,9 +790,7 @@ public class CyclicBarrierExample {
                 System.out.print("before..");
                 try {
                     cyclicBarrier.await();
-                } catch (InterruptedException e) {
-                    e.printStackTrace();
-                } catch (BrokenBarrierException e) {
+                } catch (InterruptedException | BrokenBarrierException e) {
                     e.printStackTrace();
                 }
                 System.out.print("after..");
@@ -1198,8 +1210,6 @@ volatile 关键字通过添加内存屏障的方式来禁止指令重排，即
 
 上面提到了可以用 volatile 和 synchronized 来保证有序性。除此之外，JVM 还规定了先行发生原则，让一个操作无需控制就能先于另一个操作完成。
 
-主要有以下这些原则：
-
 ### 1. 单一线程原则
 
 > Single Thread rule
@@ -1270,14 +1280,16 @@ Thread 对象的结束先行发生于 join() 方法返回。
 
 ### 1. 不可变
 
-不可变（Immutable）的对象一定是线程安全的，无论是对象的方法实现还是方法的调用者，都不需要再采取任何的线程安全保障措施，只要一个不可变的对象被正确地构建出来，那其外部的可见状态永远也不会改变，永远也不会看到它在多个线程之中处于不一致的状态。
+不可变（Immutable）的对象一定是线程安全的，不需要再采取任何的线程安全保障措施。只要一个不可变的对象被正确地构建出来，永远也不会看到它在多个线程之中处于不一致的状态。
+
+多线程环境下，应当尽量使对象成为不可变，来满足线程安全。
 
 不可变的类型：
 
-- final 关键字修饰的基本数据类型；
+- final 关键字修饰的基本数据类型
 - String
 - 枚举类型
-- Number 部分子类，如 Long 和 Double 等数值包装类型，BigInteger 和 BigDecimal 等大数据类型。但同为 Number 的子类型的原子类 AtomicInteger 和 AtomicLong 则并非不可变的。
+- Number 部分子类，如 Long 和 Double 等数值包装类型，BigInteger 和 BigDecimal 等大数据类型。但同为 Number 的原子类 AtomicInteger 和 AtomicLong 则是可变的。
 
 对于集合类型，可以使用 Collections.unmodifiableXXX() 方法来获取一个不可变的集合。
 
@@ -1305,21 +1317,19 @@ public V put(K key, V value) {
 }
 ```
 
-多线程环境下，应当尽量使对象成为不可变，来满足线程安全。
-
 ### 2. 绝对线程安全
 
 不管运行时环境如何，调用者都不需要任何额外的同步措施。
 
 ### 3. 相对线程安全
 
-相对的线程安全需要保证对这个对象单独的操作是线程安全的，在调用的时候不需要做额外的保障措施，但是对于一些特定顺序的连续调用，就可能需要在调用端使用额外的同步手段来保证调用的正确性。
+相对线程安全需要保证对这个对象单独的操作是线程安全的，在调用的时候不需要做额外的保障措施。但是对于一些特定顺序的连续调用，就可能需要在调用端使用额外的同步手段来保证调用的正确性。
 
 在 Java 语言中，大部分的线程安全类都属于这种类型，例如 Vector、HashTable、Collections 的 synchronizedCollection() 方法包装的集合等。
 
-对于下面的代码，如果删除元素的线程删除了一个元素，而获取元素的线程试图访问一个已经被删除的元素，那么就会抛出 ArrayIndexOutOfBoundsException。
+对于下面的代码，如果删除元素的线程删除了 Vector 的一个元素，而获取元素的线程试图访问一个已经被删除的元素，那么就会抛出 ArrayIndexOutOfBoundsException。
 
-```java
+```Java
 public class VectorUnsafeExample {
     private static Vector<Integer> vector = new Vector<>();
 
@@ -1389,15 +1399,17 @@ synchronized 和 ReentrantLock。
 
 ### 2. 非阻塞同步
 
-互斥同步最主要的问题就是进行线程阻塞和唤醒所带来的性能问题，因此这种同步也称为阻塞同步。
+互斥同步最主要的问题就是线程阻塞和唤醒所带来的性能问题，因此这种同步也称为阻塞同步。
 
-互斥同步属于一种悲观的并发策略，总是认为只要不去做正确的同步措施，那就肯定会出现问题。论共享数据是否真的会出现竞争，它都要进行加锁（这里讨论的是概念模型，实际上虚拟机会优化掉很大一部分不必要的加锁）、用户态核心态转换、维护锁计数器和检查是否有被阻塞的线程需要唤醒等操作。
+互斥同步属于一种悲观的并发策略，总是认为只要不去做正确的同步措施，那就肯定会出现问题。无论共享数据是否真的会出现竞争，它都要进行加锁（这里讨论的是概念模型，实际上虚拟机会优化掉很大一部分不必要的加锁）、用户态核心态转换、维护锁计数器和检查是否有被阻塞的线程需要唤醒等操作。
 
-随着硬件指令集的发展，我们可以使用基于冲突检测的乐观并发策略：先进行操作，如果没有其它线程争用共享数据，那操作就成功了，否则采取补偿措施（不断地重试，直到成功为止）。这种乐观的并发策略的许多实现都不需要把线程挂起，因此这种同步操作称为非阻塞同步。
+**（一）CAS** 
 
-乐观锁需要操作和冲突检测这两个步骤具备原子性，这里就不能再使用互斥同步来保证了，只能靠硬件来完成。
+随着硬件指令集的发展，我们可以使用基于冲突检测的乐观并发策略：先进行操作，如果没有其它线程争用共享数据，那操作就成功了，否则采取补偿措施（不断地重试，直到成功为止）。这种乐观的并发策略的许多实现都不需要将线程阻塞，因此这种同步操作称为非阻塞同步。
 
-硬件支持的原子性操作最典型的是：比较并交换（Compare-and-Swap，CAS）。CAS 指令需要有 3 个操作数，分别是内存地址 V、旧的预期值 A 和新值 B。当执行操作时，只有当 V 的值等于 A，才将 V 的值更新为 B。
+乐观锁需要操作和冲突检测这两个步骤具备原子性，这里就不能再使用互斥同步来保证了，只能靠硬件来完成。硬件支持的原子性操作最典型的是：比较并交换（Compare-and-Swap，CAS）。CAS 指令需要有 3 个操作数，分别是内存地址 V、旧的预期值 A 和新值 B。当执行操作时，只有当 V 的值等于 A，才将 V 的值更新为 B。
+
+**（二）AtomicInteger** 
 
 J.U.C 包里面的整数原子类 AtomicInteger，其中的 compareAndSet() 和 getAndIncrement() 等方法都使用了 Unsafe 类的 CAS 操作。
 
@@ -1419,7 +1431,7 @@ public final int incrementAndGet() {
 }
 ```
 
-以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果该字段内存地址中的值 ==var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。
+以下代码是 getAndAddInt() 源码，var1 指示对象内存地址，var2 指示该字段相对对象内存地址的偏移，var4 指示操作需要加的数值，这里为 1。通过 getIntVolatile(var1, var2) 得到旧的预期值，通过调用 compareAndSwapInt() 来进行 CAS 比较，如果该字段内存地址中的值等于 var5，那么就更新内存地址为 var1+var2 的变量为 var5+var4。
 
 可以看到 getAndAddInt() 在一个循环中进行，发生冲突的做法是不断的进行重试。
 
@@ -1434,23 +1446,19 @@ public final int getAndAddInt(Object var1, long var2, int var4) {
 }
 ```
 
-ABA ：如果一个变量初次读取的时候是 A 值，它的值被改成了 B，后来又被改回为 A，那 CAS 操作就会误认为它从来没有被改变过。
+**（三）ABA** 
+
+如果一个变量初次读取的时候是 A 值，它的值被改成了 B，后来又被改回为 A，那 CAS 操作就会误认为它从来没有被改变过。
 
 J.U.C 包提供了一个带有标记的原子引用类 AtomicStampedReference 来解决这个问题，它可以通过控制变量值的版本来保证 CAS 的正确性。大部分情况下 ABA 问题不会影响程序并发的正确性，如果需要解决 ABA 问题，改用传统的互斥同步可能会比原子类更高效。
 
 ### 3. 无同步方案
 
-要保证线程安全，并不是一定就要进行同步，两者没有因果关系。同步只是保证共享数据争用时的正确性的手段，如果一个方法本来就不涉及共享数据，那它自然就无须任何同步措施去保证正确性，因此会有一些代码天生就是线程安全的。
+要保证线程安全，并不是一定就要进行同步。如果一个方法本来就不涉及共享数据，那它自然就无须任何同步措施去保证正确性。
 
-**（一）可重入代码（Reentrant Code）** 
+**（一）栈封闭** 
 
-这种代码也叫做纯代码（Pure Code），可以在代码执行的任何时刻中断它，转而去执行另外一段代码（包括递归调用它本身），而在控制权返回后，原来的程序不会出现任何错误。
-
-可重入代码有一些共同的特征，例如不依赖存储在堆上的数据和公用的系统资源、用到的状态量都由参数中传入、不调用非可重入的方法等。
-
-**（二）栈封闭** 
-
-多个线程访问同一个方法的局部变量时，不会出现线程安全问题，因为局部变量存储在栈中，属于线程私有的。
+多个线程访问同一个方法的局部变量时，不会出现线程安全问题，因为局部变量存储在虚拟机栈中，属于线程私有的。
 
 ```java
 import java.util.concurrent.ExecutorService;
@@ -1482,11 +1490,11 @@ public static void main(String[] args) {
 100
 ```
 
-**（三）线程本地存储（Thread Local Storage）** 
+**（二）线程本地存储（Thread Local Storage）** 
 
 如果一段代码中所需要的数据必须与其他代码共享，那就看看这些共享数据的代码是否能保证在同一个线程中执行。如果能保证，我们就可以把共享数据的可见范围限制在同一个线程之内，这样，无须同步也能保证线程之间不出现数据争用的问题。
 
-符合这种特点的应用并不少见，大部分使用消费队列的架构模式（如“生产者-消费者”模式）都会将产品的消费过程尽量在一个线程中消费完，其中最重要的一个应用实例就是经典 Web 交互模型中的“一个请求对应一个服务器线程”（Thread-per-Request）的处理方式，这种处理方式的广泛应用使得很多 Web 服务端应用都可以使用线程本地存储来解决线程安全问题。
+符合这种特点的应用并不少见，大部分使用消费队列的架构模式（如“生产者-消费者”模式）都会将产品的消费过程尽量在一个线程中消费完。其中最重要的一个应用实例就是经典 Web 交互模型中的“一个请求对应一个服务器线程”（Thread-per-Request）的处理方式，这种处理方式的广泛应用使得很多 Web 服务端应用都可以使用线程本地存储来解决线程安全问题。
 
 可以使用 java.lang.ThreadLocal 类来实现线程本地存储功能。
 
@@ -1584,17 +1592,25 @@ public T get() {
 }
 ```
 
-ThreadLocal 从理论上讲并不是用来解决多线程并发问题的，因为根本不存在多线程竞争。在一些场景 (尤其是使用线程池) 下，由于 ThreadLocal.ThreadLocalMap 的底层数据结构导致 ThreadLocal 有内存泄漏的情况，尽可能在每次使用 ThreadLocal 后手动调用 remove()，以避免出现 ThreadLocal 经典的内存泄漏甚至是造成自身业务混乱的风险。
+ThreadLocal 从理论上讲并不是用来解决多线程并发问题的，因为根本不存在多线程竞争。
+
+在一些场景 (尤其是使用线程池) 下，由于 ThreadLocal.ThreadLocalMap 的底层数据结构导致 ThreadLocal 有内存泄漏的情况，应该尽可能在每次使用 ThreadLocal 后手动调用 remove()，以避免出现 ThreadLocal 经典的内存泄漏甚至是造成自身业务混乱的风险。
+
+**（三）可重入代码（Reentrant Code）** 
+
+这种代码也叫做纯代码（Pure Code），可以在代码执行的任何时刻中断它，转而去执行另外一段代码（包括递归调用它本身），而在控制权返回后，原来的程序不会出现任何错误。
+
+可重入代码有一些共同的特征，例如不依赖存储在堆上的数据和公用的系统资源、用到的状态量都由参数中传入、不调用非可重入的方法等。
 
 # 十二、锁优化
 
-这里的锁优化主要是指虚拟机对 synchronized 的优化。
+这里的锁优化主要是指 JVM 对 synchronized 的优化。
 
 ## 自旋锁
 
-互斥同步的进入阻塞状态的开销都很大，应该尽量避免。在许多应用中，共享数据的锁定状态只会持续很短的一段时间。自旋锁的思想是让一个线程在请求一个共享数据的锁时执行忙循环（自旋）一段时间，如果在这段时间内能获得锁，就可以避免进入阻塞状态。
+互斥同步进入阻塞状态的开销都很大，应该尽量避免。在许多应用中，共享数据的锁定状态只会持续很短的一段时间。自旋锁的思想是让一个线程在请求一个共享数据的锁时执行忙循环（自旋）一段时间，如果在这段时间内能获得锁，就可以避免进入阻塞状态。
 
-自选锁虽然能避免进入阻塞状态从而减少开销，但是它需要进行忙循环操作占用 CPU 时间，它只适用于共享数据的锁定状态很短的场景。
+自旋锁虽然能避免进入阻塞状态从而减少开销，但是它需要进行忙循环操作占用 CPU 时间，它只适用于共享数据的锁定状态很短的场景。
 
 在 JDK 1.6 中引入了自适应的自旋锁。自适应意味着自旋的次数不再固定了，而是由前一次在同一个锁上的自旋次数及锁的拥有者的状态来决定。
 
@@ -1624,7 +1640,7 @@ public static String concatString(String s1, String s2, String s3) {
 }
 ```
 
-每个 append() 方法中都有一个同步块。虚拟机观察变量 sb，很快就会发现它的动态作用域被限制在 concatString() 方法内部。也就是说，sb 的所有引用永远不会“逃逸”到 concatString() 方法之外，其他线程无法访问到它，因此可以进行消除。
+每个 append() 方法中都有一个同步块。虚拟机观察变量 sb，很快就会发现它的动态作用域被限制在 concatString() 方法内部。也就是说，sb 的所有引用永远不会逃逸到 concatString() 方法之外，其他线程无法访问到它，因此可以进行消除。
 
 ## 锁粗化
 
@@ -1636,9 +1652,9 @@ public static String concatString(String s1, String s2, String s3) {
 
 JDK 1.6 引入了偏向锁和轻量级锁，从而让锁拥有了四个状态：无锁状态（unlocked）、偏向锁状态（biasble）、轻量级锁状态（lightweight locked）和重量级锁状态（inflated）。
 
-以下是 HotSpot 虚拟机对象头的内存布局，这些数据被称为 mark word。其中 tag bits 对应了五个状态，这些状态在右侧的 state 表格中给出，应该注意的是 state 表格不是存储在对象头中的。除了 marked for gc 状态，其它四个状态已经在前面介绍过了。
+以下是 HotSpot 虚拟机对象头的内存布局，这些数据被称为 Mark Word。其中 tag bits 对应了五个状态，这些状态在右侧的 state 表格中给出。除了 marked for gc 状态，其它四个状态已经在前面介绍过了。
 
-<div align="center"> <img src="../pics//bb6a49be-00f2-4f27-a0ce-4ed764bc605c.png" width="600"/> </div><br>
+<div align="center"> <img src="../pics//bb6a49be-00f2-4f27-a0ce-4ed764bc605c.png" width="500"/> </div><br>
 
 下图左侧是一个线程的虚拟机栈，其中有一部分称为 Lock Record 的区域，这是在轻量级锁运行过程创建的，用于存放锁对象的 Mark Word。而右侧就是一个锁对象，包含了 Mark Word 和其它信息。
 
@@ -1646,9 +1662,9 @@ JDK 1.6 引入了偏向锁和轻量级锁，从而让锁拥有了四个状态：
 
 轻量级锁是相对于传统的重量级锁而言，它使用 CAS 操作来避免重量级锁使用互斥量的开销。对于绝大部分的锁，在整个同步周期内都是不存在竞争的，因此也就不需要都使用互斥量进行同步，可以先采用 CAS 操作进行同步，如果 CAS 失败了再改用互斥量进行同步。
 
-当尝试获取一个锁对象时，如果锁对象标记为 0 01，说明锁对象的锁未锁定（unlocked）状态。此时虚拟机在当前线程栈中创建 Lock Record，然后使用 CAS 操作将对象的 Mark Word 更新为 Lock Record 指针。如果 CAS 操作成功了，那么线程就获取了该对象上的锁，并且对象的 Mark Word 的锁标记变为 00，表示该对象处于轻量级锁状态。
+当尝试获取一个锁对象时，如果锁对象标记为 0 01，说明锁对象的锁未锁定（unlocked）状态。此时虚拟机在当前线程的虚拟机栈中创建 Lock Record，然后使用 CAS 操作将对象的 Mark Word 更新为 Lock Record 指针。如果 CAS 操作成功了，那么线程就获取了该对象上的锁，并且对象的 Mark Word 的锁标记变为 00，表示该对象处于轻量级锁状态。
 
-<div align="center"> <img src="../pics//baaa681f-7c52-4198-a5ae-303b9386cf47.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//baaa681f-7c52-4198-a5ae-303b9386cf47.png" width="400"/> </div><br>
 
 如果 CAS 操作失败了，虚拟机首先会检查对象的 Mark Word 是否指向当前线程的虚拟机栈，如果是的话说明当前线程已经拥有了这个锁对象，那就可以直接进入同步块继续执行，否则说明这个锁对象已经被其他线程线程抢占了。如果有两条以上的线程争用同一个锁，那轻量级锁就不再有效，要膨胀为重量级锁。
 
@@ -1666,9 +1682,9 @@ JDK 1.6 引入了偏向锁和轻量级锁，从而让锁拥有了四个状态：
 
 - 给线程起个有意义的名字，这样可以方便找 Bug。
 
-- 缩小同步范围，例如对于 synchronized，应该尽量使用同步块而不是同步方法。
+- 缩小同步范围，从而减少锁争用。例如对于 synchronized，应该尽量使用同步块而不是同步方法。
 
-- 多用同步类少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现对复杂的控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善，使用这些更高等级的同步工具你的程序可以不费吹灰之力获得优化。
+- 多用同步工具少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现复杂控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善，使用这些更高等级的同步工具你的程序可以不费吹灰之力获得优化。
 
 - 多用并发集合少用同步集合，例如应该使用 ConcurrentHashMap 而不是 Hashtable。
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 0c9bd1f9..4fcc45dc 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -2520,7 +2520,7 @@ public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
 
 ## 题目描述
 
-输入一个递增排序的数组和一个数字 S，在数组中查找两个数，使得他们的和正好是 S，如果有多对数字的和等于 S，输出两个数的乘积最小的。
+输入一个递增排序的数组和一个数字 S，在数组中查找两个数，使得他们的和正好是 S。如果有多对数字的和等于 S，输出两个数的乘积最小的。
 
 ## 解题思路
 
@@ -2596,9 +2596,13 @@ public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
 
 ## 题目描述
 
-输入："I am a student."
+```html
+Input:
+"I am a student."
 
-输出："student. a am I"
+Output:
+"student. a am I"
+```
 
 ## 解题思路
 
@@ -2640,7 +2644,14 @@ private void swap(char[] c, int i, int j) {
 
 ## 题目描述
 
-对于一个给定的字符序列 S，请你把其循环左移 K 位后的序列输出。例如，字符序列 S=”abcXYZdef”, 要求输出循环左移 3 位后的结果，即“XYZdefabc”。
+```html
+Input:
+S="abcXYZdef"
+K=3
+
+Output:
+"XYZdefabc"
+```
 
 ## 解题思路
 
@@ -2675,7 +2686,9 @@ private void swap(char[] chars, int i, int j) {
 
 ## 题目描述
 
-给定一个数组和滑动窗口的大小，找出所有滑动窗口里数值的最大值。例如，如果输入数组 {2, 3, 4, 2, 6, 2, 5, 1} 及滑动窗口的大小 3，那么一共存在 6 个滑动窗口，他们的最大值分别为 {4, 4, 6, 6, 6, 5}。
+给定一个数组和滑动窗口的大小，找出所有滑动窗口里数值的最大值。
+
+例如，如果输入数组 {2, 3, 4, 2, 6, 2, 5, 1} 及滑动窗口的大小 3，那么一共存在 6 个滑动窗口，他们的最大值分别为 {4, 4, 6, 6, 6, 5}。
 
 ## 解题思路
 
@@ -2774,7 +2787,7 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
 
 ## 题目描述
 
-五张牌，其中大小鬼为癞子，牌面大小为 0。判断是否能组成顺子。
+五张牌，其中大小鬼为癞子，牌面大小为 0。判断这五张牌是否能组成顺子。
 
 ## 解题思路
 
@@ -2803,7 +2816,7 @@ public boolean isContinuous(int[] nums) {
 
 ## 题目描述
 
-让小朋友们围成一个大圈。然后，他随机指定一个数 m，让编号为 0 的小朋友开始报数。每次喊到 m-1 的那个小朋友要出列唱首歌，然后可以在礼品箱中任意的挑选礼物，并且不再回到圈中，从他的下一个小朋友开始，继续 0...m-1 报数 .... 这样下去 .... 直到剩下最后一个小朋友，可以不用表演。
+让小朋友们围成一个大圈。然后，随机指定一个数 m，让编号为 0 的小朋友开始报数。每次喊到 m-1 的那个小朋友要出列唱首歌，然后可以在礼品箱中任意的挑选礼物，并且不再回到圈中，从他的下一个小朋友开始，继续 0...m-1 报数 .... 这样下去 .... 直到剩下最后一个小朋友，可以不用表演。
 
 ## 解题思路
 
@@ -2813,7 +2826,7 @@ public boolean isContinuous(int[] nums) {
 public int LastRemaining_Solution(int n, int m) {
     if (n == 0)     /* 特殊输入的处理 */
         return -1;
-    if (n == 1)     /* 返回条件 */
+    if (n == 1)     /* 递归返回条件 */
         return 0;
     return (LastRemaining_Solution(n - 1, m) + m) % n;
 }
@@ -2851,7 +2864,7 @@ public int maxProfit(int[] prices) {
 
 ## 题目描述
 
-求 1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case 等关键字及条件判断语句（A?B:C）。
+要求不能使用乘除法、for、while、if、else、switch、case 等关键字及条件判断语句 A ? B : C。
 
 ## 解题思路
 
@@ -2859,7 +2872,7 @@ public int maxProfit(int[] prices) {
 
 条件与 && 具有短路原则，即在第一个条件语句为 false 的情况下不会去执行第二个条件语句。利用这一特性，将递归的返回条件取非然后作为 && 的第一个条件语句，递归的主体转换为第二个条件语句，那么当递归的返回条件为 true 的情况下就不会执行递归的主体部分，递归返回。
 
-以下实现中，递归的返回条件为 n <= 0，取非后就是 n > 0，递归的主体部分为 sum += Sum_Solution(n - 1)，转换为条件语句后就是 (sum += Sum_Solution(n - 1)) > 0。
+本题的递归返回条件为 n <= 0，取非后就是 n > 0；递归的主体部分为 sum += Sum_Solution(n - 1)，转换为条件语句后就是 (sum += Sum_Solution(n - 1)) > 0。
 
 ```java
 public int Sum_Solution(int n) {
@@ -2875,7 +2888,7 @@ public int Sum_Solution(int n) {
 
 ## 题目描述
 
-写一个函数，求两个整数之和，要求在函数体内不得使用 +、-、\*、/ 四则运算符号。
+写一个函数，求两个整数之和，要求不得使用 +、-、\*、/ 四则运算符号。
 
 ## 解题思路
 
@@ -2895,7 +2908,7 @@ public int Add(int a, int b) {
 
 ## 题目描述
 
-给定一个数组 A[0, 1,..., n-1], 请构建一个数组 B[0, 1,..., n-1], 其中 B 中的元素 B[i]=A[0]\*A[1]\*...\*A[i-1]\*A[i+1]\*...\*A[n-1]。不能使用除法。
+给定一个数组 A[0, 1,..., n-1]，请构建一个数组 B[0, 1,..., n-1]，其中 B 中的元素 B[i]=A[0]\*A[1]\*...\*A[i-1]\*A[i+1]\*...\*A[n-1]。要求不能使用除法。
 
 ## 解题思路
 
@@ -2917,7 +2930,7 @@ public int[] multiply(int[] A) {
 
 ## 题目描述
 
-将一个字符串转换成一个整数，要求不能使用字符串转换整数的库函数。 数值为 0 或者字符串不是一个合法的数值则返回 0。
+将一个字符串转换成一个整数，字符串不是一个合法的数值则返回 0，要求不能使用字符串转换整数的库函数。
 
 ```html
 Iuput:
@@ -2979,7 +2992,7 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 
 [Leetcode : 236. Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/)
 
-在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是 LCA。
+在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是最低公共祖先。
 
 ```java
 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

From adda52204350ab9fe571da04383c931f21bfe3d4 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 14 Aug 2018 21:10:49 +0800
Subject: [PATCH 110/172] auto commit

---
 notes/Java 并发.md | 4 ----
 1 file changed, 4 deletions(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index ed60676b..9cd5955b 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -758,7 +758,6 @@ run..run..run..run..run..run..run..run..run..run..end
 
 用来控制多个线程互相等待，只有当多个线程都到达时，这些线程才会继续执行。
 
-<<<<<<< HEAD
 和 CountdownLatch 相似，都是通过维护计数器来实现的。线程执行 await() 方法之后计数器会减 1，并进行等待，直到计数器为 0，所有调用 awati() 方法而在等待的线程才能继续执行。
 
 CyclicBarrier 和 CountdownLatch 的一个区别是，CyclicBarrier 的计数器通过调用 reset() 方法可以循环使用，所以它才叫做循环屏障。
@@ -772,9 +771,6 @@ public CyclicBarrier(int parties, Runnable barrierAction) {
     this.count = parties;
     this.barrierCommand = barrierAction;
 }
-=======
-和 CountdownLatch 相似，都是通过维护计数器来实现的。初始化的时候计数器count与构造函数传入的参数parties相同，每次有线程调用await方法时候，计数器count不断减1，直到count减为0时等待的所有线程才会继续执行。和 CountdownLatch 的另一个区别是，CyclicBarrier 可以通过调用reset方法循环使用，所以它才叫做循环屏障。
->>>>>>> 8fffe40cb7afc5159cb5751b8906ac9cea7cc563
 
 public CyclicBarrier(int parties) {
     this(parties, null);

From f7c5303cab250ec41998b3ffbcdafa70a76d752a Mon Sep 17 00:00:00 2001
From: Rui Li <mmssddss@gmail.com>
Date: Tue, 14 Aug 2018 22:36:08 +0800
Subject: [PATCH 111/172]  fixed typo

---
 notes/MySQL.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 690425db..7ac9685f 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -129,7 +129,7 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 ### 1. 数据结构
 
-B Tree 指的是 Balance Tree，也就是平衡树。平衡树时一颗查找树，并且所有叶子节点位于同一层。
+B Tree 指的是 Balance Tree，也就是平衡树。平衡树是一颗查找树，并且所有叶子节点位于同一层。
 
 B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具有 B Tree 的平衡性，并且通过顺序访问指针来提高区间查询的性能。
 

From 7f1134e6725ed7c670d9fa864d52f7ba48824169 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 14 Aug 2018 23:09:22 +0800
Subject: [PATCH 112/172] auto commit

---
 notes/MySQL.md           |  2 +-
 notes/剑指 offer 题解.md | 76 +++++++++++++++++++++++-----------------
 2 files changed, 44 insertions(+), 34 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 690425db..7ac9685f 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -129,7 +129,7 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 ### 1. 数据结构
 
-B Tree 指的是 Balance Tree，也就是平衡树。平衡树时一颗查找树，并且所有叶子节点位于同一层。
+B Tree 指的是 Balance Tree，也就是平衡树。平衡树是一颗查找树，并且所有叶子节点位于同一层。
 
 B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具有 B Tree 的平衡性，并且通过顺序访问指针来提高区间查询的性能。
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 4fcc45dc..0ac7d7e4 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -58,8 +58,8 @@
 * [50. 第一个只出现一次的字符位置](#50-第一个只出现一次的字符位置)
 * [51. 数组中的逆序对](#51-数组中的逆序对)
 * [52. 两个链表的第一个公共结点](#52-两个链表的第一个公共结点)
-* [53 数字在排序数组中出现的次数](#53-数字在排序数组中出现的次数)
-* [54. 二叉搜索树的第 K 个结点](#54-二叉搜索树的第-k-个结点)
+* [53. 数字在排序数组中出现的次数](#53-数字在排序数组中出现的次数)
+* [54. 二叉查找树的第 K 个结点](#54-二叉查找树的第-k-个结点)
 * [55.1 二叉树的深度](#551-二叉树的深度)
 * [55.2 平衡二叉树](#552-平衡二叉树)
 * [56. 数组中只出现一次的数字](#56-数组中只出现一次的数字)
@@ -112,11 +112,11 @@ Output:
 
 要求复杂度为 O(N) + O(1)，也就是时间复杂度 O(N)，空间复杂度 O(1)。因此不能使用排序的方法，也不能使用额外的标记数组。牛客网讨论区这一题的首票答案使用 nums[i] + length 来将元素标记，这么做会有加法溢出问题。
 
-这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。
+这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素调整到第 i 个位置上。
 
 以 (2, 3, 1, 0, 2, 5) 为例：
 
-```text-html-basic
+```text
 position-0 : (2,3,1,0,2,5) // 2 <-> 1
              (1,3,2,0,2,5) // 1 <-> 3
              (3,1,2,0,2,5) // 3 <-> 0
@@ -146,7 +146,9 @@ public boolean duplicate(int[] nums, int length, int[] duplication) {
 }
 
 private void swap(int[] nums, int i, int j) {
-    int t = nums[i]; nums[i] = nums[j]; nums[j] = t;
+    int t = nums[i];
+    nums[i] = nums[j];
+    nums[j] = t;
 }
 ```
 
@@ -176,12 +178,12 @@ Given target = 20, return false.
 
 从右上角开始查找。矩阵中的一个数，它左边的数都比它小，下边的数都比它大。因此，从右上角开始查找，就可以根据 target 和当前元素的大小关系来缩小查找区间。
 
+复杂度：O(M + N) + O(1)
+
 当前元素的查找区间为左下角的所有元素，例如元素 12 的查找区间如下：
 
 <div align="center"> <img src="../pics//f94389e9-55b1-4f49-9d37-00ed05900ae0.png" width="250"/> </div><br>
 
-复杂度：O(M + N) + O(1)
-
 ```java
 public boolean Find(int target, int[][] matrix) {
     if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
@@ -193,7 +195,7 @@ public boolean Find(int target, int[][] matrix) {
             return true;
         else if (target > matrix[r][c])
             r++;
-        else 
+        else
             c--;
     }
     return false;
@@ -206,26 +208,33 @@ public boolean Find(int target, int[][] matrix) {
 
 ## 题目描述
 
-请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为 We Are Happy. 则经过替换之后的字符串为 We%20Are%20Happy。
+
+将一个字符串中的空格替换成 "%20"。
+
+```text
+Input:
+"We Are Happy"
+
+Output:
+"We%20Are%20Happy"
+```
 
 ## 解题思路
 
-在字符串尾部填充任意字符，使得字符串的长度等于字符串替换之后的长度。因为一个空格要替换成三个字符（%20），因此当遍历到一个空格时，需要在尾部填充两个任意字符。
+在字符串尾部填充任意字符，使得字符串的长度等于替换之后的长度。因为一个空格要替换成三个字符（%20），因此当遍历到一个空格时，需要在尾部填充两个任意字符。
 
 令 P1 指向字符串原来的末尾位置，P2 指向字符串现在的末尾位置。P1 和 P2从后向前遍历，当 P1 遍历到一个空格时，就需要令 P2 指向的位置依次填充 02%（注意是逆序的），否则就填充上 P1 指向字符的值。
 
 从后向前遍是为了在改变 P2 所指向的内容时，不会影响到 P1 遍历原来字符串的内容。
 
-复杂度：O(N) + O(1)
-
 ```java
 public String replaceSpace(StringBuffer str) {
-    int oldLen = str.length();
-    for (int i = 0; i < oldLen; i++)
+    int P1 = str.length() - 1;
+    for (int i = 0; i < str.length(); i++)
         if (str.charAt(i) == ' ')
             str.append("  ");
 
-    int P1 = oldLen - 1, P2 = str.length() - 1;
+    int P2 = str.length() - 1;
     while (P1 >= 0 && P2 > P1) {
         char c = str.charAt(P1--);
         if (c == ' ') {
@@ -345,23 +354,23 @@ inorder =  [9,3,15,20,7]
 前序遍历的第一个值为根节点的值，使用这个值将中序遍历结果分成两部分，左部分为树的左子树中序遍历结果，右部分为树的右子树中序遍历的结果。
 
 ```java
-// 缓存中序遍历数组的每个值对应的索引
-private Map<Integer, Integer> inOrderNumsIndexs = new HashMap<>();
+// 缓存中序遍历数组每个值对应的索引
+private Map<Integer, Integer> indexForInOrders = new HashMap<>();
 
 public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
     for (int i = 0; i < in.length; i++)
-        inOrderNumsIndexs.put(in[i], i);
-    return reConstructBinaryTree(pre, 0, pre.length - 1, 0, in.length - 1);
+        indexForInOrders.put(in[i], i);
+    return reConstructBinaryTree(pre, 0, pre.length - 1, 0);
 }
 
-private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int inL, int inR) {
+private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int inL) {
     if (preL > preR)
         return null;
     TreeNode root = new TreeNode(pre[preL]);
-    int inIndex = inOrderNumsIndexs.get(root.val);
+    int inIndex = indexForInOrders.get(root.val);
     int leftTreeSize = inIndex - inL;
-    root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, inL, inL + leftTreeSize - 1);
-    root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, inL + leftTreeSize + 1, inR);
+    root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, inL);
+    root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, inL + leftTreeSize + 1);
     return root;
 }
 ```
@@ -2259,7 +2268,7 @@ public int GetUglyNumber_Solution(int N) {
 
 ## 题目描述
 
-在一个字符串 (1 <= 字符串长度 <= 10000，全部由字母组成) 中找到第一个只出现一次的字符，并返回它的位置。
+在一个字符串 中找到第一个只出现一次的字符，并返回它的位置。
 
 ## 解题思路
 
@@ -2277,7 +2286,7 @@ public int FirstNotRepeatingChar(String str) {
 }
 ```
 
-以上实现的空间复杂度还不是最优的。考虑到只需要找到只出现一次的字符，那么我们只需要统计的次数信息只有 0,1,更大，使用两个比特位就能存储这些信息。
+以上实现的空间复杂度还不是最优的。考虑到只需要找到只出现一次的字符，那么需要统计的次数信息只有 0,1,更大，使用两个比特位就能存储这些信息。
 
 ```java
 public int FirstNotRepeatingChar2(String str) {
@@ -2304,13 +2313,13 @@ public int FirstNotRepeatingChar2(String str) {
 
 ## 题目描述
 
-在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组，求出这个数组中的逆序对的总数 P。
+在数组中的两个数字，如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对。输入一个数组，求出这个数组中的逆序对的总数。
 
 ## 解题思路
 
 ```java
 private long cnt = 0;
-private int[] tmp;  // 在这里创建辅助数组，而不是在 merge() 递归函数中创建
+private int[] tmp;  // 在这里声明辅助数组，而不是在 merge() 递归函数中声明
 
 public int InversePairs(int[] nums) {
     tmp = new int[nums.length];
@@ -2359,7 +2368,7 @@ private void merge(int[] nums, int l, int m, int h) {
 
 设 A 的长度为 a + c，B 的长度为 b + c，其中 c 为尾部公共部分长度，可知 a + c + b = b + c + a。
 
-当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部重新开始访问链表 B；同样地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部重新开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。
+当访问链表 A 的指针访问到链表尾部时，令它从链表 B 的头部重新开始访问链表 B；同样地，当访问链表 B 的指针访问到链表尾部时，令它从链表 A 的头部重新开始访问链表 A。这样就能控制访问 A 和 B 两个链表的指针能同时访问到交点。
 
 ```java
 public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
@@ -2372,7 +2381,7 @@ public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
 }
 ```
 
-# 53 数字在排序数组中出现的次数
+# 53. 数字在排序数组中出现的次数
 
 [NowCoder](https://www.nowcoder.com/practice/70610bf967994b22bb1c26f9ae901fa2?tpId=13&tqId=11190&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
 
@@ -2380,8 +2389,9 @@ public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
 
 ```html
 Input:
-1, 2, 3, 3, 3, 3, 4, 6
-3
+nums = 1, 2, 3, 3, 3, 3, 4, 6
+K = 3
+
 Output:
 4
 ```
@@ -2408,13 +2418,13 @@ private int binarySearch(int[] nums, int K) {
 }
 ```
 
-# 54. 二叉搜索树的第 K 个结点
+# 54. 二叉查找树的第 K 个结点
 
 [NowCoder](https://www.nowcoder.com/practice/ef068f602dde4d28aab2b210e859150a?tpId=13&tqId=11215&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
 
 ## 解题思路
 
-利用二叉搜索数中序遍历有序的特点。
+利用二叉查找树中序遍历有序的特点。
 
 ```java
 private TreeNode ret;

From 6c0eb1e9efcbb783e75c2322590a82c73a83d3aa Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Wed, 15 Aug 2018 13:42:58 +0800
Subject: [PATCH 113/172] Update README.md

---
 README.md | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

diff --git a/README.md b/README.md
index 42788e54..f35ee455 100644
--- a/README.md
+++ b/README.md
@@ -179,12 +179,6 @@
 
 欢迎在 Issue 中提交对本仓库的改进建议~
 
-### Authorization
-
-虽然没有加开源协议，但是允许非商业性使用。
-
-转载使用请注明出处，谢谢！
-
 ### Typesetting
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
@@ -201,6 +195,12 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
+### License
+
+在对本作品进行演绎时，请署名并以相同方式共享。
+
+<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
+
 ### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)

From 9e1c1c51c301c95c2cda81bfbe23289cc883281a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 19:39:49 +0800
Subject: [PATCH 114/172] auto commit

---
 notes/剑指 offer 题解.md | 212 +++++++++++++++++----------------------
 1 file changed, 94 insertions(+), 118 deletions(-)

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 0ac7d7e4..1bac6bbb 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -10,8 +10,8 @@
 * [9. 用两个栈实现队列](#9-用两个栈实现队列)
 * [10.1 斐波那契数列](#101-斐波那契数列)
 * [10.2 跳台阶](#102-跳台阶)
-* [10.3 变态跳台阶](#103-变态跳台阶)
-* [10.4 矩形覆盖](#104-矩形覆盖)
+* [10.3 矩形覆盖](#103-矩形覆盖)
+* [10.4 变态跳台阶](#104-变态跳台阶)
 * [11. 旋转数组的最小数字](#11-旋转数组的最小数字)
 * [12. 矩阵中的路径](#12-矩阵中的路径)
 * [13. 机器人的运动范围](#13-机器人的运动范围)
@@ -465,7 +465,7 @@ public int pop() throws Exception {
 
 ## 题目描述
 
-求菲波那契数列的第 n 项，n <= 39。
+求斐波那契数列的第 n 项，n <= 39。
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}0&&{n=0}\\1&&{n=1}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right."/></div> <br>
 
@@ -535,23 +535,6 @@ public class Solution {
 
 ## 解题思路
 
-复杂度：O(N) + O(N)
-
-```java
-public int JumpFloor(int n) {
-    if (n == 1)
-        return 1;
-    int[] dp = new int[n];
-    dp[0] = 1;
-    dp[1] = 2;
-    for (int i = 2; i < n; i++)
-        dp[i] = dp[i - 1] + dp[i - 2];
-    return dp[n - 1];
-}
-```
-
-复杂度：O(N) + O(1)
-
 ```java
 public int JumpFloor(int n) {
     if (n <= 2)
@@ -567,7 +550,32 @@ public int JumpFloor(int n) {
 }
 ```
 
-# 10.3 变态跳台阶
+# 10.3 矩形覆盖
+
+[NowCoder](https://www.nowcoder.com/practice/72a5a919508a4251859fb2cfb987a0e6?tpId=13&tqId=11163&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+我们可以用 2\*1 的小矩形横着或者竖着去覆盖更大的矩形。请问用 n 个 2\*1 的小矩形无重叠地覆盖一个 2\*n 的大矩形，总共有多少种方法？
+
+## 解题思路
+
+```java
+public int RectCover(int n) {
+    if (n <= 2)
+        return n;
+    int pre2 = 1, pre1 = 2;
+    int result = 0;
+    for (int i = 3; i <= n; i++) {
+        result = pre2 + pre1;
+        pre2 = pre1;
+        pre1 = result;
+    }
+    return result;
+}
+```
+
+# 10.4 变态跳台阶
 
 [NowCoder](https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
 
@@ -588,47 +596,6 @@ public int JumpFloorII(int target) {
 }
 ```
 
-# 10.4 矩形覆盖
-
-[NowCoder](https://www.nowcoder.com/practice/72a5a919508a4251859fb2cfb987a0e6?tpId=13&tqId=11163&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
-
-## 题目描述
-
-我们可以用 2\*1 的小矩形横着或者竖着去覆盖更大的矩形。请问用 n 个 2\*1 的小矩形无重叠地覆盖一个 2\*n 的大矩形，总共有多少种方法？
-
-## 解题思路
-
-复杂度：O(N) + O(N)
-
-```java
-public int RectCover(int n) {
-    if (n <= 2)
-        return n;
-    int[] dp = new int[n];
-    dp[0] = 1;
-    dp[1] = 2;
-    for (int i = 2; i < n; i++)
-        dp[i] = dp[i - 1] + dp[i - 2];
-    return dp[n - 1];
-}
-```
-
-复杂度：O(N) + O(1)
-
-```java
-public int RectCover(int n) {
-    if (n <= 2)
-        return n;
-    int pre2 = 1, pre1 = 2;
-    int result = 0;
-    for (int i = 3; i <= n; i++) {
-        result = pre2 + pre1;
-        pre2 = pre1;
-        pre1 = result;
-    }
-    return result;
-}
-```
 
 # 11. 旋转数组的最小数字
 
@@ -638,18 +605,34 @@ public int RectCover(int n) {
 
 把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。
 
-例如数组 {3, 4, 5, 1, 2} 为 {1, 2, 3, 4, 5} 的一个旋转，该数组的最小值为 1。NOTE：给出的所有元素都大于 0，若数组大小为 0，请返回 0。
+例如数组 {3, 4, 5, 1, 2} 为 {1, 2, 3, 4, 5} 的一个旋转，该数组的最小值为 1。
 
 ## 解题思路
 
+在一个有序数组中查找一个元素可以用二分查找，二分查找也称为折半查找，每次都能将查找区间减半，这种折半特性的算法时间复杂度都为 O(logN)。
+
+本题可以修改二分查找算法进行求解：
+
 - 当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；
 - 否则解在 [m + 1, h] 之间，令 l = m + 1。
 
-因为 h 的赋值表达式为 h = m，因此循环体的循环条件应该为 l < h，详细解释请见 [Leetcode 题解](https://github.com/CyC2018/Interview-Notebook/blob/master/notes/Leetcode%20%E9%A2%98%E8%A7%A3.md) 二分查找部分。
+```java
+public int minNumberInRotateArray(int[] nums) {
+    if (nums.length == 0)
+        return 0;
+    int l = 0, h = nums.length - 1;
+    while (l < h) {
+        int m = l + (h - l) / 2;
+        if (nums[m] <= nums[h])
+            h = m;
+        else
+            l = m + 1;
+    }
+    return nums[l];
+}
+```
 
-但是如果出现 nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，需要切换到顺序查找。
-
-复杂度：O(logN) + O(1)
+如果数组元素允许重复的话，那么就会出现一个特殊的情况：nums[l] == nums[m] == nums[h]，那么此时无法确定解在哪个区间，需要切换到顺序查找。例如对于数组 {1,1,1,0,1}，l、m 和 h 指向的数都为 1，此时无法知道最小数字 0 在哪个区间。
 
 ```java
 public int minNumberInRotateArray(int[] nums) {
@@ -737,7 +720,9 @@ private char[][] buildMatrix(char[] array) {
 
 ## 题目描述
 
-地上有一个 m 行和 n 列的方格。一个机器人从坐标 (0, 0) 的格子开始移动，每一次只能向左右上下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于 k 的格子。例如，当 k 为 18 时，机器人能够进入方格（35, 37），因为 3+5+3+7=18。但是，它不能进入方格（35, 38），因为 3+5+3+8=19。请问该机器人能够达到多少个格子？
+地上有一个 m 行和 n 列的方格。一个机器人从坐标 (0, 0) 的格子开始移动，每一次只能向左右上下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于 k 的格子。
+
+例如，当 k 为 18 时，机器人能够进入方格 (35,37)，因为 3+5+3+7=18。但是，它不能进入方格 (35,37)，因为 3+5+3+8=19。请问该机器人能够达到多少个格子？
 
 ## 解题思路
 
@@ -806,7 +791,7 @@ return 36 (10 = 3 + 3 + 4)
 
 ### 贪心
 
-尽可能多剪长度为 3 的绳子，并且不允许有长度为 1 的绳子出现，如果出现了，就从已经切好长度为 3 的绳子中拿出一段与长度为 1 的绳子重新组合，把它们切成两段长度为 2 的绳子。
+尽可能多剪长度为 3 的绳子，并且不允许有长度为 1 的绳子出现。如果出现了，就从已经切好长度为 3 的绳子中拿出一段与长度为 1 的绳子重新组合，把它们切成两段长度为 2 的绳子。
 
 证明：当 n >= 5 时，3(n - 3) - 2(n - 2) = n - 5 >= 0。因此把长度大于 5 的绳子切成两段，令其中一段长度为 3 可以使得两段的乘积最大。
 
@@ -847,19 +832,9 @@ public int integerBreak(int n) {
 
 输入一个整数，输出该数二进制表示中 1 的个数。
 
-### Integer.bitCount()
-
-```java
-public int NumberOf1(int n) {
-    return Integer.bitCount(n);
-}
-```
-
 ### n&(n-1)
 
-O(M) 时间复杂度解法，其中 M 表示 1 的个数。
-
-该位运算是去除 n 的位级表示中最低的那一位。
+该位运算去除 n 的位级表示中最低的那一位。
 
 ```
 n       : 10110100
@@ -867,6 +842,9 @@ n-1     : 10110011
 n&(n-1) : 10110000
 ```
 
+时间复杂度：O(M)，其中 M 表示 1 的个数。
+
+
 ```java
 public int NumberOf1(int n) {
     int cnt = 0;
@@ -878,13 +856,22 @@ public int NumberOf1(int n) {
 }
 ```
 
+
+### Integer.bitCount()
+
+```java
+public int NumberOf1(int n) {
+    return Integer.bitCount(n);
+}
+```
+
 # 16. 数值的整数次方
 
 [NowCoder](https://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00?tpId=13&tqId=11165&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
 
 ## 题目描述
 
-给定一个 double 类型的浮点数 base 和 int 类型的整数 exponent。求 base 的 exponent 次方。
+给定一个 double 类型的浮点数 base 和 int 类型的整数 exponent，求 base 的 exponent 次方。
 
 ## 解题思路
 
@@ -892,7 +879,7 @@ public int NumberOf1(int n) {
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?x^n=\left\{\begin{array}{rcl}(x*x)^{n/2}&&{n\%2=0}\\x*(x*x)^{n/2}&&{n\%2=1}\end{array}\right."/></div> <br>
 
-因为 (x\*x)<sup>n/2</sup> 可以通过递归求解，并且每递归一次，n 都减小一半，因此整个算法的时间复杂度为 O(logN)。
+因为 (x\*x)<sup>n/2</sup> 可以通过递归求解，并且每次递归 n 都减小一半，因此整个算法的时间复杂度为 O(logN)。
 
 ```java
 public double Power(double base, int exponent) {
@@ -1028,6 +1015,7 @@ public ListNode deleteDuplication(ListNode pHead) {
 
 ```java
 public boolean match(char[] str, char[] pattern) {
+
     int m = str.length, n = pattern.length;
     boolean[][] dp = new boolean[m + 1][n + 1];
 
@@ -1058,9 +1046,24 @@ public boolean match(char[] str, char[] pattern) {
 
 ## 题目描述
 
-请实现一个函数用来判断字符串是否表示数值（包括整数和小数）。
+```html
+true
+
+"+100"
+"5e2"
+"-123"
+"3.1416"
+"-1E-16"
+
+false
+
+"12e"
+"1a3.14"
+"1.2.3"
+"+-5"
+"12e+4.3"
+```
 
-例如，字符串 "+100","5e2","-123","3.1416" 和 "-1E-16" 都表示数值。但是 "12e","1a3.14","1.2.3","+-5" 和 "12e+4.3" 都不是。
 
 ## 解题思路
 
@@ -1124,8 +1127,7 @@ public void reOrderArray(int[] nums) {
 <div align="center"> <img src="../pics//ea2304ce-268b-4238-9486-4d8f8aea8ca4.png" width="500"/> </div><br>
 
 ```java
-public ListNode FindKthToTail(ListNode head, int k)
-{
+public ListNode FindKthToTail(ListNode head, int k) {
     if (head == null)
         return null;
     ListNode P1 = head;
@@ -1148,9 +1150,7 @@ public ListNode FindKthToTail(ListNode head, int k)
 
 ## 题目描述
 
-一个链表中包含环，请找出该链表的环的入口结点。
-
-要求不能使用额外的空间。
+一个链表中包含环，请找出该链表的环的入口结点。要求不能使用额外的空间。
 
 ## 解题思路
 
@@ -1161,8 +1161,7 @@ public ListNode FindKthToTail(ListNode head, int k)
 <div align="center"> <img src="../pics//2858f8ad-aedb-45a5-a706-e98c96d690fa.jpg" width="600"/> </div><br>
 
 ```java
-public ListNode EntryNodeOfLoop(ListNode pHead)
-{
+public ListNode EntryNodeOfLoop(ListNode pHead) {
     if (pHead == null || pHead.next == null)
         return null;
     ListNode slow = pHead, fast = pHead;
@@ -1304,8 +1303,6 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
 
 ## 解题思路
 
-### 递归
-
 ```java
 public void Mirror(TreeNode root) {
     if (root == null)
@@ -1322,29 +1319,6 @@ private void swap(TreeNode root) {
 }
 ```
 
-### 迭代
-
-```java
-public void Mirror(TreeNode root) {
-    Stack<TreeNode> stack = new Stack<>();
-    stack.push(root);
-    while (!stack.isEmpty()) {
-        TreeNode node = stack.pop();
-        if (node == null)
-            continue;
-        swap(node);
-        stack.push(node.left);
-        stack.push(node.right);
-    }
-}
-
-private void swap(TreeNode node) {
-    TreeNode t = node.left;
-    node.left = node.right;
-    node.right = t;
-}
-```
-
 # 28 对称的二叉树
 
 [NowCder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
@@ -1445,7 +1419,9 @@ public int min() {
 
 ## 题目描述
 
-输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。例如序列 1,2,3,4,5 是某栈的压入顺序，序列 4,5,3,2,1 是该压栈序列对应的一个弹出序列，但 4,3,5,1,2 就不可能是该压栈序列的弹出序列。
+输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。
+
+例如序列 1,2,3,4,5 是某栈的压入顺序，序列 4,5,3,2,1 是该压栈序列对应的一个弹出序列，但 4,3,5,1,2 就不可能是该压栈序列的弹出序列。
 
 ## 解题思路
 
@@ -1838,7 +1814,7 @@ private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s) {
 
 多数投票问题，可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。
 
-使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
+使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素相等时，令 cnt++，否则令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
 
 ```java
 public int MoreThanHalfNum_Solution(int[] nums) {

From 0fb30502e019f03667dc240d110fe8b869c8b1a4 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 21:44:50 +0800
Subject: [PATCH 115/172] auto commit

---
 notes/算法.md                                 | 1073 +++++++++--------
 pics/1a2f2998-d0da-41c8-8222-1fd95083a66b.png |  Bin 0 -> 8192 bytes
 pics/2a8e1442-2381-4439-a83f-0312c8678b1f.png |  Bin 0 -> 5120 bytes
 pics/37e79a32-95a9-4503-bdb1-159527e628b8.png |  Bin 0 -> 8192 bytes
 4 files changed, 598 insertions(+), 475 deletions(-)
 create mode 100644 pics/1a2f2998-d0da-41c8-8222-1fd95083a66b.png
 create mode 100644 pics/2a8e1442-2381-4439-a83f-0312c8678b1f.png
 create mode 100644 pics/37e79a32-95a9-4503-bdb1-159527e628b8.png

diff --git a/notes/算法.md b/notes/算法.md
index cb5d4dbe..e404a3c8 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -2,19 +2,10 @@
 * [一、前言](#一前言)
 * [二、算法分析](#二算法分析)
     * [数学模型](#数学模型)
+    * [注意事项](#注意事项)
     * [ThreeSum](#threesum)
     * [倍率实验](#倍率实验)
-    * [注意事项](#注意事项)
-* [三、栈和队列](#三栈和队列)
-    * [栈](#栈)
-    * [队列](#队列)
-* [四、并查集](#四并查集)
-    * [quick-find](#quick-find)
-    * [quick-union](#quick-union)
-    * [加权 quick-union](#加权-quick-union)
-    * [路径压缩的加权 quick-union](#路径压缩的加权-quick-union)
-    * [各种 union-find 算法的比较](#各种-union-find-算法的比较)
-* [五、排序](#五排序)
+* [三、排序](#三排序)
     * [选择排序](#选择排序)
     * [冒泡排序](#冒泡排序)
     * [插入排序](#插入排序)
@@ -23,6 +14,15 @@
     * [快速排序](#快速排序)
     * [堆排序](#堆排序)
     * [小结](#小结)
+* [四、并查集](#四并查集)
+    * [Quick Find](#quick-find)
+    * [Quick Union](#quick-union)
+    * [加权 Quick Union](#加权-quick-union)
+    * [路径压缩的加权 Quick Union](#路径压缩的加权-quick-union)
+    * [比较](#比较)
+* [五、栈和队列](#五栈和队列)
+    * [栈](#栈)
+    * [队列](#队列)
 * [六、查找](#六查找)
     * [初级实现](#初级实现)
     * [二叉查找树](#二叉查找树)
@@ -39,7 +39,8 @@
 
 # 一、前言
 
-本文实现代码以及测试代码放在 [Algorithm](https://github.com/CyC2018/Algorithm)
+- 实现代码：[Algorithm](https://github.com/CyC2018/Algorithm)
+- 绘图文件：[ProcessOn](https://www.processon.com/view/link/5a3e4c1ee4b0ce9ffea8c727)
 
 # 二、算法分析
 
@@ -61,6 +62,28 @@ N<sup>3</sup>/6-N<sup>2</sup>/2+N/3 的增长数量级为 O(N<sup>3</sup>)。增
 
 使用成本模型来评估算法，例如数组的访问次数就是一种成本模型。
 
+## 注意事项
+
+### 1. 大常数
+
+在求近似时，如果低级项的常数系数很大，那么近似的结果就是错误的。
+
+### 2. 缓存
+
+计算机系统会使用缓存技术来组织内存，访问数组相邻的元素会比访问不相邻的元素快很多。
+
+### 3. 对最坏情况下的性能的保证
+
+在核反应堆、心脏起搏器或者刹车控制器中的软件，最坏情况下的性能是十分重要的。
+
+### 4. 随机化算法
+
+通过打乱输入，去除算法对输入的依赖。
+
+### 5. 均摊分析
+
+将所有操作的总成本除于操作总数来将成本均摊。例如对一个空栈进行 N 次连续的 push() 调用需要访问数组的元素为 N+4+8+16+...+2N=5N-4（N 是向数组写入元素，其余的都是调整数组大小时进行复制需要的访问数组操作），均摊后每次操作访问数组的平均次数为常数。
+
 ## ThreeSum
 
 ThreeSum 用于统计一个数组中和为 0 的三元组数量。
@@ -71,6 +94,10 @@ public interface ThreeSum {
 }
 ```
 
+### 1. ThreeSumSlow
+
+该算法的内循环为 `if (nums[i] + nums[j] + nums[k] == 0)` 语句，总共执行的次数为 N(N-1)(N-2) = N<sup>3</sup>/6-N<sup>2</sup>/2+N/3，因此它的近似执行次数为 \~N<sup>3</sup>/6，增长数量级为 O(N<sup>3</sup>)。
+
 ```java
 public class ThreeSumSlow implements ThreeSum {
     @Override
@@ -87,9 +114,7 @@ public class ThreeSumSlow implements ThreeSum {
 }
 ```
 
-该算法的内循环为 `if (nums[i] + nums[j] + nums[k] == 0)` 语句，总共执行的次数为 N(N-1)(N-2) = N<sup>3</sup>/6-N<sup>2</sup>/2+N/3，因此它的近似执行次数为 \~N<sup>3</sup>/6，增长数量级为 O(N<sup>3</sup>)。
-
-<font size=4> **改进** </font></br>
+### 2. ThreeSumFast
 
 通过将数组先排序，对两个元素求和，并用二分查找方法查找是否存在该和的相反数，如果存在，就说明存在三元组的和为 0。
 
@@ -153,21 +178,31 @@ public class BinarySearch {
 
 ```java
 public class RatioTest {
+
     public static void main(String[] args) {
+
         int N = 500;
         int loopTimes = 7;
         double preTime = -1;
+
         while (loopTimes-- > 0) {
+
             int[] nums = new int[N];
+
             StopWatch.start();
+
             ThreeSum threeSum = new ThreeSumSlow();
+
             int cnt = threeSum.count(nums);
             System.out.println(cnt);
+
             double elapsedTime = StopWatch.elapsedTime();
             double ratio = preTime == -1 ? 0 : elapsedTime / preTime;
             System.out.println(N + "  " + elapsedTime + "  " + ratio);
+
             preTime = elapsedTime;
             N *= 2;
+
         }
     }
 }
@@ -175,12 +210,15 @@ public class RatioTest {
 
 ```java
 public class StopWatch {
+
     private static long start;
-    
-    public static void start(){
+
+
+    public static void start() {
         start = System.currentTimeMillis();
     }
-    
+
+
     public static double elapsedTime() {
         long now = System.currentTimeMillis();
         return (now - start) / 1000.0;
@@ -188,438 +226,7 @@ public class StopWatch {
 }
 ```
 
-## 注意事项
-
-### 1. 大常数
-
-在求近似时，如果低级项的常数系数很大，那么近似的结果就是错误的。
-
-### 2. 缓存
-
-计算机系统会使用缓存技术来组织内存，访问数组相邻的元素会比访问不相邻的元素快很多。
-
-### 3. 对最坏情况下的性能的保证
-
-在核反应堆、心脏起搏器或者刹车控制器中的软件，最坏情况下的性能是十分重要的。
-
-### 4. 随机化算法
-
-通过打乱输入，去除算法对输入的依赖。
-
-### 5. 均摊分析
-
-将所有操作的总成本除于操作总数来将成本均摊。例如对一个空栈进行 N 次连续的 push() 调用需要访问数组的元素为 N+4+8+16+...+2N=5N-4（N 是向数组写入元素，其余的都是调整数组大小时进行复制需要的访问数组操作），均摊后每次操作访问数组的平均次数为常数。
-
-# 三、栈和队列
-
-## 栈
-
-First-In-Last-Out
-
-```java
-public interface MyStack<Item> extends Iterable<Item> {
-    MyStack<Item> push(Item item);
-
-    Item pop() throws Exception;
-
-    boolean isEmpty();
-
-    int size();
-}
-```
-
-### 1. 数组实现
-
-```java
-public class ArrayStack<Item> implements MyStack<Item> {
-    // 栈元素数组，只能通过转型来创建泛型数组
-    private Item[] a = (Item[]) new Object[1];
-    // 元素数量
-    private int N = 0;
-
-    @Override
-    public MyStack<Item> push(Item item) {
-        check();
-        a[N++] = item;
-        return this;
-    }
-
-    @Override
-    public Item pop() throws Exception {
-        if (isEmpty())
-            throw new Exception("stack is empty");
-
-        Item item = a[--N];
-        check();
-        a[N] = null; // 避免对象游离
-        return item;
-    }
-
-    private void check() {
-        if (N >= a.length)
-            resize(2 * a.length);
-        else if (N > 0 && N <= a.length / 4)
-            resize(a.length / 2);
-    }
-
-    /**
-     * 调整数组大小，使得栈具有伸缩性
-     */
-    private void resize(int size) {
-        Item[] tmp = (Item[]) new Object[size];
-        for (int i = 0; i < N; i++)
-            tmp[i] = a[i];
-        a = tmp;
-    }
-
-    @Override
-    public boolean isEmpty() {
-        return N == 0;
-    }
-
-    @Override
-    public int size() {
-        return N;
-    }
-
-    @Override
-    public Iterator<Item> iterator() {
-        // 返回逆序遍历的迭代器
-        return new Iterator<Item>() {
-            private int i = N;
-
-            @Override
-            public boolean hasNext() {
-                return i > 0;
-            }
-
-            @Override
-            public Item next() {
-                return a[--i];
-            }
-        };
-    }
-}
-```
-
-### 2. 链表实现
-
-需要使用链表的头插法来实现，因为头插法中最后压入栈的元素在链表的开头，它的 next 指针指向前一个压入栈的元素，在弹出元素时就可以通过 next 指针遍历到前一个压入栈的元素从而让这个元素称为新的栈顶元素。
-
-```java
-public class ListStack<Item> implements MyStack<Item> {
-    private Node top = null;
-    private int N = 0;
-
-    private class Node {
-        Item item;
-        Node next;
-    }
-
-    @Override
-    public MyStack<Item> push(Item item) {
-        Node newTop = new Node();
-        newTop.item = item;
-        newTop.next = top;
-        top = newTop;
-        N++;
-        return this;
-    }
-
-    @Override
-    public Item pop() throws Exception {
-        if (isEmpty())
-            throw new Exception("stack is empty");
-        Item item = top.item;
-        top = top.next;
-        N--;
-        return item;
-    }
-
-    @Override
-    public boolean isEmpty() {
-        return N == 0;
-    }
-
-    @Override
-    public int size() {
-        return N;
-    }
-
-    @Override
-    public Iterator<Item> iterator() {
-        return new Iterator<Item>() {
-            private Node cur = top;
-
-            @Override
-            public boolean hasNext() {
-                return cur != null;
-            }
-
-            @Override
-            public Item next() {
-                Item item = cur.item;
-                cur = cur.next;
-                return item;
-            }
-        };
-    }
-}
-```
-
-## 队列
-
-First-In-First-Out
-
-下面是队列的链表实现，需要维护 first 和 last 节点指针，分别指向队首和队尾。
-
-这里需要考虑 first 和 last 指针哪个作为链表的开头。因为出队列操作需要让队首元素的下一个元素成为队首，所以需要容易获取下一个元素，而链表的头部节点的 next 指针指向下一个元素，因此可以让 first 指针链表的开头。
-
-```java
-public interface MyQueue<Item> extends Iterable<Item> {
-    int size();
-
-    boolean isEmpty();
-
-    MyQueue<Item> add(Item item);
-
-    Item remove() throws Exception;
-}
-```
-
-```java
-public class ListQueue<Item> implements MyQueue<Item> {
-    private Node first;
-    private Node last;
-    int N = 0;
-
-    private class Node {
-        Item item;
-        Node next;
-    }
-
-    @Override
-    public boolean isEmpty() {
-        return N == 0;
-    }
-
-    @Override
-    public int size() {
-        return N;
-    }
-
-    @Override
-    public MyQueue<Item> add(Item item) {
-        Node newNode = new Node();
-        newNode.item = item;
-        newNode.next = null;
-        if (isEmpty()) {
-            last = newNode;
-            first = newNode;
-        } else {
-            last.next = newNode;
-            last = newNode;
-        }
-        N++;
-        return this;
-    }
-
-    @Override
-    public Item remove() throws Exception {
-        if (isEmpty())
-            throw new Exception("queue is empty");
-        Node node = first;
-        first = first.next;
-        N--;
-        if (isEmpty())
-            last = null;
-        return node.item;
-    }
-
-    @Override
-    public Iterator<Item> iterator() {
-        return new Iterator<Item>() {
-            Node cur = first;
-
-            @Override
-            public boolean hasNext() {
-                return cur != null;
-            }
-
-            @Override
-            public Item next() {
-                Item item = cur.item;
-                cur = cur.next;
-                return item;
-            }
-        };
-    }
-}
-```
-
-# 四、并查集
-
-用于解决动态连通性问题，能动态连接两个点，并且判断两个点是否连通。
-
-<div align="center"> <img src="../pics//9d0a637c-6a8f-4f5a-99b9-fdcfa26793ff.png" width="400"/> </div><br>
-
-| 方法 | 描述 |
-| :---: | :---: |
-| UF(int N) | 构造一个大小为 N 的并查集 |
-| void union(int p, int q) | 连接 p 和 q 节点 |
-| int find(int p) | 查找 p 所在的连通分量 |
-| boolean connected(int p, int q) | 判断 p 和 q 节点是否连通 |
-
-```java
-public abstract class UF {
-    protected int[] id;
-
-    public UF(int N) {
-        id = new int[N];
-        for (int i = 0; i < N; i++)
-            id[i] = i;
-    }
-
-    public boolean connected(int p, int q) {
-        return find(p) == find(q);
-    }
-
-    public abstract int find(int p);
-
-    public abstract void union(int p, int q);
-}
-```
-
-## quick-find
-
-可以快速进行 find 操作，即可以快速判断两个节点是否连通。
-
-同一连通分量的所有节点的 id 值相等。
-
-但是 union 操作代价却很高，需要将其中一个连通分量中的所有节点 id 值都修改为另一个节点的 id 值。
-
-<div align="center"> <img src="../pics//8f0cc500-5994-4c7a-91a9-62885d658662.png" width="350"/> </div><br>
-
-```java
-public class QuickFindUF extends UF {
-    public QuickFindUF(int N) {
-        super(N);
-    }
-
-    @Override
-    public int find(int p) {
-        return id[p];
-    }
-
-    @Override
-    public void union(int p, int q) {
-        int pID = find(p);
-        int qID = find(q);
-
-        if (pID == qID)
-            return;
-
-        for (int i = 0; i < id.length; i++)
-            if (id[i] == pID)
-                id[i] = qID;
-    }
-}
-
-```
-
-## quick-union
-
-可以快速进行 union 操作，只需要修改一个节点的 id 值即可。
-
-但是 find 操作开销很大，因为同一个连通分量的节点 id 值不同，id 值只是用来指向另一个节点。因此需要一直向上查找操作，直到找到最上层的节点。
-
-<div align="center"> <img src="../pics//5d4a5181-65fb-4bf2-a9c6-899cab534b44.png" width="350"/> </div><br>
-
-```java
-public class QuickUnionUF extends UF {
-    public QuickUnionUF(int N) {
-        super(N);
-    }
-
-    @Override
-    public int find(int p) {
-        while (p != id[p])
-            p = id[p];
-        return p;
-    }
-
-    @Override
-    public void union(int p, int q) {
-        int pRoot = find(p);
-        int qRoot = find(q);
-        if (pRoot != qRoot)
-            id[pRoot] = qRoot;
-    }
-}
-```
-
-这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为触点的数目。
-
-<div align="center"> <img src="../pics//bfbb11e2-d208-4efa-b97b-24cd40467cd8.png" width="150"/> </div><br>
-
-## 加权 quick-union
-
-为了解决 quick-union 的树通常会很高的问题，加权 quick-union 在 union 操作时会让较小的树连接较大的树上面。
-
-理论研究证明，加权 quick-union 算法构造的树深度最多不超过 logN。
-
-<div align="center"> <img src="../pics//a4c17d43-fa5e-4935-b74e-147e7f7e782c.png" width="200"/> </div><br>
-
-```java
-public class WeightedQuickUnionUF extends UF {
-
-    // 保存节点的数量信息
-    private int[] sz;
-
-    public WeightedQuickUnionUF(int N) {
-        super(N);
-        this.sz = new int[N];
-        for (int i = 0; i < N; i++)
-            this.sz[i] = 1;
-    }
-
-    @Override
-    public int find(int p) {
-        while (p != id[p])
-            p = id[p];
-        return p;
-    }
-
-    @Override
-    public void union(int p, int q) {
-        int i = find(p);
-        int j = find(q);
-        if (i == j) return;
-        if (sz[i] < sz[j]) {
-            id[i] = j;
-            sz[j] += sz[i];
-        } else {
-            id[j] = i;
-            sz[i] += sz[j];
-        }
-    }
-}
-```
-
-## 路径压缩的加权 quick-union
-
-在检查节点的同时将它们直接链接到根节点，只需要在 find 中添加一个循环即可。
-
-## 各种 union-find 算法的比较
-
-| 算法 | union | find |
-| :---: | :---: | :---: |
-| quick-find | N | 1 |
-| quick-union | 树高 | 树高 |
-| 加权 quick-union | logN | logN |
-| 路径压缩的加权 quick-union | 非常接近 1 | 非常接近 1 |
-
-# 五、排序
+# 三、排序
 
 待排序的元素需要实现 Java 的 Comparable 接口，该接口有 compareTo() 方法，可以用它来判断两个元素的大小关系。
 
@@ -648,41 +255,49 @@ public abstract class Sort<T extends Comparable<T>> {
 
 选择出数组中的最小元素，将它与数组的第一个元素交换位置。再从剩下的元素中选择出最小的元素，将它与数组的第二个元素交换位置。不断进行这样的操作，直到将整个数组排序。
 
-<div align="center"> <img src="../pics//ed7b96ac-6428-4bd5-9986-674c54c2a959.png" width="250"/> </div><br>
+选择排序需要 \~N<sup>2</sup>/2 次比较和 \~N 次交换，它的运行时间与输入无关，这个特点使得它对一个已经排序的数组也需要这么多的比较和交换操作。
+
+<div align="center"> <img src="../pics//37e79a32-95a9-4503-bdb1-159527e628b8.png" width="250"/> </div><br>
 
 ```java
 public class Selection<T extends Comparable<T>> extends Sort<T> {
+
     @Override
     public void sort(T[] nums) {
         int N = nums.length;
         for (int i = 0; i < N; i++) {
             int min = i;
-            for (int j = i + 1; j < N; j++)
-                if (less(nums[j], nums[min]))
+            for (int j = i + 1; j < N; j++) {
+                if (less(nums[j], nums[min])) {
                     min = j;
+                }
+            }
             swap(nums, i, min);
         }
     }
 }
 ```
 
-选择排序需要 \~N<sup>2</sup>/2 次比较和 \~N 次交换，它的运行时间与输入无关，这个特点使得它对一个已经排序的数组也需要这么多的比较和交换操作。
-
 ## 冒泡排序
 
-通过从左到右不断交换相邻逆序的相邻元素，在一轮的交换之后，可以让未排序的元素上浮到右侧。
+从左到右不断交换相邻逆序的元素，在一轮的循环之后，可以让未排序的最大元素上浮到右侧。
 
 在一轮循环中，如果没有发生交换，就说明数组已经是有序的，此时可以直接退出。
 
+以下演示了在一轮循环中，将最大的元素 5 上浮到最右侧。
+
+<div align="center"> <img src="../pics//1a2f2998-d0da-41c8-8222-1fd95083a66b.png" width="250"/> </div><br>
+
 ```java
 public class Bubble<T extends Comparable<T>> extends Sort<T> {
+
     @Override
     public void sort(T[] nums) {
         int N = nums.length;
         boolean hasSorted = false;
-        for (int i = 0; i < N && !hasSorted; i++) {
+        for (int i = N - 1; i > 0 && !hasSorted; i--) {
             hasSorted = true;
-            for (int j = 0; j < N - i - 1; j++) {
+            for (int j = 0; j < i; j++) {
                 if (less(nums[j + 1], nums[j])) {
                     hasSorted = false;
                     swap(nums, j, j + 1);
@@ -695,23 +310,7 @@ public class Bubble<T extends Comparable<T>> extends Sort<T> {
 
 ## 插入排序
 
-插入排序从左到右进行，每次都将当前元素插入到左侧已经排序的数组中，使得插入之后左部数组依然有序。
-
-第 j 元素是通过不断向左比较并交换来实现插入过程：当第 j 元素小于第 j - 1 元素，就将它们的位置交换，然后令 j 指针向左移动一个位置，不断进行以上操作。
-
-<div align="center"> <img src="../pics//c9a1de44-b1c0-4d13-a654-827d4ef8a723.png" width="250"/> </div><br>
-
-```java
-public class Insertion<T extends Comparable<T>> extends Sort<T> {
-    @Override
-    public void sort(T[] nums) {
-        int N = nums.length;
-        for (int i = 1; i < N; i++)
-            for (int j = i; j > 0 && less(nums[j], nums[j - 1]); j--)
-                swap(nums, j, j - 1);
-    }
-}
-```
+每次都将当前元素插入到左侧已经排序的数组中，使得插入之后左侧数组依然有序。
 
 对于数组 {3, 5, 2, 4, 1}，它具有以下逆序：(3, 2), (3, 1), (5, 2), (5, 4), (5, 1), (2, 1), (4, 1)，插入排序每次只能交换相邻元素，令逆序数量减少 1，因此插入排序需要交换的次数为逆序数量。
 
@@ -721,6 +320,25 @@ public class Insertion<T extends Comparable<T>> extends Sort<T> {
 - 最坏的情况下需要 \~N<sup>2</sup>/2 比较以及 \~N<sup>2</sup>/2 次交换，最坏的情况是数组是倒序的；
 - 最好的情况下需要 N-1 次比较和 0 次交换，最好的情况就是数组已经有序了。
 
+以下演示了在一轮循环中，将元素 2 插入到左侧已经排序的数组中。
+
+<div align="center"> <img src="../pics//2a8e1442-2381-4439-a83f-0312c8678b1f.png" width="250"/> </div><br>
+
+```java
+public class Insertion<T extends Comparable<T>> extends Sort<T> {
+
+    @Override
+    public void sort(T[] nums) {
+        int N = nums.length;
+        for (int i = 1; i < N; i++) {
+            for (int j = i; j > 0 && less(nums[j], nums[j - 1]); j--) {
+                swap(nums, j, j - 1);
+            }
+        }
+    }
+}
+```
+
 ## 希尔排序
 
 对于大规模的数组，插入排序很慢，因为它只能交换相邻的元素，每次只能将逆序数量减少 1。
@@ -1137,6 +755,511 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 Java 主要排序方法为 java.util.Arrays.sort()，对于原始数据类型使用三向切分的快速排序，对于引用类型使用归并排序。
 
+# 四、并查集
+
+用于解决动态连通性问题，能动态连接两个点，并且判断两个点是否连通。
+
+<div align="center"> <img src="../pics//9d0a637c-6a8f-4f5a-99b9-fdcfa26793ff.png" width="400"/> </div><br>
+
+| 方法 | 描述 |
+| :---: | :---: |
+| UF(int N) | 构造一个大小为 N 的并查集 |
+| void union(int p, int q) | 连接 p 和 q 节点 |
+| int find(int p) | 查找 p 所在的连通分量 |
+| boolean connected(int p, int q) | 判断 p 和 q 节点是否连通 |
+
+```java
+public abstract class UF {
+
+    protected int[] id;
+
+    public UF(int N) {
+        id = new int[N];
+        for (int i = 0; i < N; i++) {
+            id[i] = i;
+        }
+    }
+
+    public boolean connected(int p, int q) {
+        return find(p) == find(q);
+    }
+
+    public abstract int find(int p);
+
+    public abstract void union(int p, int q);
+}
+```
+
+## Quick Find
+
+可以快速进行 find 操作，即可以快速判断两个节点是否连通。
+
+需要保证同一连通分量的所有节点的 id 值相等。
+
+但是 union 操作代价却很高，需要将其中一个连通分量中的所有节点 id 值都修改为另一个节点的 id 值。
+
+<div align="center"> <img src="../pics//8f0cc500-5994-4c7a-91a9-62885d658662.png" width="350"/> </div><br>
+
+```java
+public class QuickFindUF extends UF {
+
+    public QuickFindUF(int N) {
+        super(N);
+    }
+
+
+    @Override
+    public int find(int p) {
+        return id[p];
+    }
+
+
+    @Override
+    public void union(int p, int q) {
+
+        int pID = find(p);
+        int qID = find(q);
+
+        if (pID == qID) {
+            return;
+        }
+
+        for (int i = 0; i < id.length; i++) {
+            if (id[i] == pID) {
+                id[i] = qID;
+            }
+        }
+    }
+}
+```
+
+## Quick Union
+
+可以快速进行 union 操作，只需要修改一个节点的 id 值即可。
+
+但是 find 操作开销很大，因为同一个连通分量的节点 id 值不同，id 值只是用来指向另一个节点。因此需要一直向上查找操作，直到找到最上层的节点。
+
+<div align="center"> <img src="../pics//5d4a5181-65fb-4bf2-a9c6-899cab534b44.png" width="350"/> </div><br>
+
+```java
+public class QuickUnionUF extends UF {
+
+    public QuickUnionUF(int N) {
+        super(N);
+    }
+
+
+    @Override
+    public int find(int p) {
+
+        while (p != id[p]) {
+            p = id[p];
+        }
+        return p;
+    }
+
+
+    @Override
+    public void union(int p, int q) {
+
+        int pRoot = find(p);
+        int qRoot = find(q);
+
+        if (pRoot != qRoot) {
+            id[pRoot] = qRoot;
+        }
+    }
+}
+```
+
+这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为触点的数目。
+
+<div align="center"> <img src="../pics//bfbb11e2-d208-4efa-b97b-24cd40467cd8.png" width="150"/> </div><br>
+
+## 加权 Quick Union
+
+为了解决 quick-union 的树通常会很高的问题，加权 quick-union 在 union 操作时会让较小的树连接较大的树上面。
+
+理论研究证明，加权 quick-union 算法构造的树深度最多不超过 logN。
+
+<div align="center"> <img src="../pics//a4c17d43-fa5e-4935-b74e-147e7f7e782c.png" width="200"/> </div><br>
+
+```java
+public class WeightedQuickUnionUF extends UF {
+
+    // 保存节点的数量信息
+    private int[] sz;
+
+
+    public WeightedQuickUnionUF(int N) {
+        super(N);
+        this.sz = new int[N];
+        for (int i = 0; i < N; i++) {
+            this.sz[i] = 1;
+        }
+    }
+
+
+    @Override
+    public int find(int p) {
+        while (p != id[p]) {
+            p = id[p];
+        }
+        return p;
+    }
+
+
+    @Override
+    public void union(int p, int q) {
+
+        int i = find(p);
+        int j = find(q);
+
+        if (i == j) return;
+
+        if (sz[i] < sz[j]) {
+            id[i] = j;
+            sz[j] += sz[i];
+        } else {
+            id[j] = i;
+            sz[i] += sz[j];
+        }
+    }
+}
+```
+
+## 路径压缩的加权 Quick Union
+
+在检查节点的同时将它们直接链接到根节点，只需要在 find 中添加一个循环即可。
+
+## 比较
+
+| 算法 | union | find |
+| :---: | :---: | :---: |
+| Quick Find | N | 1 |
+| Quick Union | 树高 | 树高 |
+| 加权 Quick Union | logN | logN |
+| 路径压缩的加权 Quick Union | 非常接近 1 | 非常接近 1 |
+
+# 五、栈和队列
+
+## 栈
+
+```java
+public interface MyStack<Item> extends Iterable<Item> {
+
+    MyStack<Item> push(Item item);
+
+    Item pop() throws Exception;
+
+    boolean isEmpty();
+
+    int size();
+
+}
+```
+
+### 1. 数组实现
+
+```java
+public class ArrayStack<Item> implements MyStack<Item> {
+
+    // 栈元素数组，只能通过转型来创建泛型数组
+    private Item[] a = (Item[]) new Object[1];
+
+    // 元素数量
+    private int N = 0;
+
+
+    @Override
+    public MyStack<Item> push(Item item) {
+        check();
+        a[N++] = item;
+        return this;
+    }
+
+
+    @Override
+    public Item pop() throws Exception {
+
+        if (isEmpty()) {
+            throw new Exception("stack is empty");
+        }
+
+        Item item = a[--N];
+
+        check();
+
+        // 避免对象游离
+        a[N] = null;
+
+        return item;
+    }
+
+
+    private void check() {
+
+        if (N >= a.length) {
+            resize(2 * a.length);
+
+        } else if (N > 0 && N <= a.length / 4) {
+            resize(a.length / 2);
+        }
+    }
+
+
+    /**
+     * 调整数组大小，使得栈具有伸缩性
+     */
+    private void resize(int size) {
+
+        Item[] tmp = (Item[]) new Object[size];
+
+        for (int i = 0; i < N; i++) {
+            tmp[i] = a[i];
+        }
+
+        a = tmp;
+    }
+
+
+    @Override
+    public boolean isEmpty() {
+        return N == 0;
+    }
+
+
+    @Override
+    public int size() {
+        return N;
+    }
+
+
+    @Override
+    public Iterator<Item> iterator() {
+
+        // 返回逆序遍历的迭代器
+        return new Iterator<Item>() {
+
+            private int i = N;
+
+            @Override
+            public boolean hasNext() {
+                return i > 0;
+            }
+
+            @Override
+            public Item next() {
+                return a[--i];
+            }
+        };
+
+    }
+}
+```
+
+### 2. 链表实现
+
+需要使用链表的头插法来实现，因为头插法中最后压入栈的元素在链表的开头，它的 next 指针指向前一个压入栈的元素，在弹出元素时就可以通过 next 指针遍历到前一个压入栈的元素从而让这个元素称为新的栈顶元素。
+
+```java
+public class ListStack<Item> implements MyStack<Item> {
+
+    private Node top = null;
+    private int N = 0;
+
+
+    private class Node {
+        Item item;
+        Node next;
+    }
+
+
+    @Override
+    public MyStack<Item> push(Item item) {
+
+        Node newTop = new Node();
+
+        newTop.item = item;
+        newTop.next = top;
+
+        top = newTop;
+
+        N++;
+
+        return this;
+    }
+
+
+    @Override
+    public Item pop() throws Exception {
+
+        if (isEmpty()) {
+            throw new Exception("stack is empty");
+        }
+
+        Item item = top.item;
+
+        top = top.next;
+        N--;
+
+        return item;
+    }
+
+
+    @Override
+    public boolean isEmpty() {
+        return N == 0;
+    }
+
+
+    @Override
+    public int size() {
+        return N;
+    }
+
+
+    @Override
+    public Iterator<Item> iterator() {
+
+        return new Iterator<Item>() {
+
+            private Node cur = top;
+
+
+            @Override
+            public boolean hasNext() {
+                return cur != null;
+            }
+
+
+            @Override
+            public Item next() {
+                Item item = cur.item;
+                cur = cur.next;
+                return item;
+            }
+        };
+
+    }
+}
+```
+
+## 队列
+
+First-In-First-Out
+
+下面是队列的链表实现，需要维护 first 和 last 节点指针，分别指向队首和队尾。
+
+这里需要考虑 first 和 last 指针哪个作为链表的开头。因为出队列操作需要让队首元素的下一个元素成为队首，所以需要容易获取下一个元素，而链表的头部节点的 next 指针指向下一个元素，因此可以让 first 指针链表的开头。
+
+```java
+public interface MyQueue<Item> extends Iterable<Item> {
+
+    int size();
+
+    boolean isEmpty();
+
+    MyQueue<Item> add(Item item);
+
+    Item remove() throws Exception;
+}
+```
+
+```java
+public class ListQueue<Item> implements MyQueue<Item> {
+
+    private Node first;
+    private Node last;
+    int N = 0;
+
+
+    private class Node {
+        Item item;
+        Node next;
+    }
+
+
+    @Override
+    public boolean isEmpty() {
+        return N == 0;
+    }
+
+
+    @Override
+    public int size() {
+        return N;
+    }
+
+
+    @Override
+    public MyQueue<Item> add(Item item) {
+
+        Node newNode = new Node();
+        newNode.item = item;
+        newNode.next = null;
+
+        if (isEmpty()) {
+            last = newNode;
+            first = newNode;
+        } else {
+            last.next = newNode;
+            last = newNode;
+        }
+
+        N++;
+        return this;
+    }
+
+
+    @Override
+    public Item remove() throws Exception {
+
+        if (isEmpty()) {
+            throw new Exception("queue is empty");
+        }
+
+        Node node = first;
+        first = first.next;
+        N--;
+
+        if (isEmpty()) {
+            last = null;
+        }
+
+        return node.item;
+    }
+
+
+    @Override
+    public Iterator<Item> iterator() {
+
+        return new Iterator<Item>() {
+
+            Node cur = first;
+
+
+            @Override
+            public boolean hasNext() {
+                return cur != null;
+            }
+
+
+            @Override
+            public Item next() {
+                Item item = cur.item;
+                cur = cur.next;
+                return item;
+            }
+        };
+    }
+}
+```
+
+
+
+
+
 # 六、查找
 
 符号表（Symbol Table）是一种存储键值对的数据结构，可以支持快速查找操作。
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From 6478389db1ccb3cbab94c5478d3955ad6e4e7898 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 22:16:30 +0800
Subject: [PATCH 116/172] auto commit

---
 notes/算法.md                                 |  92 ++++++++++++------
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 5 files changed, 61 insertions(+), 31 deletions(-)
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diff --git a/notes/算法.md b/notes/算法.md
index e404a3c8..f0e889db 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -347,25 +347,32 @@ public class Insertion<T extends Comparable<T>> extends Sort<T> {
 
 希尔排序使用插入排序对间隔 h 的序列进行排序。通过不断减小 h，最后令 h=1，就可以使得整个数组是有序的。
 
-<div align="center"> <img src="../pics//cdbe1d12-5ad9-4acb-a717-bbc822c2acf3.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//0157d362-98dd-4e51-ac26-00ecb76beb3e.png" width="500"/> </div><br>
 
 ```java
 public class Shell<T extends Comparable<T>> extends Sort<T> {
+
     @Override
     public void sort(T[] nums) {
+
         int N = nums.length;
         int h = 1;
-        while (h < N / 3)
-            h = 3 * h + 1;  // 1, 4, 13, 40, ...
+
+        while (h < N / 3) {
+            h = 3 * h + 1; // 1, 4, 13, 40, ...
+        }
 
         while (h >= 1) {
-            for (int i = h; i < N; i++)
-                for (int j = i; j >= h && less(nums[j], nums[j - h]); j -= h)
+            for (int i = h; i < N; i++) {
+                for (int j = i; j >= h && less(nums[j], nums[j - h]); j -= h) {
                     swap(nums, j, j - h);
+                }
+            }
             h = h / 3;
         }
     }
 }
+
 ```
 
 希尔排序的运行时间达不到平方级别，使用递增序列 1, 4, 13, 40, ...  的希尔排序所需要的比较次数不会超过 N 的若干倍乘于递增序列的长度。后面介绍的高级排序算法只会比希尔排序快两倍左右。
@@ -374,7 +381,7 @@ public class Shell<T extends Comparable<T>> extends Sort<T> {
 
 归并排序的思想是将数组分成两部分，分别进行排序，然后归并起来。
 
-<div align="center"> <img src="../pics//8dfb4cc9-26da-45e7-b820-4576fa1cbb0e.png" width="350"/> </div><br>
+<div align="center"> <img src="../pics//220790c6-4377-4a2e-8686-58398afc8a18.png" width="350"/> </div><br>
 
 ### 1. 归并方法
 
@@ -385,21 +392,28 @@ public abstract class MergeSort<T extends Comparable<T>> extends Sort<T> {
 
     protected T[] aux;
 
+
     protected void merge(T[] nums, int l, int m, int h) {
+
         int i = l, j = m + 1;
 
-        for (int k = l; k <= h; k++)
-            aux[k] = nums[k];        // 将数据复制到辅助数组
+        for (int k = l; k <= h; k++) {
+            aux[k] = nums[k]; // 将数据复制到辅助数组
+        }
 
         for (int k = l; k <= h; k++) {
-            if (i > m)
+            if (i > m) {
                 nums[k] = aux[j++];
-            else if (j > h)
+
+            } else if (j > h) {
                 nums[k] = aux[i++];
-            else if (aux[i].compareTo(nums[j]) <= 0)
-                nums[k] = aux[i++];  // 先进行这一步，保证稳定性
-            else
+
+            } else if (aux[i].compareTo(nums[j]) <= 0) {
+                nums[k] = aux[i++]; // 先进行这一步，保证稳定性
+
+            } else {
                 nums[k] = aux[j++];
+            }
         }
     }
 }
@@ -407,10 +421,13 @@ public abstract class MergeSort<T extends Comparable<T>> extends Sort<T> {
 
 ### 2. 自顶向下归并排序
 
-<div align="center"> <img src="../pics//0c55e11c-d3ce-4cd8-b139-028aea6f40e3.png" width="450"/> </div><br>
+将一个大数组分成两个小数组去求解。
+
+因为每次都将问题对半分成两个子问题，而这种对半分的算法复杂度一般为 O(NlogN)，因此该归并排序方法的时间复杂度也为 O(NlogN)。
 
 ```java
 public class Up2DownMergeSort<T extends Comparable<T>> extends MergeSort<T> {
+
     @Override
     public void sort(T[] nums) {
         aux = (T[]) new Comparable[nums.length];
@@ -418,8 +435,9 @@ public class Up2DownMergeSort<T extends Comparable<T>> extends MergeSort<T> {
     }
 
     private void sort(T[] nums, int l, int h) {
-        if (h <= l)
+        if (h <= l) {
             return;
+        }
         int mid = l + (h - l) / 2;
         sort(nums, l, mid);
         sort(nums, mid + 1, h);
@@ -428,23 +446,27 @@ public class Up2DownMergeSort<T extends Comparable<T>> extends MergeSort<T> {
 }
 ```
 
-因为每次都将问题对半分成两个子问题，而这种对半分的算法复杂度一般为 O(NlogN)，因此该归并排序方法的时间复杂度也为 O(NlogN)。
-
 ### 3. 自底向上归并排序
 
 先归并那些微型数组，然后成对归并得到的微型数组。
 
 ```java
 public class Down2UpMergeSort<T extends Comparable<T>> extends MergeSort<T> {
+
     @Override
     public void sort(T[] nums) {
+
         int N = nums.length;
         aux = (T[]) new Comparable[N];
-        for (int sz = 1; sz < N; sz += sz)
-            for (int lo = 0; lo < N - sz; lo += sz + sz)
+
+        for (int sz = 1; sz < N; sz += sz) {
+            for (int lo = 0; lo < N - sz; lo += sz + sz) {
                 merge(nums, lo, lo + sz - 1, Math.min(lo + sz + sz - 1, N - 1));
+            }
+        }
     }
 }
+
 ```
 
 ## 快速排序
@@ -454,10 +476,11 @@ public class Down2UpMergeSort<T extends Comparable<T>> extends MergeSort<T> {
 - 归并排序将数组分为两个子数组分别排序，并将有序的子数组归并使得整个数组排序；
 - 快速排序通过一个切分元素将数组分为两个子数组，左子数组小于等于切分元素，右子数组大于等于切分元素，将这两个子数组排序也就将整个数组排序了。
 
-<div align="center"> <img src="../pics//ab77240d-7338-4547-9183-00215e7220ec.png" width="500"/> </div><br>
+<div align="center"> <img src="../pics//f8047846-efd4-42be-b6b7-27a7c4998b51.png" width="500"/> </div><br>
 
 ```java
 public class QuickSort<T extends Comparable<T>> extends Sort<T> {
+
     @Override
     public void sort(T[] nums) {
         shuffle(nums);
@@ -482,9 +505,9 @@ public class QuickSort<T extends Comparable<T>> extends Sort<T> {
 
 ### 2. 切分
 
-取 a[lo] 作为切分元素，然后从数组的左端向右扫描直到找到第一个大于等于它的元素，再从数组的右端向左扫描找到第一个小于等于它的元素，交换这两个元素，并不断进行这个过程，就可以保证左指针 i 的左侧元素都不大于切分元素，右指针 j 的右侧元素都不小于切分元素。当两个指针相遇时，将切分元素 a[lo] 和 a[j] 交换位置。
+取 a[l] 作为切分元素，然后从数组的左端向右扫描直到找到第一个大于等于它的元素，再从数组的右端向左扫描找到第一个小于等于它的元素，交换这两个元素。不断进行这个过程，就可以保证左指针 i 的左侧元素都不大于切分元素，右指针 j 的右侧元素都不小于切分元素。当两个指针相遇时，将切分元素 a[l] 和 a[j] 交换位置。
 
-<div align="center"> <img src="../pics//5aac64d3-2c7b-4f32-9e9a-1df2186f588b.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//864bfa7d-1149-420c-a752-f9b3d4e782ec.png" width="400"/> </div><br>
 
 ```java
 private int partition(T[] nums, int l, int h) {
@@ -528,20 +551,23 @@ private int partition(T[] nums, int l, int h) {
 
 ```java
 public class ThreeWayQuickSort<T extends Comparable<T>> extends QuickSort<T> {
+
     @Override
     protected void sort(T[] nums, int l, int h) {
-        if (h <= l)
+        if (h <= l) {
             return;
+        }
         int lt = l, i = l + 1, gt = h;
         T v = nums[l];
         while (i <= gt) {
             int cmp = nums[i].compareTo(v);
-            if (cmp < 0)
+            if (cmp < 0) {
                 swap(nums, lt++, i++);
-            else if (cmp > 0)
+            } else if (cmp > 0) {
                 swap(nums, i, gt--);
-            else
+            } else {
                 i++;
+            }
         }
         sort(nums, l, lt - 1);
         sort(nums, gt + 1, h);
@@ -555,24 +581,28 @@ public class ThreeWayQuickSort<T extends Comparable<T>> extends QuickSort<T> {
 
 可以利用这个特性找出数组的第 k 个元素。
 
+该算法是线性级别的，因为每次能将数组二分，那么比较的总次数为 (N+N/2+N/4+..)，直到找到第 k 个元素，这个和显然小于 2N。
+
 ```java
 public T select(T[] nums, int k) {
     int l = 0, h = nums.length - 1;
     while (h > l) {
         int j = partition(nums, l, h);
-        if (j == k)
+
+        if (j == k) {
             return nums[k];
-        else if (j > k)
+
+        } else if (j > k) {
             h = j - 1;
-        else
+
+        } else {
             l = j + 1;
+        }
     }
     return nums[k];
 }
 ```
 
-该算法是线性级别的。因为每次能将数组二分，那么比较的总次数为 (N+N/2+N/4+..)，直到找到第 k 个元素，这个和显然小于 2N。
-
 ## 堆排序
 
 ### 1. 堆
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From d655c65eff7d1914e961d0e414edf7df735b979c Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 22:23:46 +0800
Subject: [PATCH 117/172] auto commit

---
 notes/算法.md | 16 ++++++++--------
 1 file changed, 8 insertions(+), 8 deletions(-)

diff --git a/notes/算法.md b/notes/算法.md
index f0e889db..8ce9682d 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -770,14 +770,14 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 | 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
-| 选择排序 | no | N<sup>2</sup> | 1 | |
-| 冒泡排序 | yes |  N<sup>2</sup> | 1 | |
-| 插入排序 | yes |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
-| 希尔排序 | no |  N 的若干倍乘于递增序列的长度 | 1 | |
-| 快速排序 | no | NlogN | logN | |
-| 三向切分快速排序 | no |  N \~ NlogN | logN | 适用于有大量重复主键|
-| 归并排序 | yes |  NlogN | N | |
-| 堆排序 | no |  NlogN | 1 | | |
+| 选择排序 | :heavy_check_mark:| N<sup>2</sup> | 1 | |
+| 冒泡排序 | :heavy_check_mark: |  N<sup>2</sup> | 1 | |
+| 插入排序 | :heavy_check_mark: |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
+| 希尔排序 | :heavy_multiplication_x:  |  N 的若干倍乘于递增序列的长度 | 1 | |
+| 快速排序 | :heavy_multiplication_x:  | NlogN | logN | |
+| 三向切分快速排序 | :heavy_multiplication_x:  |  N \~ NlogN | logN | 适用于有大量重复主键|
+| 归并排序 | :heavy_check_mark: |  NlogN | N | |
+| 堆排序 | :heavy_multiplication_x:  |  NlogN | 1 | | |
 
 快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其他线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
 

From a14c6fd5fdf082be7b9307013b7b1ec7b9958003 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 22:27:28 +0800
Subject: [PATCH 118/172] auto commit

---
 notes/算法.md | 16 ++++++++--------
 1 file changed, 8 insertions(+), 8 deletions(-)

diff --git a/notes/算法.md b/notes/算法.md
index 8ce9682d..30467b42 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -770,14 +770,14 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 | 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
-| 选择排序 | :heavy_check_mark:| N<sup>2</sup> | 1 | |
-| 冒泡排序 | :heavy_check_mark: |  N<sup>2</sup> | 1 | |
-| 插入排序 | :heavy_check_mark: |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
-| 希尔排序 | :heavy_multiplication_x:  |  N 的若干倍乘于递增序列的长度 | 1 | |
-| 快速排序 | :heavy_multiplication_x:  | NlogN | logN | |
-| 三向切分快速排序 | :heavy_multiplication_x:  |  N \~ NlogN | logN | 适用于有大量重复主键|
-| 归并排序 | :heavy_check_mark: |  NlogN | N | |
-| 堆排序 | :heavy_multiplication_x:  |  NlogN | 1 | | |
+| 选择排序 | √| N<sup>2</sup> | 1 | |
+| 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
+| 插入排序 | √ |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
+| 希尔排序 | ×  |  N 的若干倍乘于递增序列的长度 | 1 | |
+| 快速排序 | ×  | NlogN | logN | |
+| 三向切分快速排序 | ×  |  N \~ NlogN | logN | 适用于有大量重复主键|
+| 归并排序 | √ |  NlogN | N | |
+| 堆排序 | ×  |  NlogN | 1 | | |
 
 快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其他线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
 

From aa4dfe2ea77394aacbd002eaffb55c03cf2575f8 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 22:38:23 +0800
Subject: [PATCH 119/172] auto commit

---
 notes/Java 虚拟机.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 891211da..a4a4ba12 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -45,7 +45,7 @@
 可以通过 -Xss 这个虚拟机参数来指定每个线程的 Java 虚拟机栈内存大小：
 
 ```java
-java -Xss=512M HackTheJava
+java -Xss512M HackTheJava
 ```
 
 该区域可能抛出以下异常：
@@ -81,7 +81,7 @@ java -Xss=512M HackTheJava
 可以通过 -Xms 和 -Xmx 两个虚拟机参数来指定一个程序的堆内存大小，第一个参数设置初始值，第二个参数设置最大值。
 
 ```java
-java -Xms=1M -Xmx=2M HackTheJava
+java -Xms1M -Xmx2M HackTheJava
 ```
 
 ## 方法区

From 3da8d737e631103d1415590d223aa2b61a920a5a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Wed, 15 Aug 2018 22:55:12 +0800
Subject: [PATCH 120/172] auto commit

---
 notes/Java 基础.md | 17 +++++++++++++++--
 1 file changed, 15 insertions(+), 2 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index adf899fa..b329e0bc 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -576,9 +576,22 @@ SuperExtendExample.func()
 
 ## 重写与重载
 
-- 重写（Override）存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法。子类的返回值类型要等于或者小于父类的返回值；
+**1. 重写（Override）** 
 
-- 重载（Overload）存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。应该注意的是，返回值不同，其它都相同不算是重载。
+存在于继承体系中，指子类实现了一个与父类在方法声明上完全相同的一个方法。
+
+为了满足里式替换原则，重写有有以下两个限制：
+
+- 子类方法的访问权限必须大于等于父类方法；
+- 子类方法的返回类型必须是父类方法返回类型或为其子类型。
+
+使用 @Override 注解，可以让编译器帮忙检查是否满足上面的两个限制条件。
+
+**2. 重载（Overload）** 
+
+存在于同一个类中，指一个方法与已经存在的方法名称上相同，但是参数类型、个数、顺序至少有一个不同。
+
+应该注意的是，返回值不同，其它都相同不算是重载。
 
 # 五、Object 通用方法
 

From 56a271b4337ec2f25f46bc3d79267acb7d62946d Mon Sep 17 00:00:00 2001
From: wangxujian-laixuzhui
 <33362969+wangxujian-laixuzhui@users.noreply.github.com>
Date: Thu, 16 Aug 2018 08:49:34 +0800
Subject: [PATCH 121/172] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E7=BA=BF=E7=A8=8B?=
 =?UTF-8?q?=E4=B8=89=E7=A7=8D=E5=AE=9E=E7=8E=B0=E6=96=B9=E5=BC=8F=E7=9A=84?=
 =?UTF-8?q?=E6=AF=94=E8=BE=83?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 5 ++++-
 1 file changed, 4 insertions(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 9cd5955b..1a1bacc0 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -182,7 +182,10 @@ public static void main(String[] args) {
     mt.start();
 }
 ```
-
+## 三种实现方式的比较
+- 实现Runnable接又可以避免Java单继承特性而带来的局限;增强程序的健壮性，代码能够被多个线程共享，代码与数据是独立的;适合多个相同程序代码的线程区处理同一资源的情况。
+- 继承Thread类和实现Runnable方法启动线 程都是使用start方法，然后JVM虚拟机将此线程放到就绪队列中，如果有处理机可用， 则执行run方法。
+- 实现Callable接又要实现call方法，并且线 程执行完毕后会有返回值。其他的两种都是 重写run方法，没有返回值。
 ## 实现接口 VS 继承 Thread
 
 实现接口会更好一些，因为：

From 9f1e252e36d7f9a35222b72ecadcd25fbb73b917 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 16 Aug 2018 22:34:21 +0800
Subject: [PATCH 122/172] auto commit

---
 notes/MySQL.md          | 314 ++++++++++++++++++++--------------------
 notes/Redis.md          |   6 +-
 notes/数据库系统原理.md |  86 +++++------
 3 files changed, 201 insertions(+), 205 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 7ac9685f..54067977 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -1,22 +1,23 @@
 <!-- GFM-TOC -->
-* [一、存储引擎](#一存储引擎)
+* [一、索引](#一索引)
+    * [B+ Tree 原理](#b-tree-原理)
+    * [MySQL 索引](#mysql-索引)
+    * [索引优化](#索引优化)
+    * [索引的优点](#索引的优点)
+    * [索引的使用场景](#索引的使用场景)
+* [二、查询性能优化](#二查询性能优化)
+    * [使用 Explain 进行分析](#使用-explain-进行分析)
+    * [优化数据访问](#优化数据访问)
+    * [重构查询方式](#重构查询方式)
+* [三、存储引擎](#三存储引擎)
     * [InnoDB](#innodb)
     * [MyISAM](#myisam)
     * [比较](#比较)
-* [二、数据类型](#二数据类型)
+* [四、数据类型](#四数据类型)
     * [整型](#整型)
     * [浮点数](#浮点数)
     * [字符串](#字符串)
     * [时间和日期](#时间和日期)
-* [三、索引](#三索引)
-    * [B+ Tree 原理](#b-tree-原理)
-    * [索引分类](#索引分类)
-    * [索引的优点](#索引的优点)
-    * [索引优化](#索引优化)
-* [四、查询性能优化](#四查询性能优化)
-    * [使用 Explain 进行分析](#使用-explain-进行分析)
-    * [优化数据访问](#优化数据访问)
-    * [重构查询方式](#重构查询方式)
 * [五、切分](#五切分)
     * [水平切分](#水平切分)
     * [垂直切分](#垂直切分)
@@ -29,101 +30,7 @@
 <!-- GFM-TOC -->
 
 
-# 一、存储引擎
-
-## InnoDB
-
-InnoDB 是 MySQL 默认的事务型存储引擎，只有在需要 InnoDB 不支持的特性时，才考虑使用其它存储引擎。
-
-实现了四个标准的隔离级别，默认级别是可重复读（REPEATABLE READ）。在可重复读隔离级别下，通过多版本并发控制（MVCC）+ 间隙锁（next-key locking）防止幻影读。
-
-主索引是聚簇索引，在索引中保存了数据，从而避免直接读取磁盘，因此对查询性能有很大的提升。
-
-内部做了很多优化，包括从磁盘读取数据时采用的可预测性读、能够加快读操作并且自动创建的自适应哈希索引、能够加速插入操作的插入缓冲区等。
-
-支持真正的在线热备份。其它存储引擎不支持在线热备份，要获取一致性视图需要停止对所有表的写入，而在读写混合场景中，停止写入可能也意味着停止读取。
-
-## MyISAM
-
-MyISAM 设计简单，数据以紧密格式存储。对于只读数据，或者表比较小、可以容忍修复操作，则依然可以使用 MyISAM。
-
-MyISAM 提供了大量的特性，包括压缩表、空间数据索引等。
-
-不支持事务。
-
-不支持行级锁，只能对整张表加锁，读取时会对需要读到的所有表加共享锁，写入时则对表加排它锁。但在表有读取操作的同时，也可以往表中插入新的记录，这被称为并发插入（CONCURRENT INSERT）。
-
-可以手工或者自动执行检查和修复操作，但是和事务恢复以及崩溃恢复不同，可能导致一些数据丢失，而且修复操作是非常慢的。
-
-如果指定了 DELAY_KEY_WRITE 选项，在每次修改执行完成时，不会立即将修改的索引数据写入磁盘，而是会写到内存中的键缓冲区，只有在清理键缓冲区或者关闭表的时候才会将对应的索引块写入磁盘。这种方式可以极大的提升写入性能，但是在数据库或者主机崩溃时会造成索引损坏，需要执行修复操作。
-
-## 比较
-
-- 事务：InnoDB 是事务型的，可以使用 Commit 和 Rollback 语句。
-
-- 并发：MyISAM 只支持表级锁，而 InnoDB 还支持行级锁。
-
-- 外键：InnoDB 支持外键。
-
-- 备份：InnoDB 支持在线热备份。
-
-- 崩溃恢复：MyISAM 崩溃后发生损坏的概率比 InnoDB 高很多，而且恢复的速度也更慢。
-
-- 其它特性：MyISAM 支持压缩表和空间数据索引。
-
-# 二、数据类型
-
-## 整型
-
-TINYINT, SMALLINT, MEDIUMINT, INT, BIGINT 分别使用 8, 16, 24, 32, 64 位存储空间，一般情况下越小的列越好。
-
-INT(11) 中的数字只是规定了交互工具显示字符的个数，对于存储和计算来说是没有意义的。
-
-## 浮点数
-
-FLOAT 和 DOUBLE 为浮点类型，DECIMAL 为高精度小数类型。CPU 原生支持浮点运算，但是不支持 DECIMAl 类型的计算，因此 DECIMAL 的计算比浮点类型需要更高的代价。
-
-FLOAT、DOUBLE 和 DECIMAL 都可以指定列宽，例如 DECIMAL(18, 9) 表示总共 18 位，取 9 位存储小数部分，剩下 9 位存储整数部分。
-
-## 字符串
-
-主要有 CHAR 和 VARCHAR 两种类型，一种是定长的，一种是变长的。
-
-VARCHAR 这种变长类型能够节省空间，因为只需要存储必要的内容。但是在执行 UPDATE 时可能会使行变得比原来长，当超出一个页所能容纳的大小时，就要执行额外的操作。MyISAM 会将行拆成不同的片段存储，而 InnoDB 则需要分裂页来使行放进页内。
-
-VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
-
-## 时间和日期
-
-MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
-
-### 1. DATETIME
-
-能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
-
-它与时区无关。
-
-默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
-
-### 2. TIMESTAMP
-
-和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年 到 2038 年。
-
-它和时区有关，也就是说一个时间戳在不同的时区所代表的具体时间是不同的。
-
-MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提供了 UNIX_TIMESTAMP() 函数把日期转换为 UNIX 时间戳。
-
-默认情况下，如果插入时没有指定 TIMESTAMP 列的值，会将这个值设置为当前时间。
-
-应该尽量使用 TIMESTAMP，因为它比 DATETIME 空间效率更高。
-
-# 三、索引
-
-索引能够轻易将查询性能提升几个数量级。
-
-对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效。对于中到大型的表，索引就非常有效。但是对于特大型的表，建立和维护索引的代价将会随之增长。这种情况下，需要用到一种技术可以直接区分出需要查询的一组数据，而不是一条记录一条记录地匹配，例如可以使用分区技术。
-
-索引是在存储引擎层实现的，而不是在服务器层实现的，所以不同存储引擎具有不同的索引类型和实现。
+# 一、索引
 
 ## B+ Tree 原理
 
@@ -141,37 +48,37 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 进行查找操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
-插入删除操作记录会破坏平衡树的平衡性，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作。
+插入删除操作记录会破坏平衡树的平衡性，因此在插入删除操作之后，需要对树进行一个分裂、合并、旋转等操作来维护平衡性。
 
 ### 3. 与红黑树的比较
 
 红黑树等平衡树也可以用来实现索引，但是文件系统及数据库系统普遍采用 B+ Tree 作为索引结构，主要有以下两个原因：
 
-（一）更少的检索次数
+（一）更少的查找次数
 
-平衡树检索数据的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
+平衡树查找操作的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
 
-红黑树的出度为 2，而 B+ Tree 的出度一般都非常大。红黑树的树高 h 很明显比 B+ Tree 大非常多，因此检索的次数也就更多。
+红黑树的出度为 2，而 B+ Tree 的出度一般都非常大，所以红黑树的树高 h 很明显比 B+ Tree 大非常多，检索的次数也就更多。
 
 （二）利用计算机预读特性
 
-为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。这样做的理论依据是计算机科学中著名的局部性原理：当一个数据被用到时，其附近的数据也通常会马上被使用。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，因此速度会非常快。
+为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，因此速度会非常快。
 
 操作系统一般将内存和磁盘分割成固态大小的块，每一块称为一页，内存与磁盘以页为单位交换数据。数据库系统将索引的一个节点的大小设置为页的大小，使得一次 I/O 就能完全载入一个节点，并且可以利用预读特性，相邻的节点也能够被预先载入。
 
-## 索引分类
+## MySQL 索引
+
+索引是在存储引擎层实现的，而不是在服务器层实现的，所以不同存储引擎具有不同的索引类型和实现。
 
 ### 1. B+Tree 索引
 
-B+Tree 索引是大多数 MySQL 存储引擎的默认索引类型。
+是大多数 MySQL 存储引擎的默认索引类型。
 
 因为不再需要进行全表扫描，只需要对树进行搜索即可，因此查找速度快很多。除了用于查找，还可以用于排序和分组。
 
 可以指定多个列作为索引列，多个索引列共同组成键。
 
-B+Tree 索引适用于全键值、键值范围和键前缀查找，其中键前缀查找只适用于最左前缀查找。
-
-如果不是按照索引列的顺序进行查找，则无法使用索引。
+适用于全键值、键值范围和键前缀查找，其中键前缀查找只适用于最左前缀查找。如果不是按照索引列的顺序进行查找，则无法使用索引。
 
 InnoDB 的 B+Tree 索引分为主索引和辅助索引。
 
@@ -185,12 +92,12 @@ InnoDB 的 B+Tree 索引分为主索引和辅助索引。
 
 ### 2. 哈希索引
 
-InnoDB 引擎有一个特殊的功能叫“自适应哈希索引”，当某个索引值被使用的非常频繁时，会在 B+Tree 索引之上再创建一个哈希索引，这样就让 B+Tree 索引具有哈希索引的一些优点，比如快速的哈希查找。
-
 哈希索引能以 O(1) 时间进行查找，但是失去了有序性，它具有以下限制：
 
 - 无法用于排序与分组；
-- 只支持精确查找，无法用于部分查找和范围查找；
+- 只支持精确查找，无法用于部分查找和范围查找。
+
+InnoDB 存储引擎有一个特殊的功能叫“自适应哈希索引”，当某个索引值被使用的非常频繁时，会在 B+Tree 索引之上再创建一个哈希索引，这样就让 B+Tree 索引具有哈希索引的一些优点，比如快速的哈希查找。
 
 ### 3. 全文索引
 
@@ -200,20 +107,12 @@ MyISAM 存储引擎支持全文索引，用于查找文本中的关键词，而
 
 InnoDB 存储引擎在 MySQL 5.6.4 版本中也开始支持全文索引。
 
-### 4. 空间数据索引（R-Tree）
+### 4. 空间数据索引
 
-MyISAM 存储引擎支持空间数据索引，可以用于地理数据存储。空间数据索引会从所有维度来索引数据，可以有效地使用任意维度来进行组合查询。
+MyISAM 存储引擎支持空间数据索引（R-Tree），可以用于地理数据存储。空间数据索引会从所有维度来索引数据，可以有效地使用任意维度来进行组合查询。
 
 必须使用 GIS 相关的函数来维护数据。
 
-## 索引的优点
-
-- 大大减少了服务器需要扫描的数据行数。
-
-- 帮助服务器避免进行排序和创建临时表（B+Tree 索引是有序的，可以用来做 ORDER BY 和 GROUP BY 操作）；
-
-- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，也就将相邻的数据都存储在一起）。
-
 ## 索引优化
 
 ### 1. 独立的列
@@ -266,11 +165,25 @@ customer_id_selectivity: 0.0373
 
 具有以下优点：
 
-- 因为索引条目通常远小于数据行的大小，所以若只读取索引，能大大减少数据访问量。
+- 索引通常远小于数据行的大小，只读取索引能大大减少数据访问量。
 - 一些存储引擎（例如 MyISAM）在内存中只缓存索引，而数据依赖于操作系统来缓存。因此，只访问索引可以不使用系统调用（通常比较费时）。
 - 对于 InnoDB 引擎，若辅助索引能够覆盖查询，则无需访问主索引。
 
-# 四、查询性能优化
+## 索引的优点
+
+- 大大减少了服务器需要扫描的数据行数。
+
+- 帮助服务器避免进行排序和分组，也就不需要创建临时表（B+Tree 索引是有序的，可以用于 ORDER BY 和 GROUP BY 操作。临时表主要是在排序和分组过程中创建，因为不需要排序和分组，也就不需要创建临时表）。
+
+- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，也就将相邻的数据都存储在一起）。
+
+## 索引的使用场景
+
+- 对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效。
+- 对于中到大型的表，索引就非常有效。
+- 但是对于特大型的表，建立和维护索引的代价将会随之增长。这种情况下，需要用到一种技术可以直接区分出需要查询的一组数据，而不是一条记录一条记录地匹配，例如可以使用分区技术。
+
+# 二、查询性能优化
 
 ## 使用 Explain 进行分析
 
@@ -282,23 +195,13 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 - key : 使用的索引
 - rows : 扫描的行数
 
-更多内容请参考：[MySQL 性能优化神器 Explain 使用分析](https://segmentfault.com/a/1190000008131735)
-
 ## 优化数据访问
 
 ### 1. 减少请求的数据量
 
-（一）只返回必要的列
-
-最好不要使用 SELECT * 语句。
-
-（二）只返回必要的行
-
-使用 WHERE 语句进行查询过滤，有时候也需要使用 LIMIT 语句来限制返回的数据。
-
-（三）缓存重复查询的数据
-
-使用缓存可以避免在数据库中进行查询，特别要查询的数据经常被重复查询，缓存可以带来的查询性能提升将会是非常明显的。
+- 只返回必要的列：最好不要使用 SELECT * 语句。
+- 只返回必要的行：使用 LIMIT 语句来限制返回的数据。
+- 缓存重复查询的数据：使用缓存可以避免在数据库中进行查询，特别在要查询的数据经常被重复查询时，缓存带来的查询性能提升将会是非常明显的。
 
 ### 2. 减少服务器端扫描的行数
 
@@ -324,12 +227,12 @@ do {
 
 ### 2. 分解大连接查询
 
-将一个大连接查询（JOIN）分解成对每一个表进行一次单表查询，然后将结果在应用程序中进行关联，这样做的好处有：
+将一个大连接查询分解成对每一个表进行一次单表查询，然后将结果在应用程序中进行关联，这样做的好处有：
 
 - 让缓存更高效。对于连接查询，如果其中一个表发生变化，那么整个查询缓存就无法使用。而分解后的多个查询，即使其中一个表发生变化，对其它表的查询缓存依然可以使用。
 - 分解成多个单表查询，这些单表查询的缓存结果更可能被其它查询使用到，从而减少冗余记录的查询。
 - 减少锁竞争；
-- 在应用层进行连接，可以更容易对数据库进行拆分，从而更容易做到高性能和可扩展。
+- 在应用层进行连接，可以更容易对数据库进行拆分，从而更容易做到高性能和可伸缩。
 - 查询本身效率也可能会有所提升。例如下面的例子中，使用 IN() 代替连接查询，可以让 MySQL 按照 ID 顺序进行查询，这可能比随机的连接要更高效。
 
 ```sql
@@ -345,23 +248,111 @@ SELECT * FROM tag_post WHERE tag_id=1234;
 SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 ```
 
+# 三、存储引擎
+
+## InnoDB
+
+是 MySQL 默认的事务型存储引擎，只有在需要它不支持的特性时，才考虑使用其它存储引擎。
+
+实现了四个标准的隔离级别，默认级别是可重复读（REPEATABLE READ）。在可重复读隔离级别下，通过多版本并发控制（MVCC）+ 间隙锁（Next-Key Locking）防止幻影读。
+
+主索引是聚簇索引，在索引中保存了数据，从而避免直接读取磁盘，因此对查询性能有很大的提升。
+
+内部做了很多优化，包括从磁盘读取数据时采用的可预测性读、能够加快读操作并且自动创建的自适应哈希索引、能够加速插入操作的插入缓冲区等。
+
+支持真正的在线热备份。其它存储引擎不支持在线热备份，要获取一致性视图需要停止对所有表的写入，而在读写混合场景中，停止写入可能也意味着停止读取。
+
+## MyISAM
+
+设计简单，数据以紧密格式存储。对于只读数据，或者表比较小、可以容忍修复操作，则依然可以使用它。
+
+提供了大量的特性，包括压缩表、空间数据索引等。
+
+不支持事务。
+
+不支持行级锁，只能对整张表加锁，读取时会对需要读到的所有表加共享锁，写入时则对表加排它锁。但在表有读取操作的同时，也可以往表中插入新的记录，这被称为并发插入（CONCURRENT INSERT）。
+
+可以手工或者自动执行检查和修复操作，但是和事务恢复以及崩溃恢复不同，可能导致一些数据丢失，而且修复操作是非常慢的。
+
+如果指定了 DELAY_KEY_WRITE 选项，在每次修改执行完成时，不会立即将修改的索引数据写入磁盘，而是会写到内存中的键缓冲区，只有在清理键缓冲区或者关闭表的时候才会将对应的索引块写入磁盘。这种方式可以极大的提升写入性能，但是在数据库或者主机崩溃时会造成索引损坏，需要执行修复操作。
+
+## 比较
+
+- 事务：InnoDB 是事务型的，可以使用 Commit 和 Rollback 语句。
+
+- 并发：MyISAM 只支持表级锁，而 InnoDB 还支持行级锁。
+
+- 外键：InnoDB 支持外键。
+
+- 备份：InnoDB 支持在线热备份。
+
+- 崩溃恢复：MyISAM 崩溃后发生损坏的概率比 InnoDB 高很多，而且恢复的速度也更慢。
+
+- 其它特性：MyISAM 支持压缩表和空间数据索引。
+
+# 四、数据类型
+
+## 整型
+
+TINYINT, SMALLINT, MEDIUMINT, INT, BIGINT 分别使用 8, 16, 24, 32, 64 位存储空间，一般情况下越小的列越好。
+
+INT(11) 中的数字只是规定了交互工具显示字符的个数，对于存储和计算来说是没有意义的。
+
+## 浮点数
+
+FLOAT 和 DOUBLE 为浮点类型，DECIMAL 为高精度小数类型。CPU 原生支持浮点运算，但是不支持 DECIMAl 类型的计算，因此 DECIMAL 的计算比浮点类型需要更高的代价。
+
+FLOAT、DOUBLE 和 DECIMAL 都可以指定列宽，例如 DECIMAL(18, 9) 表示总共 18 位，取 9 位存储小数部分，剩下 9 位存储整数部分。
+
+## 字符串
+
+主要有 CHAR 和 VARCHAR 两种类型，一种是定长的，一种是变长的。
+
+VARCHAR 这种变长类型能够节省空间，因为只需要存储必要的内容。但是在执行 UPDATE 时可能会使行变得比原来长，当超出一个页所能容纳的大小时，就要执行额外的操作。MyISAM 会将行拆成不同的片段存储，而 InnoDB 则需要分裂页来使行放进页内。
+
+VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
+
+## 时间和日期
+
+MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
+
+### 1. DATETIME
+
+能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
+
+它与时区无关。
+
+默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
+
+### 2. TIMESTAMP
+
+和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年 到 2038 年。
+
+它和时区有关，也就是说一个时间戳在不同的时区所代表的具体时间是不同的。
+
+MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提供了 UNIX_TIMESTAMP() 函数把日期转换为 UNIX 时间戳。
+
+默认情况下，如果插入时没有指定 TIMESTAMP 列的值，会将这个值设置为当前时间。
+
+应该尽量使用 TIMESTAMP，因为它比 DATETIME 空间效率更高。
+
 # 五、切分
 
 ## 水平切分
 
-<div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
-
 水平切分又称为 Sharding，它是将同一个表中的记录拆分到多个结构相同的表中。
 
 当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓存单个数据库的压力。
 
+<div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
+
 ## 垂直切分
 
 <div align="center"> <img src="../pics//e130e5b8-b19a-4f1e-b860-223040525cf6.jpg"/> </div><br>
 
 垂直切分是将一张表按列切分成多个表，通常是按照列的关系密集程度进行切分，也可以利用垂直切分将经常被使用的列和不经常被使用的列切分到不同的表中。
 
-在数据库的层面使用垂直切分将按数据库中表的密集程度部署到不同的库中，例如将原来的电商数据库垂直切分成商品数据库 payDB、用户数据库 userDB 等。
+在数据库的层面使用垂直切分将按数据库中表的密集程度部署到不同的库中，例如将原来的电商数据库垂直切分成商品数据库、用户数据库等。
 
 ## Sharding 策略
 
@@ -375,15 +366,15 @@ SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 
 使用分布式事务来解决，比如 XA 接口。
 
-### 2. JOIN
+### 2. 链接
 
-可以将原来的 JOIN 查询分解成多个单表查询，然后在用户程序中进行 JOIN。
+可以将原来的 JOIN 分解成多个单表查询，然后在用户程序中进行 JOIN。
 
 ### 3. ID 唯一性
 
-- 使用全局唯一 ID：GUID。
-- 为每个分片指定一个 ID 范围。
-- 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)。
+- 使用全局唯一 ID：GUID
+- 为每个分片指定一个 ID 范围
+- 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)
 
 更多内容请参考：
 
@@ -396,24 +387,24 @@ SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 
 主要涉及三个线程：binlog 线程、I/O 线程和 SQL 线程。
 
--  **binlog 线程** ：负责将主服务器上的数据更改写入二进制文件（binlog）中。
--  **I/O 线程** ：负责从主服务器上读取二进制日志文件，并写入从服务器的中继日志中。
+-  **binlog 线程** ：负责将主服务器上的数据更改写入二进制日志中。
+-  **I/O 线程** ：负责从主服务器上读取二进制日志，并写入从服务器的中继日志中。
 -  **SQL 线程** ：负责读取中继日志并重放其中的 SQL 语句。
 
 <div align="center"> <img src="../pics//master-slave.png"/> </div><br>
 
 ## 读写分离
 
-主服务器用来处理写操作以及实时性要求比较高的读操作，而从服务器用来处理读操作。
+主服务器处理写操作以及实时性要求比较高的读操作，而从服务器处理读操作。
 
-读写分离常用代理方式来实现，代理服务器接收应用层传来的读写请求，然后决定转发到哪个服务器。
-
-MySQL 读写分离能提高性能的原因在于：
+读写分离能提高性能的原因在于：
 
 - 主从服务器负责各自的读和写，极大程度缓解了锁的争用；
-- 从服务器可以配置 MyISAM 引擎，提升查询性能以及节约系统开销；
+- 从服务器可以使用 MyISAM，提升查询性能以及节约系统开销；
 - 增加冗余，提高可用性。
 
+读写分离常用代理方式来实现，代理服务器接收应用层传来的读写请求，然后决定转发到哪个服务器。
+
 <div align="center"> <img src="../pics//master-slave-proxy.png"/> </div><br>
 
 # 参考资料
@@ -425,3 +416,4 @@ MySQL 读写分离能提高性能的原因在于：
 - [How to create unique row ID in sharded databases?](https://stackoverflow.com/questions/788829/how-to-create-unique-row-id-in-sharded-databases)
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
 - [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
+- [MySQL 性能优化神器 Explain 使用分析](https://segmentfault.com/a/1190000008131735)
diff --git a/notes/Redis.md b/notes/Redis.md
index 0981abf0..8f231382 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -49,7 +49,7 @@
 
 Redis 是速度非常快的非关系型（NoSQL）内存键值数据库，可以存储键和五种不同类型的值之间的映射。
 
-键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、有序集合、散列表。
+键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、散列表、有序集合。
 
 Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，使用复制来扩展读性能，使用分片来扩展写性能。
 
@@ -58,7 +58,7 @@ Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，
 | 数据类型 | 可以存储的值 | 操作 |
 | :--: | :--: | :--: |
 | STRING | 字符串、整数或者浮点数 | 对整个字符串或者字符串的其中一部分执行操作</br> 对整数和浮点数执行自增或者自减操作 |
-| LIST | 列表 | 从两端压入或者弹出元素</br> 读取单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
+| LIST | 列表 | 从两端压入或者弹出元素</br> 对单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
 | SET | 无序集合 | 添加、获取、移除单个元素</br> 检查一个元素是否存在于集合中</br> 计算交集、并集、差集</br> 从集合里面随机获取元素 |
 | HASH | 包含键值对的无序散列表 | 添加、获取、移除单个键值对</br> 获取所有键值对</br> 检查某个键是否存在|
 | ZSET | 有序集合 | 添加、获取、删除元素</br> 根据分值范围或者成员来获取元素</br> 计算一个键的排名 |
@@ -555,7 +555,7 @@ Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状
 
 分片是将数据划分为多个部分的方法，可以将数据存储到多台机器里面，这种方法在解决某些问题时可以获得线性级别的性能提升。
 
-假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... 等等，有不同的方式来选择一个指定的键存储在哪个实例中。
+假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... ，有不同的方式来选择一个指定的键存储在哪个实例中。
 
 - 最简单的方式是范围分片，例如用户 id 从 0\~1000 的存储到实例 R0 中，用户 id 从 1001\~2000 的存储到实例 R1 中，等等。但是这样需要维护一张映射范围表，维护操作代价很高。
 - 还有一种方式是哈希分片，使用 CRC32 哈希函数将键转换为一个数字，再对实例数量求模就能知道应该存储的实例。
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index e7ef804b..1bcd2533 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -20,6 +20,7 @@
     * [可串行化（SERIALIZABLE）](#可串行化serializable)
 * [五、多版本并发控制](#五多版本并发控制)
     * [版本号](#版本号)
+    * [隐藏的列](#隐藏的列)
     * [Undo 日志](#undo-日志)
     * [实现过程](#实现过程)
     * [快照读与当前读](#快照读与当前读)
@@ -58,9 +59,7 @@
 
 ### 2. 一致性（Consistency）
 
-数据库在事务执行前后都保持一致性状态。
-
-在一致性状态下，所有事务对一个数据的读取结果都是相同的。
+数据库在事务执行前后都保持一致性状态。在一致性状态下，所有事务对一个数据的读取结果都是相同的。
 
 ### 3. 隔离性（Isolation）
 
@@ -78,10 +77,10 @@
 
 - 只有满足一致性，事务的执行结果才是正确的。
 - 在无并发的情况下，事务串行执行，隔离性一定能够满足。此时只要能满足原子性，就一定能满足一致性。
-- 在并发的情况下，多个事务并发执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
+- 在并发的情况下，多个事务并行执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
 - 事务满足持久化是为了能应对数据库崩溃的情况。
 
-<div align="center"> <img src="../pics//a58e294a-615d-4ea0-9fbf-064a6daec4b2.png" width="400"/> </div><br>
+<div align="center"> <img src="../pics//a58e294a-615d-4ea0-9fbf-064a6daec4b2.png" width="450"/> </div><br>
 
 ## AUTOCOMMIT
 
@@ -95,25 +94,25 @@ MySQL 默认采用自动提交模式。也就是说，如果不显式使用`STAR
 
 T<sub>1</sub> 和 T<sub>2</sub> 两个事务都对一个数据进行修改，T<sub>1</sub> 先修改，T<sub>2</sub> 随后修改，T<sub>2</sub> 的修改覆盖了 T<sub>1</sub> 的修改。
 
-<div align="center"> <img src="../pics//88ff46b3-028a-4dbb-a572-1f062b8b96d3.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//88ff46b3-028a-4dbb-a572-1f062b8b96d3.png" width="350"/> </div><br>
 
 ## 读脏数据
 
 T<sub>1</sub> 修改一个数据，T<sub>2</sub> 随后读取这个数据。如果 T<sub>1</sub> 撤销了这次修改，那么 T<sub>2</sub> 读取的数据是脏数据。
 
-<div align="center"> <img src="../pics//dd782132-d830-4c55-9884-cfac0a541b8e.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//dd782132-d830-4c55-9884-cfac0a541b8e.png" width="400"/> </div><br>
 
 ## 不可重复读
 
 T<sub>2</sub> 读取一个数据，T<sub>1</sub> 对该数据做了修改。如果 T<sub>2</sub> 再次读取这个数据，此时读取的结果和第一次读取的结果不同。
 
-<div align="center"> <img src="../pics//c8d18ca9-0b09-441a-9a0c-fb063630d708.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//c8d18ca9-0b09-441a-9a0c-fb063630d708.png" width="350"/> </div><br>
 
 ## 幻影读
 
 T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插入新的数据，T<sub>1</sub> 再次读取这个范围的数据，此时读取的结果和和第一次读取的结果不同。
 
-<div align="center"> <img src="../pics//72fe492e-f1cb-4cfc-92f8-412fb3ae6fec.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//72fe492e-f1cb-4cfc-92f8-412fb3ae6fec.png" width="350"/> </div><br>
 
 ----
 
@@ -123,7 +122,6 @@ T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插
 
 ## 封锁粒度
 
-
 MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 应该尽量只锁定需要修改的那部分数据，而不是所有的资源。锁定的数据量越少，发生锁争用的可能就越小，系统的并发程度就越高。
@@ -150,8 +148,8 @@ MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 | - | X | S |
 | :--: | :--: | :--: |
-|X|NO|NO|
-|S|NO|YES|
+|X|×|×|
+|S|×|√|
 
 ### 2. 意向锁
 
@@ -170,10 +168,10 @@ MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 | - | X | IX | S | IS |
 | :--: | :--: | :--: | :--: | :--: |
-|X     |NO    |NO    |NO   | NO|
-|IX    |NO    |YES   |NO   | YES|
-|S     |NO    |NO    |YES  | YES|
-|IS    |NO    |YES  |YES  | YES|
+|X     |×    |×    |×   | ×|
+|IX    |×    |√   |×   | √|
+|S     |×    |×    |√  | √|
+|IS    |×    |√  |√  | √|
 
 解释如下：
 
@@ -298,32 +296,30 @@ SELECT ... FOR UPDATE;
 
 | 隔离级别 | 脏读 | 不可重复读 | 幻影读 |
 | :---: | :---: | :---:| :---: |
-| 未提交读 | YES | YES | YES |
-| 提交读 | NO | YES | YES |
-| 可重复读 | NO | NO | YES |
-| 可串行化 | NO | NO | NO |
+| 未提交读 | √ | √ | √ |
+| 提交读 | × | √ | √ |
+| 可重复读 | × | × | √ |
+| 可串行化 | × | × | × |
 
 # 五、多版本并发控制
 
-多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。
-
-而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC。
-
-可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
+多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC。可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
 
 ## 版本号
 
 - 系统版本号：是一个递增的数字，每开始一个新的事务，系统版本号就会自动递增。
 - 事务版本号：事务开始时的系统版本号。
 
-InooDB 的 MVCC 在每行记录后面都保存着两个隐藏的列，用来存储两个版本号：
+## 隐藏的列
+
+MVCC 在每行记录后面都保存着两个隐藏的列，用来存储两个版本号：
 
 - 创建版本号：指示创建一个数据行的快照时的系统版本号；
 - 删除版本号：如果该快照的删除版本号大于当前事务版本号表示该快照有效，否则表示该快照已经被删除了。
 
 ## Undo 日志
 
-InnoDB 的 MVCC 使用到的快照存储在 Undo 日志中，该日志通过回滚指针把一个数据行（Record）的所有快照连接起来。
+MVCC 使用到的快照存储在 Undo 日志中，该日志通过回滚指针把一个数据行（Record）的所有快照连接起来。
 
 <div align="center"> <img src="../pics//e41405a8-7c05-4f70-8092-e961e28d3112.jpg" width=""/> </div><br>
 
@@ -331,15 +327,13 @@ InnoDB 的 MVCC 使用到的快照存储在 Undo 日志中，该日志通过回
 
 以下实现过程针对可重复读隔离级别。
 
-### 1. SELECT
-
 当开始新一个事务时，该事务的版本号肯定会大于当前所有数据行快照的创建版本号，理解这一点很关键。
 
+### 1. SELECT
+
 多个事务必须读取到同一个数据行的快照，并且这个快照是距离现在最近的一个有效快照。但是也有例外，如果有一个事务正在修改该数据行，那么它可以读取事务本身所做的修改，而不用和其它事务的读取结果一致。
 
-把没有对一个数据行做修改的事务称为 T，T 所要读取的数据行快照的创建版本号必须小于 T 的版本号，因为如果大于或者等于 T 的版本号，那么表示该数据行快照是其它事务的最新修改，因此不能去读取它。
-
-除了上面的要求，T 所要读取的数据行快照的删除版本号必须大于 T 的版本号，因为如果小于等于 T 的版本号，那么表示该数据行快照是已经被删除的，不应该去读取它。
+把没有对一个数据行做修改的事务称为 T，T 所要读取的数据行快照的创建版本号必须小于 T 的版本号，因为如果大于或者等于 T 的版本号，那么表示该数据行快照是其它事务的最新修改，因此不能去读取它。除此之外，T 所要读取的数据行快照的删除版本号必须大于 T 的版本号，因为如果小于等于 T 的版本号，那么表示该数据行快照是已经被删除的，不应该去读取它。
 
 ### 2. INSERT
 
@@ -385,7 +379,7 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 锁定一个记录上的索引，而不是记录本身。
 
-如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
+如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚簇索引，因此 Record Locks 依然可以使用。
 
 ## Gap Locks
 
@@ -397,7 +391,7 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 ## Next-Key Locks
 
-它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定范围内的索引。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
+它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定索引之间的间隙。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
 
 ```sql
 (negative infinity, 10]
@@ -432,10 +426,10 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 不符合范式的关系，会产生很多异常，主要有以下四种异常：
 
-- 冗余数据：例如 学生-2 出现了两次。
+- 冗余数据：例如 `学生-2` 出现了两次。
 - 修改异常：修改了一个记录中的信息，但是另一个记录中相同的信息却没有被修改。
-- 删除异常：删除一个信息，那么也会丢失其它信息。例如如果删除了 课程-1，需要删除第一行和第三行，那么 学生-1 的信息就会丢失。
-- 插入异常，例如想要插入一个学生的信息，如果这个学生还没选课，那么就无法插入。
+- 删除异常：删除一个信息，那么也会丢失其它信息。例如删除了 `课程-1` 需要删除第一行和第三行，那么 `学生-1` 的信息就会丢失。
+- 插入异常：例如想要插入一个学生的信息，如果这个学生还没选课，那么就无法插入。
 
 ## 范式
 
@@ -506,7 +500,11 @@ Sname, Sdept 和 Mname 都部分依赖于键码，当一个学生选修了多门
 
 非主属性不传递函数依赖于键码。
 
-上面的 关系-1 中存在以下传递函数依赖：Sno -> Sdept -> Mname，可以进行以下分解：
+上面的 关系-1 中存在以下传递函数依赖：
+
+- Sno -> Sdept -> Mname
+
+可以进行以下分解：
 
 关系-11
 
@@ -533,13 +531,19 @@ Entity-Relationship，有三个组成部分：实体、属性、联系。
 
 包含一对一，一对多，多对多三种。
 
-如果 A 到 B 是一对多关系，那么画个带箭头的线段指向 B；如果是一对一，画两个带箭头的线段；如果是多对多，画两个不带箭头的线段。下图的 Course 和 Student 是一对多的关系。
+- 如果 A 到 B 是一对多关系，那么画个带箭头的线段指向 B；
+- 如果是一对一，画两个带箭头的线段；
+- 如果是多对多，画两个不带箭头的线段。
+
+下图的 Course 和 Student 是一对多的关系。
 
 <div align="center"> <img src="../pics//292b4a35-4507-4256-84ff-c218f108ee31.jpg" width=""/> </div><br>
 
 ## 表示出现多次的关系
 
-一个实体在联系出现几次，就要用几条线连接。下图表示一个课程的先修关系，先修关系出现两个 Course 实体，第一个是先修课程，后一个是后修课程，因此需要用两条线来表示这种关系。
+一个实体在联系出现几次，就要用几条线连接。
+
+下图表示一个课程的先修关系，先修关系出现两个 Course 实体，第一个是先修课程，后一个是后修课程，因此需要用两条线来表示这种关系。
 
 <div align="center"> <img src="../pics//8b798007-e0fb-420c-b981-ead215692417.jpg" width=""/> </div><br>
 
@@ -549,7 +553,7 @@ Entity-Relationship，有三个组成部分：实体、属性、联系。
 
 <div align="center"> <img src="../pics//423f2a40-bee1-488e-b460-8e76c48ee560.png" width=""/> </div><br>
 
-一般只使用二元联系，可以把多元关系转换为二元关系。
+一般只使用二元联系，可以把多元联系转换为二元联系。
 
 <div align="center"> <img src="../pics//de9b9ea0-1327-4865-93e5-6f805c48bc9e.png" width=""/> </div><br>
 

From c6b517553a562ecec4505f453bff14fd818e3b59 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Fri, 17 Aug 2018 21:19:16 +0800
Subject: [PATCH 123/172] auto commit

---
 notes/Java 容器.md      |   2 +-
 notes/Java 并发.md      |   2 +
 notes/Leetcode 题解.md  | 180 +++++++++++------------
 notes/Linux.md          |  20 +--
 notes/MySQL.md          | 314 ++++++++++++++++++++--------------------
 notes/Redis.md          |   6 +-
 notes/数据库系统原理.md |  86 +++++------
 notes/设计模式.md       |  76 +++++-----
 notes/面向对象思想.md   |  15 +-
 9 files changed, 349 insertions(+), 352 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 9959e2ca..001f864c 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -742,7 +742,7 @@ new capacity : 00100000
 
 对于一个 Key，
 
-- 它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；
+- 它的哈希值如果在第 5 位上为 0，那么取模得到的结果和之前一样；
 - 如果为 1，那么得到的结果为原来的结果 +16。
 
 ### 7. 扩容-计算数组容量
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 9cd5955b..d3682d33 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -168,6 +168,8 @@ public static void main(String[] args) throws ExecutionException, InterruptedExc
 
 同样也是需要实现 run() 方法，因为 Thread 类也实现了 Runable 接口。
 
+当调用 start() 方法启动一个线程时，虚拟机会将该线程放入就绪队列中等待被调度，当一个线程被调度时会执行该线程的 run() 方法。
+
 ```java
 public class MyThread extends Thread {
     public void run() {
diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 44f3d9dc..68af3cb7 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -2389,8 +2389,6 @@ private void backtracking(int row) {
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]"/></div> <br>
 
-dp[N] 即为所求。
-
 考虑到 dp[i] 只与 dp[i - 1] 和 dp[i - 2] 有关，因此可以只用两个变量来存储 dp[i - 1] 和 dp[i - 2]，使得原来的 O(N) 空间复杂度优化为 O(1) 复杂度。
 
 ```java
@@ -2468,16 +2466,6 @@ private int rob(int[] nums, int first, int last) {
 }
 ```
 
-**母牛生产** 
-
-[程序员代码面试指南-P181](#)
-
-题目描述：假设农场中成熟的母牛每年都会生 1 头小母牛，并且永远不会死。第一年有 1 只小母牛，从第二年开始，母牛开始生小母牛。每只小母牛 3 年之后成熟又可以生小母牛。给定整数 N，求 N 年后牛的数量。
-
-第 i 年成熟的牛的数量为：
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]"/></div> <br>
-
 **信件错排** 
 
 题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信方式的数量。
@@ -2491,7 +2479,15 @@ private int rob(int[] nums, int first, int last) {
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=(i-1)*dp[i-2]+(i-1)*dp[i-1]"/></div> <br>
 
-dp[N] 即为所求。
+**母牛生产** 
+
+[程序员代码面试指南-P181](#)
+
+题目描述：假设农场中成熟的母牛每年都会生 1 头小母牛，并且永远不会死。第一年有 1 只小母牛，从第二年开始，母牛开始生小母牛。每只小母牛 3 年之后成熟又可以生小母牛。给定整数 N，求 N 年后牛的数量。
+
+第 i 年成熟的牛的数量为：
+
+<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]"/></div> <br>
 
 ### 矩阵路径
 
@@ -2517,10 +2513,8 @@ public int minPathSum(int[][] grid) {
     int[] dp = new int[n];
     for (int i = 0; i < m; i++) {
         for (int j = 0; j < n; j++) {
-            if (j == 0) {
-                dp[j] = dp[j];        // 只能从上侧走到该位置
-            } else if (i == 0) {
-                dp[j] = dp[j - 1];    // 只能从左侧走到该位置
+            if (i == 0) {
+                dp[j] = dp[j - 1];
             } else {
                 dp[j] = Math.min(dp[j - 1], dp[j]);
             }
@@ -2584,17 +2578,18 @@ sumRange(0, 5) -> -3
 
 ```java
 class NumArray {
+
     private int[] sums;
 
     public NumArray(int[] nums) {
-        sums = new int[nums.length];
-        for (int i = 0; i < nums.length; i++) {
-            sums[i] = i == 0 ? nums[0] : sums[i - 1] + nums[i];
+        sums = new int[nums.length + 1];
+        for (int i = 1; i <= nums.length; i++) {
+            sums[i] = sums[i - 1] + nums[i - 1];
         }
     }
 
     public int sumRange(int i, int j) {
-        return i == 0 ? sums[j] : sums[j] - sums[i - 1];
+        return sums[j + 1] - sums[i];
     }
 }
 ```
@@ -2634,7 +2629,7 @@ return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] i
 
 dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
 
-如果 A[i] - A[i - 1] == A[i - 1] - A[i - 2]，表示 [A[i - 2], A[i - 1], A[i]] 是一个等差递增子区间。如果 [A[i - 3], A[i - 2], A[i - 1]] 是一个等差递增子区间，那么 [A[i - 3], A[i - 2], A[i - 1], A[i]] 也是。因此在这个条件下，dp[i] = dp[i-1] + 1。
+在 A[i] - A[i - 1] == A[i - 1] - A[i - 2] 的条件下，{A[i - 2], A[i - 1], A[i]} 是一个等差递增子区间。如果 {A[i - 3], A[i - 2], A[i - 1]} 是一个等差递增子区间，那么 {A[i - 3], A[i - 2], A[i - 1], A[i]} 也是等差递增子区间，dp[i] = dp[i-1] + 1。
 
 ```java
 public int numberOfArithmeticSlices(int[] A) {
@@ -2747,17 +2742,17 @@ public int numDecodings(String s) {
 
 ### 最长递增子序列
 
-已知一个序列 {S<sub>1</sub>, S<sub>2</sub>,...,S<sub>n</sub>} ，取出若干数组成新的序列 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>}，其中 i1、i2 ... im 保持递增，即新序列中各个数仍然保持原数列中的先后顺序，称新序列为原序列的一个 **子序列** 。
+已知一个序列 {S<sub>1</sub>, S<sub>2</sub>,...,S<sub>n</sub>}，取出若干数组成新的序列 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>}，其中 i1、i2 ... im 保持递增，即新序列中各个数仍然保持原数列中的先后顺序，称新序列为原序列的一个 **子序列** 。
 
 如果在子序列中，当下标 ix > iy 时，S<sub>ix</sub> > S<sub>iy</sub>，称子序列为原序列的一个 **递增子序列** 。
 
-定义一个数组 dp 存储最长递增子序列的长度，dp[n] 表示以 S<sub>n</sub> 结尾的序列的最长递增子序列长度。对于一个递增子序列 {S<sub>i1</sub>, S<sub>i2</sub>,...,S<sub>im</sub>}，如果 im < n 并且 S<sub>im</sub> < S<sub>n</sub> ，此时 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>, S<sub>n</sub>} 为一个递增子序列，递增子序列的长度增加 1。满足上述条件的递增子序列中，长度最长的那个递增子序列就是要找的，在长度最长的递增子序列上加上 S<sub>n</sub> 就构成了以 S<sub>n</sub> 为结尾的最长递增子序列。因此 dp[n] = max{ dp[i]+1 | S<sub>i</sub> < S<sub>n</sub> && i < n} 。
+定义一个数组 dp 存储最长递增子序列的长度，dp[n] 表示以 S<sub>n</sub> 结尾的序列的最长递增子序列长度。对于一个递增子序列 {S<sub>i1</sub>, S<sub>i2</sub>,...,S<sub>im</sub>}，如果 im < n 并且 S<sub>im</sub> < S<sub>n</sub>，此时 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>, S<sub>n</sub>} 为一个递增子序列，递增子序列的长度增加 1。满足上述条件的递增子序列中，长度最长的那个递增子序列就是要找的，在长度最长的递增子序列上加上 S<sub>n</sub> 就构成了以 S<sub>n</sub> 为结尾的最长递增子序列。因此 dp[n] = max{ dp[i]+1 | S<sub>i</sub> < S<sub>n</sub> && i < n} 。
 
 因为在求 dp[n] 时可能无法找到一个满足条件的递增子序列，此时 {S<sub>n</sub>} 就构成了递增子序列，需要对前面的求解方程做修改，令 dp[n] 最小为 1，即：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[n]=max\{1,dp[i]+1|S_i<S_n\&\&i<n\}"/></div> <br>
 
-对于一个长度为 N 的序列，最长递增子序列并不一定会以 S<sub>N</sub> 为结尾，因此 dp[N] 不是序列的最长递增子序列的长度，需要遍历 dp 数组找出最大值才是所要的结果，即 max{ dp[i] | 1 <= i <= N} 即为所求。
+对于一个长度为 N 的序列，最长递增子序列并不一定会以 S<sub>N</sub> 为结尾，因此 dp[N] 不是序列的最长递增子序列的长度，需要遍历 dp 数组找出最大值才是所要的结果，max{ dp[i] | 1 <= i <= N} 即为所求。
 
 **最长递增子序列** 
 
@@ -2790,7 +2785,7 @@ for (int i = 0; i < n; i++) {
 return ret;
 ```
 
-以上解法的时间复杂度为 O(N<sup>2</sup>) ，可以使用二分查找将时间复杂度降低为 O(NlogN)。
+以上解法的时间复杂度为 O(N<sup>2</sup>)，可以使用二分查找将时间复杂度降低为 O(NlogN)。
 
 定义一个 tails 数组，其中 tails[i] 存储长度为 i + 1 的最长递增子序列的最后一个元素。对于一个元素 x，
 
@@ -2915,19 +2910,19 @@ public int wiggleMaxLength(int[] nums) {
 
 定义一个二维数组 dp 用来存储最长公共子序列的长度，其中 dp[i][j] 表示 S1 的前 i 个字符与 S2 的前 j 个字符最长公共子序列的长度。考虑 S1<sub>i</sub> 与 S2<sub>j</sub> 值是否相等，分为两种情况：
 
-- 当 S1<sub>i</sub>==S2<sub>j</sub> 时，那么就能在 S1 的前 i-1 个字符与 S2 的前 j-1 个字符最长公共子序列的基础上再加上 S1<sub>i</sub> 这个值，最长公共子序列长度加 1 ，即 dp[i][j] = dp[i-1][j-1] + 1。
-- 当 S1<sub>i</sub> != S2<sub>j</sub> 时，此时最长公共子序列为 S1 的前 i-1 个字符和 S2 的前 j 个字符最长公共子序列，与 S1 的前 i 个字符和 S2 的前 j-1 个字符最长公共子序列，它们的最大者，即 dp[i][j] = max{ dp[i-1][j], dp[i][j-1] }。
+- 当 S1<sub>i</sub>==S2<sub>j</sub> 时，那么就能在 S1 的前 i-1 个字符与 S2 的前 j-1 个字符最长公共子序列的基础上再加上 S1<sub>i</sub> 这个值，最长公共子序列长度加 1，即 dp[i][j] = dp[i-1][j-1] + 1。
+- 当 S1<sub>i</sub> != S2<sub>j</sub> 时，此时最长公共子序列为 S1 的前 i-1 个字符和 S2 的前 j 个字符最长公共子序列，或者 S1 的前 i 个字符和 S2 的前 j-1 个字符最长公共子序列，取它们的最大者，即 dp[i][j] = max{ dp[i-1][j], dp[i][j-1] }。
 
 综上，最长公共子序列的状态转移方程为：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=\left\{\begin{array}{rcl}dp[i-1][j-1]&&{S1_i==S2_j}\\max(dp[i-1][j],dp[i][j-1])&&{S1_i<>S2_j}\end{array}\right."/></div> <br>
 
-对于长度为 N 的序列 S<sub>1</sub> 和 长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
+对于长度为 N 的序列 S<sub>1</sub> 和长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
 
 与最长递增子序列相比，最长公共子序列有以下不同点：
 
 - 针对的是两个序列，求它们的最长公共子序列。
-- 在最长递增子序列中，dp[i] 表示以 S<sub>i</sub> 为结尾的最长递增子序列长度，子序列必须包含 S<sub>i</sub> ；在最长公共子序列中，dp[i][j] 表示 S1 中前 i 个字符与 S2 中前 j 个字符的最长公共子序列长度，不一定包含 S1<sub>i</sub> 和 S2<sub>j</sub> 。
+- 在最长递增子序列中，dp[i] 表示以 S<sub>i</sub> 为结尾的最长递增子序列长度，子序列必须包含 S<sub>i</sub> ；在最长公共子序列中，dp[i][j] 表示 S1 中前 i 个字符与 S2 中前 j 个字符的最长公共子序列长度，不一定包含 S1<sub>i</sub> 和 S2<sub>j</sub>。
 - 在求最终解时，最长公共子序列中 dp[N][M] 就是最终解，而最长递增子序列中 dp[N] 不是最终解，因为以 S<sub>N</sub> 为结尾的最长递增子序列不一定是整个序列最长递增子序列，需要遍历一遍 dp 数组找到最大者。
 
 ```java
@@ -2956,9 +2951,7 @@ public int lengthOfLCS(int[] nums1, int[] nums2) {
 - 第 i 件物品没添加到背包，总体积不超过 j 的前 i 件物品的最大价值就是总体积不超过 j 的前 i-1 件物品的最大价值，dp[i][j] = dp[i-1][j]。
 - 第 i 件物品添加到背包中，dp[i][j] = dp[i-1][j-w] + v。
 
-第 i 件物品可添加也可以不添加，取决于哪种情况下最大价值更大。
-
-综上，0-1 背包的状态转移方程为：
+第 i 件物品可添加也可以不添加，取决于哪种情况下最大价值更大。因此，0-1 背包的状态转移方程为：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=max(dp[i-1][j],dp[i-1][j-w]+v)"/></div> <br>
 
@@ -2981,11 +2974,11 @@ public int knapsack(int W, int N, int[] weights, int[] values) {
 
 **空间优化** 
 
-在程序实现时可以对 0-1 背包做优化。观察状态转移方程可以知道，前 i 件物品的状态仅由前 i-1 件物品的状态有关，因此可以将 dp 定义为一维数组，其中 dp[j] 既可以表示 dp[i-1][j] 也可以表示 dp[i][j]。此时，
+在程序实现时可以对 0-1 背包做优化。观察状态转移方程可以知道，前 i 件物品的状态仅与前 i-1 件物品的状态有关，因此可以将 dp 定义为一维数组，其中 dp[j] 既可以表示 dp[i-1][j] 也可以表示 dp[i][j]。此时，
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[j]=max(dp[j],dp[j-w]+v)"/></div> <br>
 
-因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，以防止将 dp[i-1][j-w] 覆盖。也就是说要先计算 dp[i][j] 再计算 dp[i][j-w]，在程序实现时需要按倒序来循环求解。
+因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，以防将 dp[i-1][j-w] 覆盖。也就是说要先计算 dp[i][j] 再计算 dp[i][j-w]，在程序实现时需要按倒序来循环求解。
 
 ```java
 public int knapsack(int W, int N, int[] weights, int[] values) {
@@ -3205,7 +3198,7 @@ public int findMaxForm(String[] strs, int m, int n) {
 }
 ```
 
-**找零钱的方法数** 
+**找零钱的最少硬币数** 
 
 [322. Coin Change (Medium)](https://leetcode.com/problems/coin-change/description/)
 
@@ -3286,60 +3279,6 @@ public int combinationSum4(int[] nums, int target) {
 }
 ```
 
-**只能进行 k 次的股票交易** 
-
-[188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)
-
-```java
-public int maxProfit(int k, int[] prices) {
-    int n = prices.length;
-    if (k >= n / 2) {   // 这种情况下该问题退化为普通的股票交易问题
-        int maxProfit = 0;
-        for (int i = 1; i < n; i++) {
-            if (prices[i] > prices[i - 1]) {
-                maxProfit += prices[i] - prices[i - 1];
-            }
-        }
-        return maxProfit;
-    }
-    int[][] maxProfit = new int[k + 1][n];
-    for (int i = 1; i <= k; i++) {
-        int localMax = maxProfit[i - 1][0] - prices[0];
-        for (int j = 1; j < n; j++) {
-            maxProfit[i][j] = Math.max(maxProfit[i][j - 1], prices[j] + localMax);
-            localMax = Math.max(localMax, maxProfit[i - 1][j] - prices[j]);
-        }
-    }
-    return maxProfit[k][n - 1];
-}
-```
-
-**只能进行两次的股票交易** 
-
-[123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/)
-
-```java
-public int maxProfit(int[] prices) {
-    int firstBuy = Integer.MIN_VALUE, firstSell = 0;
-    int secondBuy = Integer.MIN_VALUE, secondSell = 0;
-    for (int curPrice : prices) {
-        if (firstBuy < -curPrice) {
-            firstBuy = -curPrice;
-        }
-        if (firstSell < firstBuy + curPrice) {
-            firstSell = firstBuy + curPrice;
-        }
-        if (secondBuy < firstSell - curPrice) {
-            secondBuy = firstSell - curPrice;
-        }
-        if (secondSell < secondBuy + curPrice) {
-            secondSell = secondBuy + curPrice;
-        }
-    }
-    return secondSell;
-}
-```
-
 ### 股票交易
 
 **需要冷却期的股票交易** 
@@ -3432,6 +3371,60 @@ public int maxProfit(int[] prices) {
 }
 ```
 
+**只能进行两次的股票交易** 
+
+[123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/)
+
+```java
+public int maxProfit(int[] prices) {
+    int firstBuy = Integer.MIN_VALUE, firstSell = 0;
+    int secondBuy = Integer.MIN_VALUE, secondSell = 0;
+    for (int curPrice : prices) {
+        if (firstBuy < -curPrice) {
+            firstBuy = -curPrice;
+        }
+        if (firstSell < firstBuy + curPrice) {
+            firstSell = firstBuy + curPrice;
+        }
+        if (secondBuy < firstSell - curPrice) {
+            secondBuy = firstSell - curPrice;
+        }
+        if (secondSell < secondBuy + curPrice) {
+            secondSell = secondBuy + curPrice;
+        }
+    }
+    return secondSell;
+}
+```
+
+**只能进行 k 次的股票交易** 
+
+[188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)
+
+```java
+public int maxProfit(int k, int[] prices) {
+    int n = prices.length;
+    if (k >= n / 2) {   // 这种情况下该问题退化为普通的股票交易问题
+        int maxProfit = 0;
+        for (int i = 1; i < n; i++) {
+            if (prices[i] > prices[i - 1]) {
+                maxProfit += prices[i] - prices[i - 1];
+            }
+        }
+        return maxProfit;
+    }
+    int[][] maxProfit = new int[k + 1][n];
+    for (int i = 1; i <= k; i++) {
+        int localMax = maxProfit[i - 1][0] - prices[0];
+        for (int j = 1; j < n; j++) {
+            maxProfit[i][j] = Math.max(maxProfit[i][j - 1], prices[j] + localMax);
+            localMax = Math.max(localMax, maxProfit[i - 1][j] - prices[j]);
+        }
+    }
+    return maxProfit[k][n - 1];
+}
+```
+
 ### 字符串编辑
 
 **删除两个字符串的字符使它们相等** 
@@ -3450,11 +3443,8 @@ Explanation: You need one step to make "sea" to "ea" and another step to make "e
 public int minDistance(String word1, String word2) {
     int m = word1.length(), n = word2.length();
     int[][] dp = new int[m + 1][n + 1];
-    for (int i = 0; i <= m; i++) {
-        for (int j = 0; j <= n; j++) {
-            if (i == 0 || j == 0) {
-                continue;
-            }
+    for (int i = 1; i <= m; i++) {
+        for (int j = 1; j <= n; j++) {
             if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                 dp[i][j] = dp[i - 1][j - 1] + 1;
             } else {
diff --git a/notes/Linux.md b/notes/Linux.md
index 14d3b317..8ff3e2c6 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -129,15 +129,16 @@ info 与 man 类似，但是 info 将文档分成一个个页面，每个页面
 /usr/local/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/home/dmtsai/.local/bin:/home/dmtsai/bin
 ```
 
-env 命令可以获取当前终端的环境变量。
-
 ## sudo
 
 sudo 允许一般用户使用 root 可执行的命令，不过只有在 /etc/sudoers 配置文件中添加的用户才能使用该指令。
 
 ## 包管理工具
 
-RPM 和 DPKG 为最常见的两类软件包管理工具。RPM 全称为 Redhat Package Manager，最早由 Red Hat 公司制定实施，随后被 GNU 开源操作系统接受并成为很多 Linux 系统 (RHEL) 的既定软件标准。与 RPM 进行竞争的是基于 Debian 操作系统 (UBUNTU) 的 DEB 软件包管理工具 DPKG，全称为 Debian Package，功能方面与 RPM 相似。
+RPM 和 DPKG 为最常见的两类软件包管理工具：
+
+- RPM 全称为 Redhat Package Manager，最早由 Red Hat 公司制定实施，随后被 GNU 开源操作系统接受并成为很多 Linux 系统 (RHEL) 的既定软件标准。
+- 与 RPM 进行竞争的是基于 Debian 操作系统 (Ubuntu) 的 DEB 软件包管理工具 DPKG，全称为 Debian Package，功能方面与 RPM 相似。
 
 YUM 基于 RPM，具有依赖管理功能，并具有软件升级的功能。
 
@@ -194,13 +195,13 @@ IDE（ATA）全称 Advanced Technology Attachment，接口速度最大为 133MB/
 
 ### 2. SATA
 
-SATA 全称 Serial ATA，也就是使用串口的 ATA 接口，因抗干扰性强，且对数据线的长度要求比 ATA 低很多，支持热插拔等功能，SATA-II 的接口速度为 300MiB/s，而新的 SATA-III 标准可达到 600MiB/s 的传输速度。SATA 的数据线也比 ATA 的细得多，有利于机箱内的空气流通，整理线材也比较方便。
+SATA 全称 Serial ATA，也就是使用串口的 ATA 接口，抗干扰性强，且对数据线的长度要求比 ATA 低很多，支持热插拔等功能。SATA-II 的接口速度为 300MiB/s，而新的 SATA-III 标准可达到 600MiB/s 的传输速度。SATA 的数据线也比 ATA 的细得多，有利于机箱内的空气流通，整理线材也比较方便。
 
 <div align="center"> <img src="../pics//f9f2a16b-4843-44d1-9759-c745772e9bcf.jpg" width=""/> </div><br>
 
 ### 3. SCSI
 
-SCSI 全称是 Small Computer System Interface（小型机系统接口），经历多代的发展，从早期的 SCSI-II，到目前的 Ultra320 SCSI 以及 Fiber-Channel（光纤通道），接口型式也多种多样。SCSI 硬盘广为工作站级个人电脑以及服务器所使用，因此会使用较为先进的技术，如碟片转速 15000rpm 的高转速，且资料传输时 CPU 占用率较低，但是单价也比相同容量的 ATA 及 SATA 硬盘更加昂贵。
+SCSI 全称是 Small Computer System Interface（小型机系统接口），经历多代的发展，从早期的 SCSI-II 到目前的 Ultra320 SCSI 以及 Fiber-Channel（光纤通道），接口型式也多种多样。SCSI 硬盘广为工作站级个人电脑以及服务器所使用，因此会使用较为先进的技术，如碟片转速 15000rpm 的高转速，且传输时 CPU 占用率较低，但是单价也比相同容量的 ATA 及 SATA 硬盘更加昂贵。
 
 <div align="center"> <img src="../pics//f0574025-c514-49f5-a591-6d6a71f271f7.jpg" width=""/> </div><br>
 
@@ -229,7 +230,7 @@ Linux 中每个硬件都被当做一个文件，包括磁盘。磁盘以磁盘
 
 MBR 中，第一个扇区最重要，里面有主要开机记录（Master boot record, MBR）及分区表（partition table），其中主要开机记录占 446 bytes，分区表占 64 bytes。
 
-分区表只有 64 bytes，最多只能存储 4 个分区，这 4 个分区为主分区（Primary）和扩展分区（Extended）。其中扩展分区只有一个，它将其它扇区用来记录分区表，因此通过扩展分区可以分出更多分区，这些分区称为逻辑分区。
+分区表只有 64 bytes，最多只能存储 4 个分区，这 4 个分区为主分区（Primary）和扩展分区（Extended）。其中扩展分区只有一个，它使用其它扇区用记录额外的分区表，因此通过扩展分区可以分出更多分区，这些分区称为逻辑分区。
 
 Linux 也把分区当成文件，分区文件的命名方式为：磁盘文件名 + 编号，例如 /dev/sda1。注意，逻辑分区的编号从 5 开始。
 
@@ -251,10 +252,10 @@ MBR 不支持 2.2 TB 以上的硬盘，GPT 则最多支持到 2<sup>33</sup> TB
 
 BIOS（Basic Input/Output System，基本输入输出系统），它是一个固件（嵌入在硬件中的软件），BIOS 程序存放在断电后内容不会丢失的只读内存中。
 
-BIOS 是开机的时候计算机执行的第一个程序，这个程序知道可以开机的磁盘，并读取磁盘第一个扇区的主要开机记录（MBR），由主要开机记录（MBR）执行其中的开机管理程序，这个开机管理程序会加载操作系统的核心文件。
-
 <div align="center"> <img src="../pics//50831a6f-2777-46ea-a571-29f23c85cc21.jpg"/> </div><br>
 
+BIOS 是开机的时候计算机执行的第一个程序，这个程序知道可以开机的磁盘，并读取磁盘第一个扇区的主要开机记录（MBR），由主要开机记录（MBR）执行其中的开机管理程序，这个开机管理程序会加载操作系统的核心文件。
+
 主要开机记录（MBR）中的开机管理程序提供以下功能：选单、载入核心文件以及转交其它开机管理程序。转交这个功能可以用来实现了多重引导，只需要将另一个操作系统的开机管理程序安装在其它分区的启动扇区上，在启动开机管理程序时，就可以通过选单选择启动当前的操作系统或者转交给其它开机管理程序从而启动另一个操作系统。
 
 下图中，第一扇区的主要开机记录（MBR）中的开机管理程序提供了两个选单：M1、M2，M1 指向了 Windows 操作系统，而 M2 指向其它分区的启动扇区，里面包含了另外一个开机管理程序，提供了一个指向 Linux 的选单。
@@ -283,11 +284,10 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 除此之外还包括：
 
 - superblock：记录文件系统的整体信息，包括 inode 和 block 的总量、使用量、剩余量，以及文件系统的格式与相关信息等；
-- block bitmap：记录 block 是否被使用的位域；
+- block bitmap：记录 block 是否被使用的位域。
 
 <div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
 
-
 ## 文件读取
 
 对于 Ext2 文件系统，当要读取一个文件的内容时，先在 inode 中去查找文件内容所在的所有 block，然后把所有 block 的内容读出来。
diff --git a/notes/MySQL.md b/notes/MySQL.md
index 7ac9685f..54067977 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -1,22 +1,23 @@
 <!-- GFM-TOC -->
-* [一、存储引擎](#一存储引擎)
+* [一、索引](#一索引)
+    * [B+ Tree 原理](#b-tree-原理)
+    * [MySQL 索引](#mysql-索引)
+    * [索引优化](#索引优化)
+    * [索引的优点](#索引的优点)
+    * [索引的使用场景](#索引的使用场景)
+* [二、查询性能优化](#二查询性能优化)
+    * [使用 Explain 进行分析](#使用-explain-进行分析)
+    * [优化数据访问](#优化数据访问)
+    * [重构查询方式](#重构查询方式)
+* [三、存储引擎](#三存储引擎)
     * [InnoDB](#innodb)
     * [MyISAM](#myisam)
     * [比较](#比较)
-* [二、数据类型](#二数据类型)
+* [四、数据类型](#四数据类型)
     * [整型](#整型)
     * [浮点数](#浮点数)
     * [字符串](#字符串)
     * [时间和日期](#时间和日期)
-* [三、索引](#三索引)
-    * [B+ Tree 原理](#b-tree-原理)
-    * [索引分类](#索引分类)
-    * [索引的优点](#索引的优点)
-    * [索引优化](#索引优化)
-* [四、查询性能优化](#四查询性能优化)
-    * [使用 Explain 进行分析](#使用-explain-进行分析)
-    * [优化数据访问](#优化数据访问)
-    * [重构查询方式](#重构查询方式)
 * [五、切分](#五切分)
     * [水平切分](#水平切分)
     * [垂直切分](#垂直切分)
@@ -29,101 +30,7 @@
 <!-- GFM-TOC -->
 
 
-# 一、存储引擎
-
-## InnoDB
-
-InnoDB 是 MySQL 默认的事务型存储引擎，只有在需要 InnoDB 不支持的特性时，才考虑使用其它存储引擎。
-
-实现了四个标准的隔离级别，默认级别是可重复读（REPEATABLE READ）。在可重复读隔离级别下，通过多版本并发控制（MVCC）+ 间隙锁（next-key locking）防止幻影读。
-
-主索引是聚簇索引，在索引中保存了数据，从而避免直接读取磁盘，因此对查询性能有很大的提升。
-
-内部做了很多优化，包括从磁盘读取数据时采用的可预测性读、能够加快读操作并且自动创建的自适应哈希索引、能够加速插入操作的插入缓冲区等。
-
-支持真正的在线热备份。其它存储引擎不支持在线热备份，要获取一致性视图需要停止对所有表的写入，而在读写混合场景中，停止写入可能也意味着停止读取。
-
-## MyISAM
-
-MyISAM 设计简单，数据以紧密格式存储。对于只读数据，或者表比较小、可以容忍修复操作，则依然可以使用 MyISAM。
-
-MyISAM 提供了大量的特性，包括压缩表、空间数据索引等。
-
-不支持事务。
-
-不支持行级锁，只能对整张表加锁，读取时会对需要读到的所有表加共享锁，写入时则对表加排它锁。但在表有读取操作的同时，也可以往表中插入新的记录，这被称为并发插入（CONCURRENT INSERT）。
-
-可以手工或者自动执行检查和修复操作，但是和事务恢复以及崩溃恢复不同，可能导致一些数据丢失，而且修复操作是非常慢的。
-
-如果指定了 DELAY_KEY_WRITE 选项，在每次修改执行完成时，不会立即将修改的索引数据写入磁盘，而是会写到内存中的键缓冲区，只有在清理键缓冲区或者关闭表的时候才会将对应的索引块写入磁盘。这种方式可以极大的提升写入性能，但是在数据库或者主机崩溃时会造成索引损坏，需要执行修复操作。
-
-## 比较
-
-- 事务：InnoDB 是事务型的，可以使用 Commit 和 Rollback 语句。
-
-- 并发：MyISAM 只支持表级锁，而 InnoDB 还支持行级锁。
-
-- 外键：InnoDB 支持外键。
-
-- 备份：InnoDB 支持在线热备份。
-
-- 崩溃恢复：MyISAM 崩溃后发生损坏的概率比 InnoDB 高很多，而且恢复的速度也更慢。
-
-- 其它特性：MyISAM 支持压缩表和空间数据索引。
-
-# 二、数据类型
-
-## 整型
-
-TINYINT, SMALLINT, MEDIUMINT, INT, BIGINT 分别使用 8, 16, 24, 32, 64 位存储空间，一般情况下越小的列越好。
-
-INT(11) 中的数字只是规定了交互工具显示字符的个数，对于存储和计算来说是没有意义的。
-
-## 浮点数
-
-FLOAT 和 DOUBLE 为浮点类型，DECIMAL 为高精度小数类型。CPU 原生支持浮点运算，但是不支持 DECIMAl 类型的计算，因此 DECIMAL 的计算比浮点类型需要更高的代价。
-
-FLOAT、DOUBLE 和 DECIMAL 都可以指定列宽，例如 DECIMAL(18, 9) 表示总共 18 位，取 9 位存储小数部分，剩下 9 位存储整数部分。
-
-## 字符串
-
-主要有 CHAR 和 VARCHAR 两种类型，一种是定长的，一种是变长的。
-
-VARCHAR 这种变长类型能够节省空间，因为只需要存储必要的内容。但是在执行 UPDATE 时可能会使行变得比原来长，当超出一个页所能容纳的大小时，就要执行额外的操作。MyISAM 会将行拆成不同的片段存储，而 InnoDB 则需要分裂页来使行放进页内。
-
-VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
-
-## 时间和日期
-
-MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
-
-### 1. DATETIME
-
-能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
-
-它与时区无关。
-
-默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
-
-### 2. TIMESTAMP
-
-和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年 到 2038 年。
-
-它和时区有关，也就是说一个时间戳在不同的时区所代表的具体时间是不同的。
-
-MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提供了 UNIX_TIMESTAMP() 函数把日期转换为 UNIX 时间戳。
-
-默认情况下，如果插入时没有指定 TIMESTAMP 列的值，会将这个值设置为当前时间。
-
-应该尽量使用 TIMESTAMP，因为它比 DATETIME 空间效率更高。
-
-# 三、索引
-
-索引能够轻易将查询性能提升几个数量级。
-
-对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效。对于中到大型的表，索引就非常有效。但是对于特大型的表，建立和维护索引的代价将会随之增长。这种情况下，需要用到一种技术可以直接区分出需要查询的一组数据，而不是一条记录一条记录地匹配，例如可以使用分区技术。
-
-索引是在存储引擎层实现的，而不是在服务器层实现的，所以不同存储引擎具有不同的索引类型和实现。
+# 一、索引
 
 ## B+ Tree 原理
 
@@ -141,37 +48,37 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 进行查找操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
-插入删除操作记录会破坏平衡树的平衡性，因此在插入删除时，需要对树进行一个分裂、合并、旋转等操作。
+插入删除操作记录会破坏平衡树的平衡性，因此在插入删除操作之后，需要对树进行一个分裂、合并、旋转等操作来维护平衡性。
 
 ### 3. 与红黑树的比较
 
 红黑树等平衡树也可以用来实现索引，但是文件系统及数据库系统普遍采用 B+ Tree 作为索引结构，主要有以下两个原因：
 
-（一）更少的检索次数
+（一）更少的查找次数
 
-平衡树检索数据的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
+平衡树查找操作的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
 
-红黑树的出度为 2，而 B+ Tree 的出度一般都非常大。红黑树的树高 h 很明显比 B+ Tree 大非常多，因此检索的次数也就更多。
+红黑树的出度为 2，而 B+ Tree 的出度一般都非常大，所以红黑树的树高 h 很明显比 B+ Tree 大非常多，检索的次数也就更多。
 
 （二）利用计算机预读特性
 
-为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。这样做的理论依据是计算机科学中著名的局部性原理：当一个数据被用到时，其附近的数据也通常会马上被使用。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，因此速度会非常快。
+为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，因此速度会非常快。
 
 操作系统一般将内存和磁盘分割成固态大小的块，每一块称为一页，内存与磁盘以页为单位交换数据。数据库系统将索引的一个节点的大小设置为页的大小，使得一次 I/O 就能完全载入一个节点，并且可以利用预读特性，相邻的节点也能够被预先载入。
 
-## 索引分类
+## MySQL 索引
+
+索引是在存储引擎层实现的，而不是在服务器层实现的，所以不同存储引擎具有不同的索引类型和实现。
 
 ### 1. B+Tree 索引
 
-B+Tree 索引是大多数 MySQL 存储引擎的默认索引类型。
+是大多数 MySQL 存储引擎的默认索引类型。
 
 因为不再需要进行全表扫描，只需要对树进行搜索即可，因此查找速度快很多。除了用于查找，还可以用于排序和分组。
 
 可以指定多个列作为索引列，多个索引列共同组成键。
 
-B+Tree 索引适用于全键值、键值范围和键前缀查找，其中键前缀查找只适用于最左前缀查找。
-
-如果不是按照索引列的顺序进行查找，则无法使用索引。
+适用于全键值、键值范围和键前缀查找，其中键前缀查找只适用于最左前缀查找。如果不是按照索引列的顺序进行查找，则无法使用索引。
 
 InnoDB 的 B+Tree 索引分为主索引和辅助索引。
 
@@ -185,12 +92,12 @@ InnoDB 的 B+Tree 索引分为主索引和辅助索引。
 
 ### 2. 哈希索引
 
-InnoDB 引擎有一个特殊的功能叫“自适应哈希索引”，当某个索引值被使用的非常频繁时，会在 B+Tree 索引之上再创建一个哈希索引，这样就让 B+Tree 索引具有哈希索引的一些优点，比如快速的哈希查找。
-
 哈希索引能以 O(1) 时间进行查找，但是失去了有序性，它具有以下限制：
 
 - 无法用于排序与分组；
-- 只支持精确查找，无法用于部分查找和范围查找；
+- 只支持精确查找，无法用于部分查找和范围查找。
+
+InnoDB 存储引擎有一个特殊的功能叫“自适应哈希索引”，当某个索引值被使用的非常频繁时，会在 B+Tree 索引之上再创建一个哈希索引，这样就让 B+Tree 索引具有哈希索引的一些优点，比如快速的哈希查找。
 
 ### 3. 全文索引
 
@@ -200,20 +107,12 @@ MyISAM 存储引擎支持全文索引，用于查找文本中的关键词，而
 
 InnoDB 存储引擎在 MySQL 5.6.4 版本中也开始支持全文索引。
 
-### 4. 空间数据索引（R-Tree）
+### 4. 空间数据索引
 
-MyISAM 存储引擎支持空间数据索引，可以用于地理数据存储。空间数据索引会从所有维度来索引数据，可以有效地使用任意维度来进行组合查询。
+MyISAM 存储引擎支持空间数据索引（R-Tree），可以用于地理数据存储。空间数据索引会从所有维度来索引数据，可以有效地使用任意维度来进行组合查询。
 
 必须使用 GIS 相关的函数来维护数据。
 
-## 索引的优点
-
-- 大大减少了服务器需要扫描的数据行数。
-
-- 帮助服务器避免进行排序和创建临时表（B+Tree 索引是有序的，可以用来做 ORDER BY 和 GROUP BY 操作）；
-
-- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，也就将相邻的数据都存储在一起）。
-
 ## 索引优化
 
 ### 1. 独立的列
@@ -266,11 +165,25 @@ customer_id_selectivity: 0.0373
 
 具有以下优点：
 
-- 因为索引条目通常远小于数据行的大小，所以若只读取索引，能大大减少数据访问量。
+- 索引通常远小于数据行的大小，只读取索引能大大减少数据访问量。
 - 一些存储引擎（例如 MyISAM）在内存中只缓存索引，而数据依赖于操作系统来缓存。因此，只访问索引可以不使用系统调用（通常比较费时）。
 - 对于 InnoDB 引擎，若辅助索引能够覆盖查询，则无需访问主索引。
 
-# 四、查询性能优化
+## 索引的优点
+
+- 大大减少了服务器需要扫描的数据行数。
+
+- 帮助服务器避免进行排序和分组，也就不需要创建临时表（B+Tree 索引是有序的，可以用于 ORDER BY 和 GROUP BY 操作。临时表主要是在排序和分组过程中创建，因为不需要排序和分组，也就不需要创建临时表）。
+
+- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，也就将相邻的数据都存储在一起）。
+
+## 索引的使用场景
+
+- 对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效。
+- 对于中到大型的表，索引就非常有效。
+- 但是对于特大型的表，建立和维护索引的代价将会随之增长。这种情况下，需要用到一种技术可以直接区分出需要查询的一组数据，而不是一条记录一条记录地匹配，例如可以使用分区技术。
+
+# 二、查询性能优化
 
 ## 使用 Explain 进行分析
 
@@ -282,23 +195,13 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 - key : 使用的索引
 - rows : 扫描的行数
 
-更多内容请参考：[MySQL 性能优化神器 Explain 使用分析](https://segmentfault.com/a/1190000008131735)
-
 ## 优化数据访问
 
 ### 1. 减少请求的数据量
 
-（一）只返回必要的列
-
-最好不要使用 SELECT * 语句。
-
-（二）只返回必要的行
-
-使用 WHERE 语句进行查询过滤，有时候也需要使用 LIMIT 语句来限制返回的数据。
-
-（三）缓存重复查询的数据
-
-使用缓存可以避免在数据库中进行查询，特别要查询的数据经常被重复查询，缓存可以带来的查询性能提升将会是非常明显的。
+- 只返回必要的列：最好不要使用 SELECT * 语句。
+- 只返回必要的行：使用 LIMIT 语句来限制返回的数据。
+- 缓存重复查询的数据：使用缓存可以避免在数据库中进行查询，特别在要查询的数据经常被重复查询时，缓存带来的查询性能提升将会是非常明显的。
 
 ### 2. 减少服务器端扫描的行数
 
@@ -324,12 +227,12 @@ do {
 
 ### 2. 分解大连接查询
 
-将一个大连接查询（JOIN）分解成对每一个表进行一次单表查询，然后将结果在应用程序中进行关联，这样做的好处有：
+将一个大连接查询分解成对每一个表进行一次单表查询，然后将结果在应用程序中进行关联，这样做的好处有：
 
 - 让缓存更高效。对于连接查询，如果其中一个表发生变化，那么整个查询缓存就无法使用。而分解后的多个查询，即使其中一个表发生变化，对其它表的查询缓存依然可以使用。
 - 分解成多个单表查询，这些单表查询的缓存结果更可能被其它查询使用到，从而减少冗余记录的查询。
 - 减少锁竞争；
-- 在应用层进行连接，可以更容易对数据库进行拆分，从而更容易做到高性能和可扩展。
+- 在应用层进行连接，可以更容易对数据库进行拆分，从而更容易做到高性能和可伸缩。
 - 查询本身效率也可能会有所提升。例如下面的例子中，使用 IN() 代替连接查询，可以让 MySQL 按照 ID 顺序进行查询，这可能比随机的连接要更高效。
 
 ```sql
@@ -345,23 +248,111 @@ SELECT * FROM tag_post WHERE tag_id=1234;
 SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 ```
 
+# 三、存储引擎
+
+## InnoDB
+
+是 MySQL 默认的事务型存储引擎，只有在需要它不支持的特性时，才考虑使用其它存储引擎。
+
+实现了四个标准的隔离级别，默认级别是可重复读（REPEATABLE READ）。在可重复读隔离级别下，通过多版本并发控制（MVCC）+ 间隙锁（Next-Key Locking）防止幻影读。
+
+主索引是聚簇索引，在索引中保存了数据，从而避免直接读取磁盘，因此对查询性能有很大的提升。
+
+内部做了很多优化，包括从磁盘读取数据时采用的可预测性读、能够加快读操作并且自动创建的自适应哈希索引、能够加速插入操作的插入缓冲区等。
+
+支持真正的在线热备份。其它存储引擎不支持在线热备份，要获取一致性视图需要停止对所有表的写入，而在读写混合场景中，停止写入可能也意味着停止读取。
+
+## MyISAM
+
+设计简单，数据以紧密格式存储。对于只读数据，或者表比较小、可以容忍修复操作，则依然可以使用它。
+
+提供了大量的特性，包括压缩表、空间数据索引等。
+
+不支持事务。
+
+不支持行级锁，只能对整张表加锁，读取时会对需要读到的所有表加共享锁，写入时则对表加排它锁。但在表有读取操作的同时，也可以往表中插入新的记录，这被称为并发插入（CONCURRENT INSERT）。
+
+可以手工或者自动执行检查和修复操作，但是和事务恢复以及崩溃恢复不同，可能导致一些数据丢失，而且修复操作是非常慢的。
+
+如果指定了 DELAY_KEY_WRITE 选项，在每次修改执行完成时，不会立即将修改的索引数据写入磁盘，而是会写到内存中的键缓冲区，只有在清理键缓冲区或者关闭表的时候才会将对应的索引块写入磁盘。这种方式可以极大的提升写入性能，但是在数据库或者主机崩溃时会造成索引损坏，需要执行修复操作。
+
+## 比较
+
+- 事务：InnoDB 是事务型的，可以使用 Commit 和 Rollback 语句。
+
+- 并发：MyISAM 只支持表级锁，而 InnoDB 还支持行级锁。
+
+- 外键：InnoDB 支持外键。
+
+- 备份：InnoDB 支持在线热备份。
+
+- 崩溃恢复：MyISAM 崩溃后发生损坏的概率比 InnoDB 高很多，而且恢复的速度也更慢。
+
+- 其它特性：MyISAM 支持压缩表和空间数据索引。
+
+# 四、数据类型
+
+## 整型
+
+TINYINT, SMALLINT, MEDIUMINT, INT, BIGINT 分别使用 8, 16, 24, 32, 64 位存储空间，一般情况下越小的列越好。
+
+INT(11) 中的数字只是规定了交互工具显示字符的个数，对于存储和计算来说是没有意义的。
+
+## 浮点数
+
+FLOAT 和 DOUBLE 为浮点类型，DECIMAL 为高精度小数类型。CPU 原生支持浮点运算，但是不支持 DECIMAl 类型的计算，因此 DECIMAL 的计算比浮点类型需要更高的代价。
+
+FLOAT、DOUBLE 和 DECIMAL 都可以指定列宽，例如 DECIMAL(18, 9) 表示总共 18 位，取 9 位存储小数部分，剩下 9 位存储整数部分。
+
+## 字符串
+
+主要有 CHAR 和 VARCHAR 两种类型，一种是定长的，一种是变长的。
+
+VARCHAR 这种变长类型能够节省空间，因为只需要存储必要的内容。但是在执行 UPDATE 时可能会使行变得比原来长，当超出一个页所能容纳的大小时，就要执行额外的操作。MyISAM 会将行拆成不同的片段存储，而 InnoDB 则需要分裂页来使行放进页内。
+
+VARCHAR 会保留字符串末尾的空格，而 CHAR 会删除。
+
+## 时间和日期
+
+MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
+
+### 1. DATETIME
+
+能够保存从 1001 年到 9999 年的日期和时间，精度为秒，使用 8 字节的存储空间。
+
+它与时区无关。
+
+默认情况下，MySQL 以一种可排序的、无歧义的格式显示 DATETIME 值，例如“2008-01-16 22:37:08”，这是 ANSI 标准定义的日期和时间表示方法。
+
+### 2. TIMESTAMP
+
+和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年 到 2038 年。
+
+它和时区有关，也就是说一个时间戳在不同的时区所代表的具体时间是不同的。
+
+MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提供了 UNIX_TIMESTAMP() 函数把日期转换为 UNIX 时间戳。
+
+默认情况下，如果插入时没有指定 TIMESTAMP 列的值，会将这个值设置为当前时间。
+
+应该尽量使用 TIMESTAMP，因为它比 DATETIME 空间效率更高。
+
 # 五、切分
 
 ## 水平切分
 
-<div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
-
 水平切分又称为 Sharding，它是将同一个表中的记录拆分到多个结构相同的表中。
 
 当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓存单个数据库的压力。
 
+<div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
+
 ## 垂直切分
 
 <div align="center"> <img src="../pics//e130e5b8-b19a-4f1e-b860-223040525cf6.jpg"/> </div><br>
 
 垂直切分是将一张表按列切分成多个表，通常是按照列的关系密集程度进行切分，也可以利用垂直切分将经常被使用的列和不经常被使用的列切分到不同的表中。
 
-在数据库的层面使用垂直切分将按数据库中表的密集程度部署到不同的库中，例如将原来的电商数据库垂直切分成商品数据库 payDB、用户数据库 userDB 等。
+在数据库的层面使用垂直切分将按数据库中表的密集程度部署到不同的库中，例如将原来的电商数据库垂直切分成商品数据库、用户数据库等。
 
 ## Sharding 策略
 
@@ -375,15 +366,15 @@ SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 
 使用分布式事务来解决，比如 XA 接口。
 
-### 2. JOIN
+### 2. 链接
 
-可以将原来的 JOIN 查询分解成多个单表查询，然后在用户程序中进行 JOIN。
+可以将原来的 JOIN 分解成多个单表查询，然后在用户程序中进行 JOIN。
 
 ### 3. ID 唯一性
 
-- 使用全局唯一 ID：GUID。
-- 为每个分片指定一个 ID 范围。
-- 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)。
+- 使用全局唯一 ID：GUID
+- 为每个分片指定一个 ID 范围
+- 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)
 
 更多内容请参考：
 
@@ -396,24 +387,24 @@ SELECT * FROM post WHERE post.id IN (123,456,567,9098,8904);
 
 主要涉及三个线程：binlog 线程、I/O 线程和 SQL 线程。
 
--  **binlog 线程** ：负责将主服务器上的数据更改写入二进制文件（binlog）中。
--  **I/O 线程** ：负责从主服务器上读取二进制日志文件，并写入从服务器的中继日志中。
+-  **binlog 线程** ：负责将主服务器上的数据更改写入二进制日志中。
+-  **I/O 线程** ：负责从主服务器上读取二进制日志，并写入从服务器的中继日志中。
 -  **SQL 线程** ：负责读取中继日志并重放其中的 SQL 语句。
 
 <div align="center"> <img src="../pics//master-slave.png"/> </div><br>
 
 ## 读写分离
 
-主服务器用来处理写操作以及实时性要求比较高的读操作，而从服务器用来处理读操作。
+主服务器处理写操作以及实时性要求比较高的读操作，而从服务器处理读操作。
 
-读写分离常用代理方式来实现，代理服务器接收应用层传来的读写请求，然后决定转发到哪个服务器。
-
-MySQL 读写分离能提高性能的原因在于：
+读写分离能提高性能的原因在于：
 
 - 主从服务器负责各自的读和写，极大程度缓解了锁的争用；
-- 从服务器可以配置 MyISAM 引擎，提升查询性能以及节约系统开销；
+- 从服务器可以使用 MyISAM，提升查询性能以及节约系统开销；
 - 增加冗余，提高可用性。
 
+读写分离常用代理方式来实现，代理服务器接收应用层传来的读写请求，然后决定转发到哪个服务器。
+
 <div align="center"> <img src="../pics//master-slave-proxy.png"/> </div><br>
 
 # 参考资料
@@ -425,3 +416,4 @@ MySQL 读写分离能提高性能的原因在于：
 - [How to create unique row ID in sharded databases?](https://stackoverflow.com/questions/788829/how-to-create-unique-row-id-in-sharded-databases)
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
 - [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
+- [MySQL 性能优化神器 Explain 使用分析](https://segmentfault.com/a/1190000008131735)
diff --git a/notes/Redis.md b/notes/Redis.md
index 0981abf0..8f231382 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -49,7 +49,7 @@
 
 Redis 是速度非常快的非关系型（NoSQL）内存键值数据库，可以存储键和五种不同类型的值之间的映射。
 
-键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、有序集合、散列表。
+键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、散列表、有序集合。
 
 Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，使用复制来扩展读性能，使用分片来扩展写性能。
 
@@ -58,7 +58,7 @@ Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，
 | 数据类型 | 可以存储的值 | 操作 |
 | :--: | :--: | :--: |
 | STRING | 字符串、整数或者浮点数 | 对整个字符串或者字符串的其中一部分执行操作</br> 对整数和浮点数执行自增或者自减操作 |
-| LIST | 列表 | 从两端压入或者弹出元素</br> 读取单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
+| LIST | 列表 | 从两端压入或者弹出元素</br> 对单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
 | SET | 无序集合 | 添加、获取、移除单个元素</br> 检查一个元素是否存在于集合中</br> 计算交集、并集、差集</br> 从集合里面随机获取元素 |
 | HASH | 包含键值对的无序散列表 | 添加、获取、移除单个键值对</br> 获取所有键值对</br> 检查某个键是否存在|
 | ZSET | 有序集合 | 添加、获取、删除元素</br> 根据分值范围或者成员来获取元素</br> 计算一个键的排名 |
@@ -555,7 +555,7 @@ Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状
 
 分片是将数据划分为多个部分的方法，可以将数据存储到多台机器里面，这种方法在解决某些问题时可以获得线性级别的性能提升。
 
-假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... 等等，有不同的方式来选择一个指定的键存储在哪个实例中。
+假设有 4 个 Reids 实例 R0，R1，R2，R3，还有很多表示用户的键 user:1，user:2，... ，有不同的方式来选择一个指定的键存储在哪个实例中。
 
 - 最简单的方式是范围分片，例如用户 id 从 0\~1000 的存储到实例 R0 中，用户 id 从 1001\~2000 的存储到实例 R1 中，等等。但是这样需要维护一张映射范围表，维护操作代价很高。
 - 还有一种方式是哈希分片，使用 CRC32 哈希函数将键转换为一个数字，再对实例数量求模就能知道应该存储的实例。
diff --git a/notes/数据库系统原理.md b/notes/数据库系统原理.md
index e7ef804b..1bcd2533 100644
--- a/notes/数据库系统原理.md
+++ b/notes/数据库系统原理.md
@@ -20,6 +20,7 @@
     * [可串行化（SERIALIZABLE）](#可串行化serializable)
 * [五、多版本并发控制](#五多版本并发控制)
     * [版本号](#版本号)
+    * [隐藏的列](#隐藏的列)
     * [Undo 日志](#undo-日志)
     * [实现过程](#实现过程)
     * [快照读与当前读](#快照读与当前读)
@@ -58,9 +59,7 @@
 
 ### 2. 一致性（Consistency）
 
-数据库在事务执行前后都保持一致性状态。
-
-在一致性状态下，所有事务对一个数据的读取结果都是相同的。
+数据库在事务执行前后都保持一致性状态。在一致性状态下，所有事务对一个数据的读取结果都是相同的。
 
 ### 3. 隔离性（Isolation）
 
@@ -78,10 +77,10 @@
 
 - 只有满足一致性，事务的执行结果才是正确的。
 - 在无并发的情况下，事务串行执行，隔离性一定能够满足。此时只要能满足原子性，就一定能满足一致性。
-- 在并发的情况下，多个事务并发执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
+- 在并发的情况下，多个事务并行执行，事务不仅要满足原子性，还需要满足隔离性，才能满足一致性。
 - 事务满足持久化是为了能应对数据库崩溃的情况。
 
-<div align="center"> <img src="../pics//a58e294a-615d-4ea0-9fbf-064a6daec4b2.png" width="400"/> </div><br>
+<div align="center"> <img src="../pics//a58e294a-615d-4ea0-9fbf-064a6daec4b2.png" width="450"/> </div><br>
 
 ## AUTOCOMMIT
 
@@ -95,25 +94,25 @@ MySQL 默认采用自动提交模式。也就是说，如果不显式使用`STAR
 
 T<sub>1</sub> 和 T<sub>2</sub> 两个事务都对一个数据进行修改，T<sub>1</sub> 先修改，T<sub>2</sub> 随后修改，T<sub>2</sub> 的修改覆盖了 T<sub>1</sub> 的修改。
 
-<div align="center"> <img src="../pics//88ff46b3-028a-4dbb-a572-1f062b8b96d3.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//88ff46b3-028a-4dbb-a572-1f062b8b96d3.png" width="350"/> </div><br>
 
 ## 读脏数据
 
 T<sub>1</sub> 修改一个数据，T<sub>2</sub> 随后读取这个数据。如果 T<sub>1</sub> 撤销了这次修改，那么 T<sub>2</sub> 读取的数据是脏数据。
 
-<div align="center"> <img src="../pics//dd782132-d830-4c55-9884-cfac0a541b8e.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//dd782132-d830-4c55-9884-cfac0a541b8e.png" width="400"/> </div><br>
 
 ## 不可重复读
 
 T<sub>2</sub> 读取一个数据，T<sub>1</sub> 对该数据做了修改。如果 T<sub>2</sub> 再次读取这个数据，此时读取的结果和第一次读取的结果不同。
 
-<div align="center"> <img src="../pics//c8d18ca9-0b09-441a-9a0c-fb063630d708.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//c8d18ca9-0b09-441a-9a0c-fb063630d708.png" width="350"/> </div><br>
 
 ## 幻影读
 
 T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插入新的数据，T<sub>1</sub> 再次读取这个范围的数据，此时读取的结果和和第一次读取的结果不同。
 
-<div align="center"> <img src="../pics//72fe492e-f1cb-4cfc-92f8-412fb3ae6fec.png" width="300"/> </div><br>
+<div align="center"> <img src="../pics//72fe492e-f1cb-4cfc-92f8-412fb3ae6fec.png" width="350"/> </div><br>
 
 ----
 
@@ -123,7 +122,6 @@ T<sub>1</sub> 读取某个范围的数据，T<sub>2</sub> 在这个范围内插
 
 ## 封锁粒度
 
-
 MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 应该尽量只锁定需要修改的那部分数据，而不是所有的资源。锁定的数据量越少，发生锁争用的可能就越小，系统的并发程度就越高。
@@ -150,8 +148,8 @@ MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 | - | X | S |
 | :--: | :--: | :--: |
-|X|NO|NO|
-|S|NO|YES|
+|X|×|×|
+|S|×|√|
 
 ### 2. 意向锁
 
@@ -170,10 +168,10 @@ MySQL 中提供了两种封锁粒度：行级锁以及表级锁。
 
 | - | X | IX | S | IS |
 | :--: | :--: | :--: | :--: | :--: |
-|X     |NO    |NO    |NO   | NO|
-|IX    |NO    |YES   |NO   | YES|
-|S     |NO    |NO    |YES  | YES|
-|IS    |NO    |YES  |YES  | YES|
+|X     |×    |×    |×   | ×|
+|IX    |×    |√   |×   | √|
+|S     |×    |×    |√  | √|
+|IS    |×    |√  |√  | √|
 
 解释如下：
 
@@ -298,32 +296,30 @@ SELECT ... FOR UPDATE;
 
 | 隔离级别 | 脏读 | 不可重复读 | 幻影读 |
 | :---: | :---: | :---:| :---: |
-| 未提交读 | YES | YES | YES |
-| 提交读 | NO | YES | YES |
-| 可重复读 | NO | NO | YES |
-| 可串行化 | NO | NO | NO |
+| 未提交读 | √ | √ | √ |
+| 提交读 | × | √ | √ |
+| 可重复读 | × | × | √ |
+| 可串行化 | × | × | × |
 
 # 五、多版本并发控制
 
-多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。
-
-而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC。
-
-可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
+多版本并发控制（Multi-Version Concurrency Control, MVCC）是 MySQL 的 InnoDB 存储引擎实现隔离级别的一种具体方式，用于实现提交读和可重复读这两种隔离级别。而未提交读隔离级别总是读取最新的数据行，无需使用 MVCC。可串行化隔离级别需要对所有读取的行都加锁，单纯使用 MVCC 无法实现。
 
 ## 版本号
 
 - 系统版本号：是一个递增的数字，每开始一个新的事务，系统版本号就会自动递增。
 - 事务版本号：事务开始时的系统版本号。
 
-InooDB 的 MVCC 在每行记录后面都保存着两个隐藏的列，用来存储两个版本号：
+## 隐藏的列
+
+MVCC 在每行记录后面都保存着两个隐藏的列，用来存储两个版本号：
 
 - 创建版本号：指示创建一个数据行的快照时的系统版本号；
 - 删除版本号：如果该快照的删除版本号大于当前事务版本号表示该快照有效，否则表示该快照已经被删除了。
 
 ## Undo 日志
 
-InnoDB 的 MVCC 使用到的快照存储在 Undo 日志中，该日志通过回滚指针把一个数据行（Record）的所有快照连接起来。
+MVCC 使用到的快照存储在 Undo 日志中，该日志通过回滚指针把一个数据行（Record）的所有快照连接起来。
 
 <div align="center"> <img src="../pics//e41405a8-7c05-4f70-8092-e961e28d3112.jpg" width=""/> </div><br>
 
@@ -331,15 +327,13 @@ InnoDB 的 MVCC 使用到的快照存储在 Undo 日志中，该日志通过回
 
 以下实现过程针对可重复读隔离级别。
 
-### 1. SELECT
-
 当开始新一个事务时，该事务的版本号肯定会大于当前所有数据行快照的创建版本号，理解这一点很关键。
 
+### 1. SELECT
+
 多个事务必须读取到同一个数据行的快照，并且这个快照是距离现在最近的一个有效快照。但是也有例外，如果有一个事务正在修改该数据行，那么它可以读取事务本身所做的修改，而不用和其它事务的读取结果一致。
 
-把没有对一个数据行做修改的事务称为 T，T 所要读取的数据行快照的创建版本号必须小于 T 的版本号，因为如果大于或者等于 T 的版本号，那么表示该数据行快照是其它事务的最新修改，因此不能去读取它。
-
-除了上面的要求，T 所要读取的数据行快照的删除版本号必须大于 T 的版本号，因为如果小于等于 T 的版本号，那么表示该数据行快照是已经被删除的，不应该去读取它。
+把没有对一个数据行做修改的事务称为 T，T 所要读取的数据行快照的创建版本号必须小于 T 的版本号，因为如果大于或者等于 T 的版本号，那么表示该数据行快照是其它事务的最新修改，因此不能去读取它。除此之外，T 所要读取的数据行快照的删除版本号必须大于 T 的版本号，因为如果小于等于 T 的版本号，那么表示该数据行快照是已经被删除的，不应该去读取它。
 
 ### 2. INSERT
 
@@ -385,7 +379,7 @@ MVCC 不能解决幻读的问题，Next-Key Locks 就是为了解决这个问题
 
 锁定一个记录上的索引，而不是记录本身。
 
-如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚集索引，因此 Record Locks 依然可以使用。
+如果表没有设置索引，InnoDB 会自动在主键上创建隐藏的聚簇索引，因此 Record Locks 依然可以使用。
 
 ## Gap Locks
 
@@ -397,7 +391,7 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 ## Next-Key Locks
 
-它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定范围内的索引。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
+它是 Record Locks 和 Gap Locks 的结合，不仅锁定一个记录上的索引，也锁定索引之间的间隙。例如一个索引包含以下值：10, 11, 13, and 20，那么就需要锁定以下区间：
 
 ```sql
 (negative infinity, 10]
@@ -432,10 +426,10 @@ SELECT c FROM t WHERE c BETWEEN 10 and 20 FOR UPDATE;
 
 不符合范式的关系，会产生很多异常，主要有以下四种异常：
 
-- 冗余数据：例如 学生-2 出现了两次。
+- 冗余数据：例如 `学生-2` 出现了两次。
 - 修改异常：修改了一个记录中的信息，但是另一个记录中相同的信息却没有被修改。
-- 删除异常：删除一个信息，那么也会丢失其它信息。例如如果删除了 课程-1，需要删除第一行和第三行，那么 学生-1 的信息就会丢失。
-- 插入异常，例如想要插入一个学生的信息，如果这个学生还没选课，那么就无法插入。
+- 删除异常：删除一个信息，那么也会丢失其它信息。例如删除了 `课程-1` 需要删除第一行和第三行，那么 `学生-1` 的信息就会丢失。
+- 插入异常：例如想要插入一个学生的信息，如果这个学生还没选课，那么就无法插入。
 
 ## 范式
 
@@ -506,7 +500,11 @@ Sname, Sdept 和 Mname 都部分依赖于键码，当一个学生选修了多门
 
 非主属性不传递函数依赖于键码。
 
-上面的 关系-1 中存在以下传递函数依赖：Sno -> Sdept -> Mname，可以进行以下分解：
+上面的 关系-1 中存在以下传递函数依赖：
+
+- Sno -> Sdept -> Mname
+
+可以进行以下分解：
 
 关系-11
 
@@ -533,13 +531,19 @@ Entity-Relationship，有三个组成部分：实体、属性、联系。
 
 包含一对一，一对多，多对多三种。
 
-如果 A 到 B 是一对多关系，那么画个带箭头的线段指向 B；如果是一对一，画两个带箭头的线段；如果是多对多，画两个不带箭头的线段。下图的 Course 和 Student 是一对多的关系。
+- 如果 A 到 B 是一对多关系，那么画个带箭头的线段指向 B；
+- 如果是一对一，画两个带箭头的线段；
+- 如果是多对多，画两个不带箭头的线段。
+
+下图的 Course 和 Student 是一对多的关系。
 
 <div align="center"> <img src="../pics//292b4a35-4507-4256-84ff-c218f108ee31.jpg" width=""/> </div><br>
 
 ## 表示出现多次的关系
 
-一个实体在联系出现几次，就要用几条线连接。下图表示一个课程的先修关系，先修关系出现两个 Course 实体，第一个是先修课程，后一个是后修课程，因此需要用两条线来表示这种关系。
+一个实体在联系出现几次，就要用几条线连接。
+
+下图表示一个课程的先修关系，先修关系出现两个 Course 实体，第一个是先修课程，后一个是后修课程，因此需要用两条线来表示这种关系。
 
 <div align="center"> <img src="../pics//8b798007-e0fb-420c-b981-ead215692417.jpg" width=""/> </div><br>
 
@@ -549,7 +553,7 @@ Entity-Relationship，有三个组成部分：实体、属性、联系。
 
 <div align="center"> <img src="../pics//423f2a40-bee1-488e-b460-8e76c48ee560.png" width=""/> </div><br>
 
-一般只使用二元联系，可以把多元关系转换为二元关系。
+一般只使用二元联系，可以把多元联系转换为二元联系。
 
 <div align="center"> <img src="../pics//de9b9ea0-1327-4865-93e5-6f805c48bc9e.png" width=""/> </div><br>
 
diff --git a/notes/设计模式.md b/notes/设计模式.md
index b8f4dc9b..531595e3 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -60,9 +60,9 @@
 
 （一）懒汉式-线程不安全
 
-以下实现中，私有静态变量 uniqueInstance 被延迟化实例化，这样做的好处是，如果没有用到该类，那么就不会实例化 uniqueInstance，从而节约资源。
+以下实现中，私有静态变量 uniqueInstance 被延迟实例化，这样做的好处是，如果没有用到该类，那么就不会实例化 uniqueInstance，从而节约资源。
 
-这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么多个线程会执行 `uniqueInstance = new Singleton();` 语句，这将导致多次实例化 uniqueInstance。
+这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么会有多个线程执行 `uniqueInstance = new Singleton();` 语句，这将导致多次实例化 uniqueInstance。
 
 ```java
 public class Singleton {
@@ -81,11 +81,21 @@ public class Singleton {
 }
 ```
 
-（二）懒汉式-线程安全
+（二）饿汉式-线程安全
 
-只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了对 uniqueInstance 进行多次实例化的问题。
+线程不安全问题主要是由于 uniqueInstance 被多次实例化，采取直接实例化 uniqueInstance 的方式就不会产生线程不安全问题。
 
-但是这样有一个问题，就是当一个线程进入该方法之后，其它线程试图进入该方法都必须等待，因此性能上有一定的损耗。
+但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
+
+```java
+private static Singleton uniqueInstance = new Singleton();
+```
+
+（三）懒汉式-线程安全
+
+只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了多次实例化 uniqueInstance 的问题。
+
+但是当一个线程进入该方法之后，其它试图进入该方法的线程都必须等待，因此性能上有一定的损耗。
 
 ```java
 public static synchronized Singleton getUniqueInstance() {
@@ -96,16 +106,6 @@ public static synchronized Singleton getUniqueInstance() {
 }
 ```
 
-（三）饿汉式-线程安全
-
-线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。
-
-但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
-
-```java
-private static Singleton uniqueInstance = new Singleton();
-```
-
 （四）双重校验锁-线程安全
 
 uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
@@ -175,7 +175,7 @@ public class Singleton {
 }
 ```
 
-（五）枚举实现
+（六）枚举实现
 
 这是单例模式的最佳实践，它实现简单，并且在面对复杂的序列化或者反射攻击的时候，能够防止实例化多次。
 
@@ -231,27 +231,9 @@ public class Singleton implements Serializable {
 
 简单工厂不是设计模式，更像是一种编程习惯。它把实例化的操作单独放到一个类中，这个类就成为简单工厂类，让简单工厂类来决定应该用哪个具体子类来实例化。
 
-<div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
-
 这样做能把客户类和具体子类的实现解耦，客户类不再需要知道有哪些子类以及应当实例化哪个子类。因为客户类往往有多个，如果不使用简单工厂，所有的客户类都要知道所有子类的细节。而且一旦子类发生改变，例如增加子类，那么所有的客户类都要进行修改。
 
-如果存在下面这种代码，就需要使用简单工厂将对象实例化的部分放到简单工厂中。
-
-```java
-public class Client {
-    public static void main(String[] args) {
-        int type = 1;
-        Product product;
-        if (type == 1) {
-            product = new ConcreteProduct1();
-        } else if (type == 2) {
-            product = new ConcreteProduct2();
-        } else {
-            product = new ConcreteProduct();
-        }
-    }
-}
-```
+<div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
 
 ### 实现
 
@@ -275,6 +257,27 @@ public class ConcreteProduct2 implements Product {
 }
 ```
 
+以下的 Client 类中包含了实例化的代码，这是一种错误的实现，如果在客户类中存在实例化代码，就需要将代码放到简单工厂中。
+
+```java
+public class Client {
+    public static void main(String[] args) {
+        int type = 1;
+        Product product;
+        if (type == 1) {
+            product = new ConcreteProduct1();
+        } else if (type == 2) {
+            product = new ConcreteProduct2();
+        } else {
+            product = new ConcreteProduct();
+        }
+        // do something with the product
+    }
+}
+```
+
+以下的 SimpleFactory 是简单工厂实现，它被所有需要进行实例化的客户类调用。
+
 ```java
 public class SimpleFactory {
     public Product createProduct(int type) {
@@ -293,6 +296,7 @@ public class Client {
     public static void main(String[] args) {
         SimpleFactory simpleFactory = new SimpleFactory();
         Product product = simpleFactory.createProduct(1);
+        // do something with the product
     }
 }
 ```
@@ -2906,7 +2910,7 @@ Java 利用缓存来加速大量小对象的访问时间。
 - 远程代理（Remote Proxy）：控制对远程对象（不同地址空间）的访问，它负责将请求及其参数进行编码，并向不同地址空间中的对象发送已经编码的请求。
 - 虚拟代理（Virtual Proxy）：根据需要创建开销很大的对象，它可以缓存实体的附加信息，以便延迟对它的访问，例如在网站加载一个很大图片时，不能马上完成，可以用虚拟代理缓存图片的大小信息，然后生成一张临时图片代替原始图片。
 - 保护代理（Protection Proxy）：按权限控制对象的访问，它负责检查调用者是否具有实现一个请求所必须的访问权限。
-- 智能代理（Smart Reference）：取代了简单的指针，它在访问对象时执行一些附加操作：记录对象的引用次数，比如智能智能；当第一次引用一个持久化对象时，将它装入内存；在访问一个实际对象前，检查是否已经锁定了它，以确保其它对象不能改变它。
+- 智能代理（Smart Reference）：取代了简单的指针，它在访问对象时执行一些附加操作：记录对象的引用次数；当第一次引用一个持久化对象时，将它装入内存；在访问一个实际对象前，检查是否已经锁定了它，以确保其它对象不能改变它。
 
 <div align="center"> <img src="../pics//a6c20f60-5eba-427d-9413-352ada4b40fe.png"/> </div><br>
 
diff --git a/notes/面向对象思想.md b/notes/面向对象思想.md
index 9b59e43d..203c7ebf 100644
--- a/notes/面向对象思想.md
+++ b/notes/面向对象思想.md
@@ -37,6 +37,7 @@
 
 ```java
 public class Person {
+
     private String name;
     private int gender;
     private int age;
@@ -63,17 +64,20 @@ public class Person {
 
 继承实现了  **IS-A**  关系，例如 Cat 和 Animal 就是一种 IS-A 关系，因此 Cat 可以继承自 Animal，从而获得 Animal 非 private 的属性和方法。
 
+继承应该遵循里氏替换原则，子类对象必须能够替换掉所有父类对象。
+
 Cat 可以当做 Animal 来使用，也就是说可以使用 Animal 引用 Cat 对象。父类引用指向子类对象称为  **向上转型** 。
 
 ```java
 Animal animal = new Cat();
 ```
 
-继承应该遵循里氏替换原则，子类对象必须能够替换掉所有父类对象。
-
 ## 多态
 
-多态分为编译时多态和运行时多态。编译时多态主要指方法的重载，运行时多态指程序中定义的对象引用所指向的具体类型在运行期间才确定。
+多态分为编译时多态和运行时多态：
+
+- 编译时多态主要指方法的重载
+- 运行时多态指程序中定义的对象引用所指向的具体类型在运行期间才确定
 
 运行时多态有三个条件：
 
@@ -116,7 +120,7 @@ public class Music {
 
 # 二、类图
 
-以下类图使用 [PlantUML](https://www.planttext.com/) 绘制，更多语法及使用请参考：http://plantuml.com/
+以下类图使用 [PlantUML](https://www.planttext.com/) 绘制，更多语法及使用请参考：http://plantuml.com/ 。
 
 ## 泛化关系 (Generalization)
 
@@ -327,7 +331,7 @@ Vihicle .. N
 
 ### 2. 合成复用原则
 
-尽量使用对象组合，而不是继承来达到复用的目的。
+尽量使用对象组合，而不是通过继承来达到复用的目的。
 
 ### 3. 共同封闭原则
 
@@ -349,3 +353,4 @@ Vihicle .. N
 - [看懂 UML 类图和时序图](http://design-patterns.readthedocs.io/zh_CN/latest/read_uml.html#generalization)
 - [UML 系列——时序图（顺序图）sequence diagram](http://www.cnblogs.com/wolf-sun/p/UML-Sequence-diagram.html)
 - [面向对象编程三大特性 ------ 封装、继承、多态](http://blog.csdn.net/jianyuerensheng/article/details/51602015)
+

From 99a1aeb51d11c22ee2221d201e584cb7955b4138 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Fri, 17 Aug 2018 21:36:50 +0800
Subject: [PATCH 124/172] Update README.md

---
 README.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/README.md b/README.md
index f35ee455..942d6096 100644
--- a/README.md
+++ b/README.md
@@ -13,6 +13,10 @@
 
 ## 算法 :pencil2:
 
+- [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
+
+  排序、并查集、栈和队列、红黑树、散列表。
+
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
   目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
@@ -21,10 +25,6 @@
 
   对题目做了一个大致分类，并对每种题型的解题思路做了总结。
 
-- [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
-
-  排序、并查集、栈和队列、红黑树、散列表。
-
 ## 操作系统 :computer:
 
 - [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)

From d33f9a31119bf9703f78d144a97b2533f80995e5 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Sat, 18 Aug 2018 16:29:59 +0800
Subject: [PATCH 125/172] Update README.md

---
 README.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/README.md b/README.md
index 942d6096..3209d82f 100644
--- a/README.md
+++ b/README.md
@@ -13,10 +13,6 @@
 
 ## 算法 :pencil2:
 
-- [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
-
-  排序、并查集、栈和队列、红黑树、散列表。
-
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
   目录根据原书第二版进行编排，代码和原书有所不同，尽量比原书更简洁。
@@ -24,6 +20,10 @@
 - [Leetcode 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Leetcode%20题解.md)
 
   对题目做了一个大致分类，并对每种题型的解题思路做了总结。
+  
+  - [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
+
+  排序、并查集、栈和队列、红黑树、散列表。
 
 ## 操作系统 :computer:
 

From f100a61a61dea9f7c09ee9493fd0fc8be89568bb Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Sat, 18 Aug 2018 16:30:19 +0800
Subject: [PATCH 126/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 3209d82f..f9c604d9 100644
--- a/README.md
+++ b/README.md
@@ -21,7 +21,7 @@
 
   对题目做了一个大致分类，并对每种题型的解题思路做了总结。
   
-  - [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
+ - [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
 
   排序、并查集、栈和队列、红黑树、散列表。
 

From fad52dbef4167ace6a352e7c9dd9bf57f52ceb68 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Sat, 18 Aug 2018 16:30:33 +0800
Subject: [PATCH 127/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index f9c604d9..ce7257e5 100644
--- a/README.md
+++ b/README.md
@@ -23,7 +23,7 @@
   
  - [算法](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/算法.md)
 
-  排序、并查集、栈和队列、红黑树、散列表。
+   排序、并查集、栈和队列、红黑树、散列表。
 
 ## 操作系统 :computer:
 

From 7186c28cc5a3c42d228188c579f2d7161b4634ac Mon Sep 17 00:00:00 2001
From: zhang xuelong <35026038+zhangxuelong10@users.noreply.github.com>
Date: Sun, 19 Aug 2018 18:33:49 +0800
Subject: [PATCH 128/172] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E6=8B=BC=E5=86=99?=
 =?UTF-8?q?=E9=94=99=E8=AF=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 基础.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index b329e0bc..44ca873a 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -707,7 +707,7 @@ public class EqualExample {
 
 ## hashCode()
 
-hasCode() 返回散列值，而 equals() 是用来判断两个对象是否等价。等价的两个对象散列值一定相同，但是散列值相同的两个对象不一定等价。
+hashCode() 返回散列值，而 equals() 是用来判断两个对象是否等价。等价的两个对象散列值一定相同，但是散列值相同的两个对象不一定等价。
 
 在覆盖 equals() 方法时应当总是覆盖 hashCode() 方法，保证等价的两个对象散列值也相等。
 

From d10ec622b02fff0a465ad8e93ce21a32df3dfe2e Mon Sep 17 00:00:00 2001
From: RocXing <120695313@qq.com>
Date: Mon, 20 Aug 2018 18:46:03 +0800
Subject: [PATCH 129/172] =?UTF-8?q?=E8=A1=A5=E5=AD=97?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 64f9e7d6..ea8b2898 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -185,7 +185,7 @@ public static void main(String[] args) {
 }
 ```
 ## 三种实现方式的比较
-- 实现Runnable接又可以避免Java单继承特性而带来的局限;增强程序的健壮性，代码能够被多个线程共享，代码与数据是独立的;适合多个相同程序代码的线程区处理同一资源的情况。
+- 实现Runnable接口又可以避免Java单继承特性而带来的局限;增强程序的健壮性，代码能够被多个线程共享，代码与数据是独立的;适合多个相同程序代码的线程区处理同一资源的情况。
 - 继承Thread类和实现Runnable方法启动线 程都是使用start方法，然后JVM虚拟机将此线程放到就绪队列中，如果有处理机可用， 则执行run方法。
 - 实现Callable接又要实现call方法，并且线 程执行完毕后会有返回值。其他的两种都是 重写run方法，没有返回值。
 ## 实现接口 VS 继承 Thread

From 67322b7a32ddf156827a0977a17493e3a3000509 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Mon, 20 Aug 2018 20:32:42 +0800
Subject: [PATCH 130/172] auto commit

---
 notes/Docker.md        | 11 ++++++-----
 notes/Java 并发.md     |  5 +----
 notes/Leetcode 题解.md |  9 ++++-----
 3 files changed, 11 insertions(+), 14 deletions(-)

diff --git a/notes/Docker.md b/notes/Docker.md
index f230f762..80108896 100644
--- a/notes/Docker.md
+++ b/notes/Docker.md
@@ -30,15 +30,15 @@ Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程
 
 ## 启动速度
 
-启动虚拟机需要启动虚拟机的操作系统，再启动相应的应用，这个过程会非常慢；
+启动虚拟机需要启动虚拟机的操作系统，再启动应用，这个过程非常慢；
 
 而启动 Docker 相当于启动宿主操作系统上的一个进程。
 
 ## 占用资源
 
-虚拟机是一个完整的操作系统，需要占用大量的磁盘空间、内存和 CPU，一台机器只能开启几十个的虚拟机。
+虚拟机是一个完整的操作系统，需要占用大量的磁盘、内存和 CPU，一台机器只能开启几十个的虚拟机。
 
-而 Docker 只是一个进程，只需要将应用以及相应的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
+而 Docker 只是一个进程，只需要将应用以及相关的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
 
 参考资料：
 
@@ -58,7 +58,7 @@ Docker 使用分层技术和镜像，使得应用可以更容易复用重复部
 
 ## 更容易扩展
 
-可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像得到我们想要的镜像非常容易。
+可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像可以非常容易得到我们想要的镜像。
 
 参考资料：
 
@@ -91,7 +91,7 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 
 镜像包含着容器运行时所需要的代码以及其它组件，它是一种分层结构，每一层都是只读的（read-only layers）。构建镜像时，会一层一层构建，前一层是后一层的基础。镜像的这种分层存储结构很适合镜像的复用以及定制。
 
-在构建容器时，通过在镜像的基础上添加一个可写层（writable layer），用来保存着容器运行过程中的修改。
+构建容器时，通过在镜像的基础上添加一个可写层（writable layer），用来保存着容器运行过程中的修改。
 
 <div align="center"> <img src="../pics//docker-filesystems-busyboxrw.png"/> </div><br>
 
@@ -100,3 +100,4 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 - [How to Create Docker Container using Dockerfile](https://linoxide.com/linux-how-to/dockerfile-create-docker-container/)
 - [理解 Docker（2）：Docker 镜像](http://www.cnblogs.com/sammyliu/p/5877964.html)
 
+
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index ea8b2898..d3682d33 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -184,10 +184,7 @@ public static void main(String[] args) {
     mt.start();
 }
 ```
-## 三种实现方式的比较
-- 实现Runnable接口又可以避免Java单继承特性而带来的局限;增强程序的健壮性，代码能够被多个线程共享，代码与数据是独立的;适合多个相同程序代码的线程区处理同一资源的情况。
-- 继承Thread类和实现Runnable方法启动线 程都是使用start方法，然后JVM虚拟机将此线程放到就绪队列中，如果有处理机可用， 则执行run方法。
-- 实现Callable接又要实现call方法，并且线 程执行完毕后会有返回值。其他的两种都是 重写run方法，没有返回值。
+
 ## 实现接口 VS 继承 Thread
 
 实现接口会更好一些，因为：
diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 68af3cb7..04a751c7 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -3602,7 +3602,7 @@ public int countPrimes(int n) {
 
 ```java
 int gcd(int a, int b) {
-    return b == 0 ? a : gcd(b, a% b);
+    return b == 0 ? a : gcd(b, a % b);
 }
 ```
 
@@ -3673,7 +3673,7 @@ public String convertToBase7(int num) {
 }
 ```
 
-Java 中 static String toString(int num, int radix) 可以将一个整数转换为 redix 进制表示的字符串。
+Java 中 static String toString(int num, int radix) 可以将一个整数转换为 radix 进制表示的字符串。
 
 ```java
 public String convertToBase7(int num) {
@@ -6403,7 +6403,6 @@ public int maxChunksToSorted(int[] arr) {
 }
 ```
 
-
 ## 图
 
 ### 二分图
@@ -6620,6 +6619,7 @@ public int[] findRedundantConnection(int[][] edges) {
 }
 
 private class UF {
+
     private int[] id;
 
     UF(int N) {
@@ -6665,7 +6665,7 @@ x ^ x = 0       x & x = x       x | x = x
 ```
 
 - 利用 x ^ 1s = \~x 的特点，可以将位级表示翻转；利用 x ^ x = 0 的特点，可以将三个数中重复的两个数去除，只留下另一个数。
-- 利用 x & 0s = 0 和 x & 1s = x 的特点，可以实现掩码操作。一个数 num 与 mask ：00111100 进行位与操作，只保留 num 中与 mask 的 1 部分相对应的位。
+- 利用 x & 0s = 0 和 x & 1s = x 的特点，可以实现掩码操作。一个数 num 与 mask：00111100 进行位与操作，只保留 num 中与 mask 的 1 部分相对应的位。
 - 利用 x | 0s = x 和 x | 1s = 1s 的特点，可以实现设值操作。一个数 num 与 mask：00111100 进行位或操作，将 num 中与 mask 的 1 部分相对应的位都设置为 1。
 
 位与运算技巧：
@@ -6910,7 +6910,6 @@ public boolean isPowerOfFour(int num) {
 }
 ```
 
-
 **判断一个数的位级表示是否不会出现连续的 0 和 1** 
 
 [693. Binary Number with Alternating Bits (Easy)](https://leetcode.com/problems/binary-number-with-alternating-bits/description/)

From f7f68ba1f86b7af5d938dbad8630f7dbfae124a2 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Mon, 20 Aug 2018 20:50:41 +0800
Subject: [PATCH 131/172] Update README.md

---
 README.md | 36 ++++++++++++++++++------------------
 1 file changed, 18 insertions(+), 18 deletions(-)

diff --git a/README.md b/README.md
index ce7257e5..bf186a04 100644
--- a/README.md
+++ b/README.md
@@ -11,7 +11,7 @@
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->
 
-## 算法 :pencil2:
+### 算法 :pencil2:
 
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
@@ -25,7 +25,7 @@
 
    排序、并查集、栈和队列、红黑树、散列表。
 
-## 操作系统 :computer:
+### 操作系统 :computer:
 
 - [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
@@ -35,7 +35,7 @@
 
   基本实现原理以及基本操作。
 
-## 网络 :cloud:
+### 网络 :cloud:
 
 - [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
@@ -49,7 +49,7 @@
 
   I/O 模型、I/O 多路复用。
 
-## 面向对象 :couple:
+### 面向对象 :couple:
 
 - [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
@@ -59,7 +59,7 @@
 
   三大原则（继承、封装、多态）、类图、设计原则。
 
-## 数据库 :floppy_disk:
+### 数据库 :floppy_disk:
 
 - [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
@@ -81,7 +81,7 @@
 
   五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
-## Java :coffee:
+### Java :coffee:
 
 - [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
@@ -103,7 +103,7 @@
 
   NIO 的原理以及实例。
 
-## 系统设计 :bulb:
+### 系统设计 :bulb:
 
 - [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
 
@@ -129,7 +129,7 @@
 
   消息处理模型、使用场景、可靠性
 
-## 工具 :hammer:
+### 工具 :hammer:
 
 - [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
 
@@ -147,7 +147,7 @@
 
   构建工具的基本概念、主流构建工具介绍。
 
-## 编码实践 :speak_no_evil:
+### 编码实践 :speak_no_evil:
 
 - [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
 
@@ -161,9 +161,9 @@
 
   Google 开源项目的代码风格规范。
 
-## 后记 :memo:
+### 后记 :memo:
 
-### About
+#### About
 
 这个仓库是笔者的一个学习笔记，主要总结一些比较重要的知识点，希望对大家有所帮助。
 
@@ -171,7 +171,7 @@
 
 [BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)，这个书单是笔者至今看的一些比较好的技术书籍，虽然没有全都看完，但每本书多多少少都看了一部分。
 
-### How To Contribute
+#### How To Contribute
 
 笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
 
@@ -179,7 +179,7 @@
 
 欢迎在 Issue 中提交对本仓库的改进建议~
 
-### Typesetting
+#### Typesetting
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
 
@@ -187,7 +187,7 @@
 
 笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
 
-### Uploading
+#### Uploading
 
 笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
 
@@ -195,21 +195,21 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
-### License
+#### License
 
 在对本作品进行演绎时，请署名并以相同方式共享。
 
 <a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
 
-### Statement
+#### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)
 
-### Logo
+#### Logo
 
 Power by [logomakr](https://logomakr.com/).
 
-### Acknowledgements
+#### Acknowledgements
 
 感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与笔者联系。
 

From 13e069030293de666b6c99fd928dfdaeb51f505e Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Mon, 20 Aug 2018 20:58:21 +0800
Subject: [PATCH 132/172] Update README.md

---
 README.md | 38 +++++++++++++++++++-------------------
 1 file changed, 19 insertions(+), 19 deletions(-)

diff --git a/README.md b/README.md
index bf186a04..3e829c1b 100644
--- a/README.md
+++ b/README.md
@@ -1,6 +1,6 @@
 | Ⅰ | Ⅱ | Ⅲ | Ⅳ | Ⅴ | Ⅵ | Ⅶ | Ⅷ | Ⅸ | Ⅹ |
 | :--------: | :---------: | :---------: | :---------: | :---------: | :---------:| :---------: | :-------: | :-------:| :------:|
-| 算法[:pencil2:](#算法-pencil2) | 操作系统[:computer:](#操作系统-computer)|网络[:cloud:](#网络-cloud) | 面向对象[:couple:](#面向对象-couple) |数据库[:floppy_disk:](#数据库-floppy_disk)| Java [:coffee:](#java-coffee)| 系统设计[:bulb:](#系统设计-bulb)| 工具[:hammer:](#工具-hammer)| 编码实践[:speak_no_evil:](#编码实践-speak_no_evil)| 后记[:memo:](#后记-memo) |
+| 算法[:pencil2:](#pencil2-算法) | 操作系统[:computer:](#computer-操作系统)|网络[:cloud:](#cloud-网络) | 面向对象[:couple:](#couple-面向对象) |数据库[:floppy_disk:](#floppy_disk-数据库)| Java [:coffee:](#coffee-java)| 系统设计[:bulb:](#bulb-系统设计)| 工具[:hammer:](#hammer-工具)| 编码实践[:speak_no_evil:](#speak_no_evil-编码实践)| 后记[:memo:](#memo-后记) |
 
 <br>
 <div align="center">
@@ -11,7 +11,7 @@
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->
 
-### 算法 :pencil2:
+## :pencil2: 算法
 
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
@@ -25,7 +25,7 @@
 
    排序、并查集、栈和队列、红黑树、散列表。
 
-### 操作系统 :computer:
+## :computer: 操作系统
 
 - [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
@@ -35,7 +35,7 @@
 
   基本实现原理以及基本操作。
 
-### 网络 :cloud:
+## :cloud: 网络 
 
 - [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
@@ -49,7 +49,7 @@
 
   I/O 模型、I/O 多路复用。
 
-### 面向对象 :couple:
+## :couple: 面向对象
 
 - [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
@@ -59,7 +59,7 @@
 
   三大原则（继承、封装、多态）、类图、设计原则。
 
-### 数据库 :floppy_disk:
+## :floppy_disk: 数据库 
 
 - [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
@@ -81,7 +81,7 @@
 
   五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
-### Java :coffee:
+## :coffee: Java
 
 - [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
@@ -103,7 +103,7 @@
 
   NIO 的原理以及实例。
 
-### 系统设计 :bulb:
+## :bulb: 系统设计 
 
 - [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
 
@@ -129,7 +129,7 @@
 
   消息处理模型、使用场景、可靠性
 
-### 工具 :hammer:
+## :hammer: 工具 
 
 - [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
 
@@ -147,7 +147,7 @@
 
   构建工具的基本概念、主流构建工具介绍。
 
-### 编码实践 :speak_no_evil:
+## :speak_no_evil: 编码实践 
 
 - [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
 
@@ -161,9 +161,9 @@
 
   Google 开源项目的代码风格规范。
 
-### 后记 :memo:
+## :memo: 后记 
 
-#### About
+### About
 
 这个仓库是笔者的一个学习笔记，主要总结一些比较重要的知识点，希望对大家有所帮助。
 
@@ -171,7 +171,7 @@
 
 [BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)，这个书单是笔者至今看的一些比较好的技术书籍，虽然没有全都看完，但每本书多多少少都看了一部分。
 
-#### How To Contribute
+### How To Contribute
 
 笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
 
@@ -179,7 +179,7 @@
 
 欢迎在 Issue 中提交对本仓库的改进建议~
 
-#### Typesetting
+### Typesetting
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
 
@@ -187,7 +187,7 @@
 
 笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
 
-#### Uploading
+### Uploading
 
 笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
 
@@ -195,21 +195,21 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
-#### License
+### License
 
 在对本作品进行演绎时，请署名并以相同方式共享。
 
 <a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
 
-#### Statement
+### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)
 
-#### Logo
+### Logo
 
 Power by [logomakr](https://logomakr.com/).
 
-#### Acknowledgements
+### Acknowledgements
 
 感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与笔者联系。
 

From ded6dee5cd049d22fc558b8e5e755c33c0b71b4b Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Mon, 20 Aug 2018 20:59:44 +0800
Subject: [PATCH 133/172] Update README.md

---
 README.md | 10 ++++------
 1 file changed, 4 insertions(+), 6 deletions(-)

diff --git a/README.md b/README.md
index 3e829c1b..d511fa7c 100644
--- a/README.md
+++ b/README.md
@@ -195,12 +195,6 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
-### License
-
-在对本作品进行演绎时，请署名并以相同方式共享。
-
-<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
-
 ### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)
@@ -235,5 +229,9 @@ Power by [logomakr](https://logomakr.com/).
     <img src="other/10072416.jpg" width="50px">
 </a>
 
+### License
 
+在对本作品进行演绎时，请署名并以相同方式共享。
+
+<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
 

From db08ad2cda7d49e8d93e34c9b6397380cf27bae0 Mon Sep 17 00:00:00 2001
From: Elong <peielcn@gmail.com>
Date: Tue, 21 Aug 2018 15:11:03 +0800
Subject: [PATCH 134/172] =?UTF-8?q?Update=20Java=20=E5=B9=B6=E5=8F=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

文字描述修改的更加严谨一些
---
 notes/Java 并发.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index d3682d33..16e20a24 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1020,7 +1020,7 @@ ForkJoinPool 实现了工作窃取算法来提高 CPU 的利用率。每个线
 
 如果多个线程对同一个共享数据进行访问而不采取同步操作的话，那么操作的结果是不一致的。
 
-以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值为 997 而不是 1000。
+以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值一般情况下小于 1000。
 
 ```java
 public class ThreadUnsafeExample {

From cd354fb46cd472b9b20a206df89ae7b83bb0b890 Mon Sep 17 00:00:00 2001
From: wangxujian-laixuzhui
 <33362969+wangxujian-laixuzhui@users.noreply.github.com>
Date: Tue, 21 Aug 2018 18:29:57 +0800
Subject: [PATCH 135/172] =?UTF-8?q?=E5=BF=AB=E6=8E=92=E4=B8=8D=E7=A8=B3?=
 =?UTF-8?q?=E5=AE=9A=EF=BC=8C=E5=A6=82221=EF=BC=8C=E7=AC=AC=E4=B8=80?=
 =?UTF-8?q?=E4=B8=AA2=E4=BC=9A=E5=92=8C1=E4=BA=A4=E6=8D=A2?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/算法.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/算法.md b/notes/算法.md
index 30467b42..8007375b 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -770,7 +770,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 | 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
-| 选择排序 | √| N<sup>2</sup> | 1 | |
+| 选择排序 | ×| N<sup>2</sup> | 1 | |
 | 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
 | 插入排序 | √ |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
 | 希尔排序 | ×  |  N 的若干倍乘于递增序列的长度 | 1 | |

From d4610276542bc6ebd4940a0f0845cc27ce4b74ea Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 21 Aug 2018 20:05:55 +0800
Subject: [PATCH 136/172] auto commit

---
 notes/Java 并发.md |  2 +-
 notes/攻击技术.md  |  2 +-
 notes/消息队列.md  | 13 +++++++------
 notes/算法.md      |  2 +-
 notes/缓存.md      | 30 +++++++++---------------------
 5 files changed, 19 insertions(+), 30 deletions(-)

diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 16e20a24..1403f764 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -1020,7 +1020,7 @@ ForkJoinPool 实现了工作窃取算法来提高 CPU 的利用率。每个线
 
 如果多个线程对同一个共享数据进行访问而不采取同步操作的话，那么操作的结果是不一致的。
 
-以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值一般情况下小于 1000。
+以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值有可能小于 1000。
 
 ```java
 public class ThreadUnsafeExample {
diff --git a/notes/攻击技术.md b/notes/攻击技术.md
index d5a735ba..46dd4616 100644
--- a/notes/攻击技术.md
+++ b/notes/攻击技术.md
@@ -207,7 +207,7 @@ ResultSet rs = stmt.executeQuery();
 
 拒绝服务攻击（denial-of-service attack，DoS），亦称洪水攻击，其目的在于使目标电脑的网络或系统资源耗尽，使服务暂时中断或停止，导致其正常用户无法访问。
 
-分布式拒绝服务攻击（distributed denial-of-service attack，DDoS），指攻击者使用网络上两个或以上被攻陷的电脑作为“僵尸”向特定的目标发动“拒绝服务”式攻击。
+分布式拒绝服务攻击（distributed denial-of-service attack，DDoS），指攻击者使用两个或以上被攻陷的电脑作为“僵尸”向特定的目标发动“拒绝服务”式攻击。
 
 # 参考资料
 
diff --git a/notes/消息队列.md b/notes/消息队列.md
index c5687e8a..209d962a 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -9,6 +9,7 @@
 * [三、可靠性](#三可靠性)
     * [发送端的可靠性](#发送端的可靠性)
     * [接收端的可靠性](#接收端的可靠性)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -29,15 +30,10 @@
 发布与订阅模式和观察者模式有以下不同：
 
 - 观察者模式中，观察者和主题都知道对方的存在；而在发布与订阅模式中，发布者与订阅者不知道对方的存在，它们之间通过频道进行通信。
-- 观察者模式是同步的，当事件触发时，主题会去调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息。
+- 观察者模式是同步的，当事件触发时，主题会调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息，可以立即返回。
 
 <div align="center"> <img src="../pics//bee1ff1d-c80f-4b3c-b58c-7073a8896ab2.jpg"/> </div><br>
 
-参考：
-
-- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
-- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
-
 # 二、使用场景
 
 ## 异步处理
@@ -78,3 +74,8 @@
 
 - 保证接收端处理消息的业务逻辑具有幂等性：只要具有幂等性，那么消费多少次消息，最后处理的结果都是一样的。
 - 保证消息具有唯一编号，并使用一张日志表来记录已经消费的消息编号。
+
+# 参考资料
+
+- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
+- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
diff --git a/notes/算法.md b/notes/算法.md
index 8007375b..152adfda 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -770,7 +770,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 | 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
-| 选择排序 | ×| N<sup>2</sup> | 1 | |
+| 选择排序 | × | N<sup>2</sup> | 1 | |
 | 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
 | 插入排序 | √ |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
 | 希尔排序 | ×  |  N 的若干倍乘于递增序列的长度 | 1 | |
diff --git a/notes/缓存.md b/notes/缓存.md
index 9a9f2810..ed834253 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -6,6 +6,7 @@
 * [五、缓存问题](#五缓存问题)
 * [六、数据分布](#六数据分布)
 * [七、一致性哈希](#七一致性哈希)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -29,10 +30,6 @@
 
 - LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最远的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
 
-参考资料：
-
-- [缓存那些事](https://tech.meituan.com/cache_about.html)
-
 # 二、LRU
 
 以下是一个基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
@@ -143,10 +140,6 @@ public class LRU<K, V> implements Iterable<K> {
 }
 ```
 
-源代码：
-
-- [CyC2018/Algorithm](https://github.com/CyC2018/Algorithm/tree/master/Caching)
-
 # 三、缓存位置
 
 ## 浏览器
@@ -169,7 +162,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。
 
-相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存，而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
 
 ## 数据库缓存
 
@@ -177,7 +170,7 @@ MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查
 
 # 四、CDN
 
-内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
+内容分发网络（Content distribution network，CDN）是一种互连的网络系统，它利用更靠近用户的服务器从而更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
 
 CDN 主要有以下优点：
 
@@ -187,11 +180,6 @@ CDN 主要有以下优点：
 
 <div align="center"> <img src="../pics//15313ed8-a520-4799-a300-2b6b36be314f.jpg"/> </div><br>
 
-参考资料：
-
-- [内容分发网络](https://zh.wikipedia.org/wiki/%E5%85%A7%E5%AE%B9%E5%82%B3%E9%81%9E%E7%B6%B2%E8%B7%AF)
-- [How Aspiration CDN helps to improve your website loading speed?](https://www.aspirationhosting.com/aspiration-cdn/)
-
 # 五、缓存问题
 
 ## 缓存穿透
@@ -213,7 +201,7 @@ CDN 主要有以下优点：
 
 - 为了防止缓存在同一时间大面积过期导致的缓存雪崩，可以通过观察用户行为，合理设置缓存过期时间来实现；
 - 为了防止缓存服务器宕机出现的缓存雪崩，可以使用分布式缓存，分布式缓存中每一个节点只缓存部分的数据，当某个节点宕机时可以保证其它节点的缓存仍然可用。
-- 也可以在进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
+- 也可以进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
 
 ## 缓存一致性
 
@@ -243,10 +231,6 @@ CDN 主要有以下优点：
 - 能保持数据原有的顺序；
 - 并且能够准确控制每台服务器存储的数据量，从而使得存储空间的利用率最大。
 
-参考资料：
-
-- 大规模分布式存储系统
-
 # 七、一致性哈希
 
 Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据失效的问题。
@@ -267,6 +251,10 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
 
-参考资料：
+# 参考资料
 
+- 大规模分布式存储系统
+- [缓存那些事](https://tech.meituan.com/cache_about.html)
 - [一致性哈希算法](https://my.oschina.net/jayhu/blog/732849)
+- [内容分发网络](https://zh.wikipedia.org/wiki/%E5%85%A7%E5%AE%B9%E5%82%B3%E9%81%9E%E7%B6%B2%E8%B7%AF)
+- [How Aspiration CDN helps to improve your website loading speed?](https://www.aspirationhosting.com/aspiration-cdn/)

From 25cd99a8d03637bbc624e58e3991a41c36810c1a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 21 Aug 2018 20:14:05 +0800
Subject: [PATCH 137/172] auto commit

---
 notes/算法.md                                 |   2 +-
 pics/766aedd0-1b00-4065-aa2b-7d31138df84f.png | Bin 0 -> 19456 bytes
 2 files changed, 1 insertion(+), 1 deletion(-)
 create mode 100644 pics/766aedd0-1b00-4065-aa2b-7d31138df84f.png

diff --git a/notes/算法.md b/notes/算法.md
index 152adfda..25d1550e 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -507,7 +507,7 @@ public class QuickSort<T extends Comparable<T>> extends Sort<T> {
 
 取 a[l] 作为切分元素，然后从数组的左端向右扫描直到找到第一个大于等于它的元素，再从数组的右端向左扫描找到第一个小于等于它的元素，交换这两个元素。不断进行这个过程，就可以保证左指针 i 的左侧元素都不大于切分元素，右指针 j 的右侧元素都不小于切分元素。当两个指针相遇时，将切分元素 a[l] 和 a[j] 交换位置。
 
-<div align="center"> <img src="../pics//864bfa7d-1149-420c-a752-f9b3d4e782ec.png" width="400"/> </div><br>
+<div align="center"> <img src="../pics//766aedd0-1b00-4065-aa2b-7d31138df84f.png" width="400"/> </div><br>
 
 ```java
 private int partition(T[] nums, int l, int h) {
diff --git a/pics/766aedd0-1b00-4065-aa2b-7d31138df84f.png b/pics/766aedd0-1b00-4065-aa2b-7d31138df84f.png
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From 673c9aa04a4ea67badff80df67cdd8aa3d8ad2df Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 21 Aug 2018 20:44:53 +0800
Subject: [PATCH 138/172] auto commit

---
 notes/分布式.md       | 105 ++++++++++++++++++------------------------
 notes/系统设计基础.md |   8 ++--
 2 files changed, 48 insertions(+), 65 deletions(-)

diff --git a/notes/分布式.md b/notes/分布式.md
index c048b43e..e98f2860 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -19,10 +19,11 @@
 * [五、Paxos](#五paxos)
     * [执行过程](#执行过程)
     * [约束条件](#约束条件)
-* [五、Raft](#五raft)
+* [六、Raft](#六raft)
     * [单个 Candidate 的竞选](#单个-candidate-的竞选)
     * [多个 Candidate 竞选](#多个-candidate-竞选)
-    * [日志复制](#日志复制)
+    * [数据同步](#数据同步)
+* [参考](#参考)
 <!-- GFM-TOC -->
 
 
@@ -35,7 +36,7 @@
 - 互斥量为 1 表示有其它进程在使用锁，此时处于锁定状态；
 - 互斥量为 0 表示未锁定状态。
 
-1 和 0 可以用一个整型值表示，也可以用某个数据存在或者不存在表示，存在表示互斥量为 1。
+1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示，存在表示互斥量为 1。
 
 ## 数据库的唯一索引
 
@@ -43,7 +44,7 @@
 
 存在以下几个问题：
 
-- 锁没有失效时间，解锁失败的话其它进程无法再获得锁。
+- 锁没有失效时间，解锁失败的话其它进程无法再获得该锁。
 - 只能是非阻塞锁，插入失败直接就报错了，无法重试。
 - 不可重入，已经获得锁的进程也必须重新获取锁。
 
@@ -59,7 +60,7 @@ EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了
 
 使用了多个 Redis 实例来实现分布式锁，这是为了保证在发生单点故障时仍然可用。
 
-- 尝试从 N 个相互独立 Redis 实例获取锁，如果一个实例不可用，应该尽快尝试下一个；
+- 尝试从 N 个相互独立 Redis 实例获取锁；
 - 计算获取锁消耗的时间，只有当这个时间小于锁的过期时间，并且从大多数（N / 2 + 1）实例上获取了锁，那么就认为锁获取成功了；
 - 如果锁获取失败，就到每个实例上释放锁。
 
@@ -67,7 +68,7 @@ EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了
 
 ### 1. Zookeeper 抽象模型
 
-Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示它的父节点为 /app1。
+Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父节点为 /app1。
 
 <div align="center"> <img src="../pics//31d99967-1171-448e-8531-bccf5c14cffe.jpg" width="400"/> </div><br>
 
@@ -96,21 +97,15 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 一个节点未获得锁，需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
 
-参考：
-
-- [Distributed locks with Redis](https://redis.io/topics/distlock)
-- [浅谈分布式锁](http://www.linkedkeeper.com/detail/blog.action?bid=1023)
-- [基于 Zookeeper 的分布式锁](http://www.dengshenyu.com/java/%E5%88%86%E5%B8%83%E5%BC%8F%E7%B3%BB%E7%BB%9F/2017/10/23/zookeeper-distributed-lock.html)
-
 # 二、分布式事务
 
-指事务的操作位于不同的节点上，需要保证事务的 AICD 特性。例如在下单场景下，库存和订单如果不在同一个节点上，就涉及分布式事务。
+指事务的操作位于不同的节点上，需要保证事务的 AICD 特性。
+
+例如在下单场景下，库存和订单如果不在同一个节点上，就涉及分布式事务。
 
 ## 本地消息表
 
-### 1. 原理
-
-本地消息表与业务数据表处于同一个数据库中，这样就能利用本地事务来保证在对这两个表的操作满足事务特性。
+本地消息表与业务数据表处于同一个数据库中，这样就能利用本地事务来保证在对这两个表的操作满足事务特性，并且使用了消息队列来保证最终一致性。
 
 1. 在分布式事务操作的一方完成写业务数据的操作之后向本地消息表发送一个消息，本地事务能保证这个消息一定会被写入本地消息表中。
 2. 之后将本地消息表中的消息转发到 Kafka 等消息队列中，如果转发成功则将消息从本地消息表中删除，否则继续重新转发。
@@ -118,23 +113,19 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 <div align="center"> <img src="../pics//e3bf5de4-ab1e-4a9b-896d-4b0ad7e9220a.jpg"/> </div><br>
 
-### 2. 分析
-
-本地消息表利用了本地事务来实现分布式事务，并且使用了消息队列来保证最终一致性。
-
 ## 2PC
 
 两阶段提交（Two-phase Commit，2PC），通过引入协调者（Coordinator）来协调参与者的行为，并最终决定这些参与者是否要真正执行事务。
 
 ### 1. 运行过程
 
-（一）准备阶段
+#### 1.1 准备阶段
 
 协调者询问参与者事务是否执行成功，参与者发回事务执行结果。
 
 <div align="center"> <img src="../pics//04f41228-375d-4b7d-bfef-738c5a7c8f07.jpg"/> </div><br>
 
-（二）提交阶段
+#### 1.2 提交阶段
 
 如果事务在每个参与者上都执行成功，事务协调者发送通知让参与者提交事务；否则，协调者发送通知让参与者回滚事务。
 
@@ -144,28 +135,22 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 ### 2. 存在的问题
 
-（一）同步阻塞
+#### 2.1 同步阻塞
 
 所有事务参与者在等待其它参与者响应的时候都处于同步阻塞状态，无法进行其它操作。
 
-（二）单点问题
+#### 2.2 单点问题
 
 协调者在 2PC 中起到非常大的作用，发生故障将会造成很大影响。特别是在阶段二发生故障，所有参与者会一直等待状态，无法完成其它操作。
 
-（三）数据不一致
+#### 2.3 数据不一致
 
 在阶段二，如果协调者只发送了部分 Commit 消息，此时网络发生异常，那么只有部分参与者接收到 Commit 消息，也就是说只有部分参与者提交了事务，使得系统数据不一致。
 
-（四）太过保守
+#### 2.4 太过保守
 
 任意一个节点失败就会导致整个事务失败，没有完善的容错机制。
 
-参考：
-
-- [聊聊分布式事务，再说说解决方案](https://www.cnblogs.com/savorboard/p/distributed-system-transaction-consistency.html)
-- [分布式系统的事务处理](https://coolshell.cn/articles/10910.html)
-- [深入理解分布式事务](https://juejin.im/entry/577c6f220a2b5800573492be)
-
 # 三、CAP
 
 分布式系统不可能同时满足一致性（C：Consistency）、可用性（A：Availability）和分区容忍性（P：Partition Tolerance），最多只能同时满足其中两项。
@@ -184,7 +169,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 可用性指分布式系统在面对各种异常时可以提供正常服务的能力，可以用系统可用时间占总时间的比值来衡量，4 个 9 的可用性表示系统 99.99% 的时间是可用的。
 
-在可用性条件下，系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
+在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
 
 ## 分区容忍性
 
@@ -198,16 +183,11 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 可用性和一致性往往是冲突的，很难都使它们同时满足。在多个节点之间进行数据同步时，
 
-*   为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
-*   为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
+- 为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
+- 为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
 
 <div align="center"> <img src="../pics//0b587744-c0a8-46f2-8d72-e8f070d67b4b.jpg"/> </div><br>
 
-参考：
-
-- 倪超. 从 Paxos 到 ZooKeeper : 分布式一致性原理与实践 [M]. 电子工业出版社, 2015.
-- [What is CAP theorem in distributed database system?](http://www.colooshiki.com/index.php/2017/04/20/what-is-cap-theorem-in-distributed-database-system/)
-
 # 四、BASE
 
 BASE 是基本可用（Basically Available）、软状态（Soft State）和最终一致性（Eventually Consistent）三个短语的缩写。
@@ -292,66 +272,69 @@ Acceptor 接收到接受请求时，如果序号大于等于该 Acceptor 承诺
 
 Paxos 协议能够让 Proposer 发送的提议朝着能被大多数 Acceptor 接受的那个提议靠拢，因此能够保证可终止性。
 
-参考：
+# 六、Raft
 
-- [NEAT ALGORITHMS - PAXOS](http://harry.me/blog/2014/12/27/neat-algorithms-paxos/)
-- [Paxos By Example](https://angus.nyc/2012/paxos-by-example/)
-
-# 五、Raft
-
-Raft 和 Paxos 类似，但是更容易理解，也更容易实现。
-
-Raft 主要是用来竞选主节点。
+Raft 也是分布式一致性协议，主要是用来竞选主节点。
 
 ## 单个 Candidate 的竞选
 
 有三种节点：Follower、Candidate 和 Leader。Leader 会周期性的发送心跳包给 Follower。每个 Follower 都设置了一个随机的竞选超时时间，一般为 150ms\~300ms，如果在这个时间内没有收到 Leader 的心跳包，就会变成 Candidate，进入竞选阶段。
 
-*   下图表示一个分布式系统的最初阶段，此时只有 Follower，没有 Leader。Follower A 等待一个随机的竞选超时时间之后，没收到 Leader 发来的心跳包，因此进入竞选阶段。
+- 下图展示一个分布式系统的最初阶段，此时只有 Follower 没有 Leader。Node A 等待一个随机的竞选超时时间之后，没收到 Leader 发来的心跳包，因此进入竞选阶段。
 
 <div align="center"> <img src="../pics//111521118015898.gif"/> </div><br>
 
-*   此时 A 发送投票请求给其它所有节点。
+- 此时 Node A 发送投票请求给其它所有节点。
 
 <div align="center"> <img src="../pics//111521118445538.gif"/> </div><br>
 
-*   其它节点会对请求进行回复，如果超过一半的节点回复了，那么该 Candidate 就会变成 Leader。
+- 其它节点会对请求进行回复，如果超过一半的节点回复了，那么该 Candidate 就会变成 Leader。
 
 <div align="center"> <img src="../pics//111521118483039.gif"/> </div><br>
 
-*   之后 Leader 会周期性地发送心跳包给 Follower，Follower 接收到心跳包，会重新开始计时。
+- 之后 Leader 会周期性地发送心跳包给 Follower，Follower 接收到心跳包，会重新开始计时。
 
 <div align="center"> <img src="../pics//111521118640738.gif"/> </div><br>
 
 ## 多个 Candidate 竞选
 
-*   如果有多个 Follower 成为 Candidate，并且所获得票数相同，那么就需要重新开始投票，例如下图中 Candidate B 和 Candidate D 都获得两票，因此需要重新开始投票。
+- 如果有多个 Follower 成为 Candidate，并且所获得票数相同，那么就需要重新开始投票。例如下图中 Node B 和 Node D 都获得两票，需要重新开始投票。
 
 <div align="center"> <img src="../pics//111521119203347.gif"/> </div><br>
 
-*   当重新开始投票时，由于每个节点设置的随机竞选超时时间不同，因此能下一次再次出现多个 Candidate 并获得同样票数的概率很低。
+- 由于每个节点设置的随机竞选超时时间不同，因此下一次再次出现多个 Candidate 并获得同样票数的概率很低。
 
 <div align="center"> <img src="../pics//111521119368714.gif"/> </div><br>
 
-## 日志复制
+## 数据同步
 
-*   来自客户端的修改都会被传入 Leader。注意该修改还未被提交，只是写入日志中。
+- 来自客户端的修改都会被传入 Leader。注意该修改还未被提交，只是写入日志中。
 
 <div align="center"> <img src="../pics//7.gif"/> </div><br>
 
-*   Leader 会把修改复制到所有 Follower。
+- Leader 会把修改复制到所有 Follower。
 
 <div align="center"> <img src="../pics//9.gif"/> </div><br>
 
-*   Leader 会等待大多数的 Follower 也进行了修改，然后才将修改提交。
+- Leader 会等待大多数的 Follower 也进行了修改，然后才将修改提交。
 
 <div align="center"> <img src="../pics//10.gif"/> </div><br>
 
-*   此时 Leader 会通知的所有 Follower 让它们也提交修改，此时所有节点的值达成一致。
+- 此时 Leader 会通知的所有 Follower 让它们也提交修改，此时所有节点的值达成一致。
 
 <div align="center"> <img src="../pics//11.gif"/> </div><br>
 
-参考：
+# 参考
 
+- 倪超. 从 Paxos 到 ZooKeeper : 分布式一致性原理与实践 [M]. 电子工业出版社, 2015.
+- [Distributed locks with Redis](https://redis.io/topics/distlock)
+- [浅谈分布式锁](http://www.linkedkeeper.com/detail/blog.action?bid=1023)
+- [基于 Zookeeper 的分布式锁](http://www.dengshenyu.com/java/%E5%88%86%E5%B8%83%E5%BC%8F%E7%B3%BB%E7%BB%9F/2017/10/23/zookeeper-distributed-lock.html)
 - [Raft: Understandable Distributed Consensus](http://thesecretlivesofdata.com/raft)
+- [聊聊分布式事务，再说说解决方案](https://www.cnblogs.com/savorboard/p/distributed-system-transaction-consistency.html)
+- [分布式系统的事务处理](https://coolshell.cn/articles/10910.html)
+- [深入理解分布式事务](https://juejin.im/entry/577c6f220a2b5800573492be)
+- [What is CAP theorem in distributed database system?](http://www.colooshiki.com/index.php/2017/04/20/what-is-cap-theorem-in-distributed-database-system/)
+- [NEAT ALGORITHMS - PAXOS](http://harry.me/blog/2014/12/27/neat-algorithms-paxos/)
+- [Paxos By Example](https://angus.nyc/2012/paxos-by-example/)
 
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index 805764a9..b6a3ddf5 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -68,7 +68,7 @@
 
 应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
 
-关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
+关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器的存储空间限制。
 
 对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
 
@@ -78,7 +78,7 @@
 
 实现可扩展主要有两种方式：
 
-- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
+- 使用消息队列进行解耦，应用之间通过消息传递进行通信；
 - 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以通过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
 
 # 四、可用性
@@ -87,7 +87,7 @@
 
 保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
 
-应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
+应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上，不会对用户有任何影响。
 
 存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
 
@@ -97,7 +97,7 @@
 
 ## 服务降级
 
-服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
+服务降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
 
 # 五、安全性
 

From 762c3c698c936492dbc03dec7bf821da4ad5364b Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 23 Aug 2018 21:37:51 +0800
Subject: [PATCH 139/172] auto commit

---
 notes/Redis.md           |  2 +-
 notes/剑指 offer 题解.md |  2 +-
 notes/构建工具.md        |  2 +-
 notes/算法.md            | 10 +++++-----
 4 files changed, 8 insertions(+), 8 deletions(-)

diff --git a/notes/Redis.md b/notes/Redis.md
index 8f231382..629fcdd8 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -549,7 +549,7 @@ def main():
 
 # 十二、Sentinel
 
-Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状态时，自动从从服务器中选举出新的主服务器。
+Sentinel（哨兵）可以监听集群中的服务器，并在主服务器进入下线状态时，自动从从服务器中选举出新的主服务器。
 
 # 十三、分片
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 1bac6bbb..72c20173 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1082,7 +1082,7 @@ false
 
 ```java
 public boolean isNumeric(char[] str) {
-    if (str == null)
+    if (str == null || str.length == 0)
         return false;
     return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
 }
diff --git a/notes/构建工具.md b/notes/构建工具.md
index c0e4acd5..ce157934 100644
--- a/notes/构建工具.md
+++ b/notes/构建工具.md
@@ -15,7 +15,7 @@
 
 ## 运行单元测试
 
-不再需要在项目代码中添加测试代码，从而污染项目代码。
+不再需要在项目代码中添加测试代码，从而避免了污染项目代码。
 
 ## 将源代码转化为可执行文件
 
diff --git a/notes/算法.md b/notes/算法.md
index 25d1550e..a9f9e689 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -23,7 +23,7 @@
 * [五、栈和队列](#五栈和队列)
     * [栈](#栈)
     * [队列](#队列)
-* [六、查找](#六查找)
+* [六、符号表](#六符号表)
     * [初级实现](#初级实现)
     * [二叉查找树](#二叉查找树)
     * [2-3 查找树](#2-3-查找树)
@@ -1290,7 +1290,7 @@ public class ListQueue<Item> implements MyQueue<Item> {
 
 
 
-# 六、查找
+# 六、符号表
 
 符号表（Symbol Table）是一种存储键值对的数据结构，可以支持快速查找操作。
 
@@ -1411,9 +1411,9 @@ public class ListUnorderedST<Key, Value> implements UnorderedST<Key, Value> {
 
 使用一对平行数组，一个存储键一个存储值。
 
-rank() 方法至关重要，当键在表中时，它能够知道该键的位置；当键不在表中时，它也能知道在何处插入新键。
+二分查找的 rank() 方法至关重要，当键在表中时，它能够知道该键的位置；当键不在表中时，它也能知道在何处插入新键。
 
-复杂度：二分查找最多需要 logN+1 次比较，使用二分查找实现的符号表的查找操作所需要的时间最多是对数级别的。但是插入操作需要移动数组元素，是线性级别的。
+二分查找最多需要 logN+1 次比较，使用二分查找实现的符号表的查找操作所需要的时间最多是对数级别的。但是插入操作需要移动数组元素，是线性级别的。
 
 ```java
 public class BinarySearchOrderedST<Key extends Comparable<Key>, Value> implements OrderedST<Key, Value> {
@@ -2254,7 +2254,7 @@ from H1 to H3
 - c : 40
 - d : 80
 
-可以将每种字符转换成二进制编码，例如将 a 转换为 00，b 转换为 01，c 转换为 10，d 转换为 11。这是最简单的一种编码方式，没有考虑各个字符的权值（出现频率）。而哈夫曼编码能让出现频率最高的字符的编码最短，从而保证整体的编码长度最短。
+可以将每种字符转换成二进制编码，例如将 a 转换为 00，b 转换为 01，c 转换为 10，d 转换为 11。这是最简单的一种编码方式，没有考虑各个字符的权值（出现频率）。而哈夫曼编码采用了贪心策略，使出现频率最高的字符的编码最短，从而保证整体的编码长度最短。
 
 首先生成一颗哈夫曼树，每次生成过程中选取频率最少的两个节点，生成一个新节点作为它们的父节点，并且新节点的频率为两个节点的和。选取频率最少的原因是，生成过程使得先选取的节点在树的最底层，那么需要的编码长度更长，频率更少可以使得总编码长度更少。
 

From cf2e79d8329c53f7036594cf459be689ab45eb94 Mon Sep 17 00:00:00 2001
From: newbieof410 <wangpc2015@163.com>
Date: Fri, 24 Aug 2018 10:35:57 +0800
Subject: [PATCH 140/172] Update MySQL.md
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

修改笔误
---
 notes/MySQL.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index 54067977..bb465ff3 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -214,7 +214,7 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 一个大查询如果一次性执行的话，可能一次锁住很多数据、占满整个事务日志、耗尽系统资源、阻塞很多小的但重要的查询。
 
 ```sql
-DELEFT FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
+DELETE FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
 ```
 
 ```sql
@@ -342,7 +342,7 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 水平切分又称为 Sharding，它是将同一个表中的记录拆分到多个结构相同的表中。
 
-当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓存单个数据库的压力。
+当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓解单个数据库的压力。
 
 <div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
 

From 4f3fa83fc27e18a371b5940e937cdc34c0ef0bea Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 25 Aug 2018 23:12:21 +0800
Subject: [PATCH 141/172] auto commit

---
 notes/Docker.md          |  11 +--
 notes/Java 基础.md       |   2 +-
 notes/Java 容器.md       |   2 +-
 notes/Java 并发.md       |   4 +-
 notes/Java 虚拟机.md     |   2 +-
 notes/Leetcode 题解.md   | 189 ++++++++++++++++++---------------------
 notes/Linux.md           |  26 +++---
 notes/Redis.md           |   2 +-
 notes/分布式.md          | 105 +++++++++-------------
 notes/剑指 offer 题解.md |  14 ++-
 notes/攻击技术.md        |   2 +-
 notes/构建工具.md        |   2 +-
 notes/消息队列.md        |  13 +--
 notes/算法.md            |  14 +--
 notes/系统设计基础.md    |   8 +-
 notes/缓存.md            |  30 ++-----
 notes/设计模式.md        |  76 ++++++++--------
 notes/面向对象思想.md    |  15 ++--
 18 files changed, 248 insertions(+), 269 deletions(-)

diff --git a/notes/Docker.md b/notes/Docker.md
index f230f762..80108896 100644
--- a/notes/Docker.md
+++ b/notes/Docker.md
@@ -30,15 +30,15 @@ Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程
 
 ## 启动速度
 
-启动虚拟机需要启动虚拟机的操作系统，再启动相应的应用，这个过程会非常慢；
+启动虚拟机需要启动虚拟机的操作系统，再启动应用，这个过程非常慢；
 
 而启动 Docker 相当于启动宿主操作系统上的一个进程。
 
 ## 占用资源
 
-虚拟机是一个完整的操作系统，需要占用大量的磁盘空间、内存和 CPU，一台机器只能开启几十个的虚拟机。
+虚拟机是一个完整的操作系统，需要占用大量的磁盘、内存和 CPU，一台机器只能开启几十个的虚拟机。
 
-而 Docker 只是一个进程，只需要将应用以及相应的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
+而 Docker 只是一个进程，只需要将应用以及相关的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
 
 参考资料：
 
@@ -58,7 +58,7 @@ Docker 使用分层技术和镜像，使得应用可以更容易复用重复部
 
 ## 更容易扩展
 
-可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像得到我们想要的镜像非常容易。
+可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像可以非常容易得到我们想要的镜像。
 
 参考资料：
 
@@ -91,7 +91,7 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 
 镜像包含着容器运行时所需要的代码以及其它组件，它是一种分层结构，每一层都是只读的（read-only layers）。构建镜像时，会一层一层构建，前一层是后一层的基础。镜像的这种分层存储结构很适合镜像的复用以及定制。
 
-在构建容器时，通过在镜像的基础上添加一个可写层（writable layer），用来保存着容器运行过程中的修改。
+构建容器时，通过在镜像的基础上添加一个可写层（writable layer），用来保存着容器运行过程中的修改。
 
 <div align="center"> <img src="../pics//docker-filesystems-busyboxrw.png"/> </div><br>
 
@@ -100,3 +100,4 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 - [How to Create Docker Container using Dockerfile](https://linoxide.com/linux-how-to/dockerfile-create-docker-container/)
 - [理解 Docker（2）：Docker 镜像](http://www.cnblogs.com/sammyliu/p/5877964.html)
 
+
diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index b329e0bc..44ca873a 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -707,7 +707,7 @@ public class EqualExample {
 
 ## hashCode()
 
-hasCode() 返回散列值，而 equals() 是用来判断两个对象是否等价。等价的两个对象散列值一定相同，但是散列值相同的两个对象不一定等价。
+hashCode() 返回散列值，而 equals() 是用来判断两个对象是否等价。等价的两个对象散列值一定相同，但是散列值相同的两个对象不一定等价。
 
 在覆盖 equals() 方法时应当总是覆盖 hashCode() 方法，保证等价的两个对象散列值也相等。
 
diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 9959e2ca..001f864c 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -742,7 +742,7 @@ new capacity : 00100000
 
 对于一个 Key，
 
-- 它的哈希值如果在第 6 位上为 0，那么取模得到的结果和之前一样；
+- 它的哈希值如果在第 5 位上为 0，那么取模得到的结果和之前一样；
 - 如果为 1，那么得到的结果为原来的结果 +16。
 
 ### 7. 扩容-计算数组容量
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 9cd5955b..1403f764 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -168,6 +168,8 @@ public static void main(String[] args) throws ExecutionException, InterruptedExc
 
 同样也是需要实现 run() 方法，因为 Thread 类也实现了 Runable 接口。
 
+当调用 start() 方法启动一个线程时，虚拟机会将该线程放入就绪队列中等待被调度，当一个线程被调度时会执行该线程的 run() 方法。
+
 ```java
 public class MyThread extends Thread {
     public void run() {
@@ -1018,7 +1020,7 @@ ForkJoinPool 实现了工作窃取算法来提高 CPU 的利用率。每个线
 
 如果多个线程对同一个共享数据进行访问而不采取同步操作的话，那么操作的结果是不一致的。
 
-以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值为 997 而不是 1000。
+以下代码演示了 1000 个线程同时对 cnt 执行自增操作，操作结束之后它的值有可能小于 1000。
 
 ```java
 public class ThreadUnsafeExample {
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index a4a4ba12..1bba557b 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -92,7 +92,7 @@ java -Xms1M -Xmx2M HackTheJava
 
 对这块区域进行垃圾回收的主要目标是对常量池的回收和对类的卸载，但是一般比较难实现。
 
-JDK 1.7 之前，HotSpot 虚拟机把它当成永久代来进行垃圾回收。但是从 JDK 1.7 开始，已经把原本放在永久代的字符串常量池移到 Native Method 中。
+HotSpot 虚拟机把它当成永久代来进行垃圾回收。但是很难确定永久代的大小，因为它受到很多因素影响，并且每次 Full GC 之后永久代的大小都会改变，所以经常会抛出 OutOfMemoryError 异常。为了更容易管理方法区，从 JDK 1.8 开始，移除永久代，并把方法区移至元空间，它位于本地内存中，而不是虚拟机内存中。
 
 ## 运行时常量池
 
diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md
index 44f3d9dc..04a751c7 100644
--- a/notes/Leetcode 题解.md	
+++ b/notes/Leetcode 题解.md	
@@ -2389,8 +2389,6 @@ private void backtracking(int row) {
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]"/></div> <br>
 
-dp[N] 即为所求。
-
 考虑到 dp[i] 只与 dp[i - 1] 和 dp[i - 2] 有关，因此可以只用两个变量来存储 dp[i - 1] 和 dp[i - 2]，使得原来的 O(N) 空间复杂度优化为 O(1) 复杂度。
 
 ```java
@@ -2468,16 +2466,6 @@ private int rob(int[] nums, int first, int last) {
 }
 ```
 
-**母牛生产** 
-
-[程序员代码面试指南-P181](#)
-
-题目描述：假设农场中成熟的母牛每年都会生 1 头小母牛，并且永远不会死。第一年有 1 只小母牛，从第二年开始，母牛开始生小母牛。每只小母牛 3 年之后成熟又可以生小母牛。给定整数 N，求 N 年后牛的数量。
-
-第 i 年成熟的牛的数量为：
-
-<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]"/></div> <br>
-
 **信件错排** 
 
 题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信方式的数量。
@@ -2491,7 +2479,15 @@ private int rob(int[] nums, int first, int last) {
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=(i-1)*dp[i-2]+(i-1)*dp[i-1]"/></div> <br>
 
-dp[N] 即为所求。
+**母牛生产** 
+
+[程序员代码面试指南-P181](#)
+
+题目描述：假设农场中成熟的母牛每年都会生 1 头小母牛，并且永远不会死。第一年有 1 只小母牛，从第二年开始，母牛开始生小母牛。每只小母牛 3 年之后成熟又可以生小母牛。给定整数 N，求 N 年后牛的数量。
+
+第 i 年成熟的牛的数量为：
+
+<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]"/></div> <br>
 
 ### 矩阵路径
 
@@ -2517,10 +2513,8 @@ public int minPathSum(int[][] grid) {
     int[] dp = new int[n];
     for (int i = 0; i < m; i++) {
         for (int j = 0; j < n; j++) {
-            if (j == 0) {
-                dp[j] = dp[j];        // 只能从上侧走到该位置
-            } else if (i == 0) {
-                dp[j] = dp[j - 1];    // 只能从左侧走到该位置
+            if (i == 0) {
+                dp[j] = dp[j - 1];
             } else {
                 dp[j] = Math.min(dp[j - 1], dp[j]);
             }
@@ -2584,17 +2578,18 @@ sumRange(0, 5) -> -3
 
 ```java
 class NumArray {
+
     private int[] sums;
 
     public NumArray(int[] nums) {
-        sums = new int[nums.length];
-        for (int i = 0; i < nums.length; i++) {
-            sums[i] = i == 0 ? nums[0] : sums[i - 1] + nums[i];
+        sums = new int[nums.length + 1];
+        for (int i = 1; i <= nums.length; i++) {
+            sums[i] = sums[i - 1] + nums[i - 1];
         }
     }
 
     public int sumRange(int i, int j) {
-        return i == 0 ? sums[j] : sums[j] - sums[i - 1];
+        return sums[j + 1] - sums[i];
     }
 }
 ```
@@ -2634,7 +2629,7 @@ return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] i
 
 dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
 
-如果 A[i] - A[i - 1] == A[i - 1] - A[i - 2]，表示 [A[i - 2], A[i - 1], A[i]] 是一个等差递增子区间。如果 [A[i - 3], A[i - 2], A[i - 1]] 是一个等差递增子区间，那么 [A[i - 3], A[i - 2], A[i - 1], A[i]] 也是。因此在这个条件下，dp[i] = dp[i-1] + 1。
+在 A[i] - A[i - 1] == A[i - 1] - A[i - 2] 的条件下，{A[i - 2], A[i - 1], A[i]} 是一个等差递增子区间。如果 {A[i - 3], A[i - 2], A[i - 1]} 是一个等差递增子区间，那么 {A[i - 3], A[i - 2], A[i - 1], A[i]} 也是等差递增子区间，dp[i] = dp[i-1] + 1。
 
 ```java
 public int numberOfArithmeticSlices(int[] A) {
@@ -2747,17 +2742,17 @@ public int numDecodings(String s) {
 
 ### 最长递增子序列
 
-已知一个序列 {S<sub>1</sub>, S<sub>2</sub>,...,S<sub>n</sub>} ，取出若干数组成新的序列 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>}，其中 i1、i2 ... im 保持递增，即新序列中各个数仍然保持原数列中的先后顺序，称新序列为原序列的一个 **子序列** 。
+已知一个序列 {S<sub>1</sub>, S<sub>2</sub>,...,S<sub>n</sub>}，取出若干数组成新的序列 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>}，其中 i1、i2 ... im 保持递增，即新序列中各个数仍然保持原数列中的先后顺序，称新序列为原序列的一个 **子序列** 。
 
 如果在子序列中，当下标 ix > iy 时，S<sub>ix</sub> > S<sub>iy</sub>，称子序列为原序列的一个 **递增子序列** 。
 
-定义一个数组 dp 存储最长递增子序列的长度，dp[n] 表示以 S<sub>n</sub> 结尾的序列的最长递增子序列长度。对于一个递增子序列 {S<sub>i1</sub>, S<sub>i2</sub>,...,S<sub>im</sub>}，如果 im < n 并且 S<sub>im</sub> < S<sub>n</sub> ，此时 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>, S<sub>n</sub>} 为一个递增子序列，递增子序列的长度增加 1。满足上述条件的递增子序列中，长度最长的那个递增子序列就是要找的，在长度最长的递增子序列上加上 S<sub>n</sub> 就构成了以 S<sub>n</sub> 为结尾的最长递增子序列。因此 dp[n] = max{ dp[i]+1 | S<sub>i</sub> < S<sub>n</sub> && i < n} 。
+定义一个数组 dp 存储最长递增子序列的长度，dp[n] 表示以 S<sub>n</sub> 结尾的序列的最长递增子序列长度。对于一个递增子序列 {S<sub>i1</sub>, S<sub>i2</sub>,...,S<sub>im</sub>}，如果 im < n 并且 S<sub>im</sub> < S<sub>n</sub>，此时 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>, S<sub>n</sub>} 为一个递增子序列，递增子序列的长度增加 1。满足上述条件的递增子序列中，长度最长的那个递增子序列就是要找的，在长度最长的递增子序列上加上 S<sub>n</sub> 就构成了以 S<sub>n</sub> 为结尾的最长递增子序列。因此 dp[n] = max{ dp[i]+1 | S<sub>i</sub> < S<sub>n</sub> && i < n} 。
 
 因为在求 dp[n] 时可能无法找到一个满足条件的递增子序列，此时 {S<sub>n</sub>} 就构成了递增子序列，需要对前面的求解方程做修改，令 dp[n] 最小为 1，即：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[n]=max\{1,dp[i]+1|S_i<S_n\&\&i<n\}"/></div> <br>
 
-对于一个长度为 N 的序列，最长递增子序列并不一定会以 S<sub>N</sub> 为结尾，因此 dp[N] 不是序列的最长递增子序列的长度，需要遍历 dp 数组找出最大值才是所要的结果，即 max{ dp[i] | 1 <= i <= N} 即为所求。
+对于一个长度为 N 的序列，最长递增子序列并不一定会以 S<sub>N</sub> 为结尾，因此 dp[N] 不是序列的最长递增子序列的长度，需要遍历 dp 数组找出最大值才是所要的结果，max{ dp[i] | 1 <= i <= N} 即为所求。
 
 **最长递增子序列** 
 
@@ -2790,7 +2785,7 @@ for (int i = 0; i < n; i++) {
 return ret;
 ```
 
-以上解法的时间复杂度为 O(N<sup>2</sup>) ，可以使用二分查找将时间复杂度降低为 O(NlogN)。
+以上解法的时间复杂度为 O(N<sup>2</sup>)，可以使用二分查找将时间复杂度降低为 O(NlogN)。
 
 定义一个 tails 数组，其中 tails[i] 存储长度为 i + 1 的最长递增子序列的最后一个元素。对于一个元素 x，
 
@@ -2915,19 +2910,19 @@ public int wiggleMaxLength(int[] nums) {
 
 定义一个二维数组 dp 用来存储最长公共子序列的长度，其中 dp[i][j] 表示 S1 的前 i 个字符与 S2 的前 j 个字符最长公共子序列的长度。考虑 S1<sub>i</sub> 与 S2<sub>j</sub> 值是否相等，分为两种情况：
 
-- 当 S1<sub>i</sub>==S2<sub>j</sub> 时，那么就能在 S1 的前 i-1 个字符与 S2 的前 j-1 个字符最长公共子序列的基础上再加上 S1<sub>i</sub> 这个值，最长公共子序列长度加 1 ，即 dp[i][j] = dp[i-1][j-1] + 1。
-- 当 S1<sub>i</sub> != S2<sub>j</sub> 时，此时最长公共子序列为 S1 的前 i-1 个字符和 S2 的前 j 个字符最长公共子序列，与 S1 的前 i 个字符和 S2 的前 j-1 个字符最长公共子序列，它们的最大者，即 dp[i][j] = max{ dp[i-1][j], dp[i][j-1] }。
+- 当 S1<sub>i</sub>==S2<sub>j</sub> 时，那么就能在 S1 的前 i-1 个字符与 S2 的前 j-1 个字符最长公共子序列的基础上再加上 S1<sub>i</sub> 这个值，最长公共子序列长度加 1，即 dp[i][j] = dp[i-1][j-1] + 1。
+- 当 S1<sub>i</sub> != S2<sub>j</sub> 时，此时最长公共子序列为 S1 的前 i-1 个字符和 S2 的前 j 个字符最长公共子序列，或者 S1 的前 i 个字符和 S2 的前 j-1 个字符最长公共子序列，取它们的最大者，即 dp[i][j] = max{ dp[i-1][j], dp[i][j-1] }。
 
 综上，最长公共子序列的状态转移方程为：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=\left\{\begin{array}{rcl}dp[i-1][j-1]&&{S1_i==S2_j}\\max(dp[i-1][j],dp[i][j-1])&&{S1_i<>S2_j}\end{array}\right."/></div> <br>
 
-对于长度为 N 的序列 S<sub>1</sub> 和 长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
+对于长度为 N 的序列 S<sub>1</sub> 和长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
 
 与最长递增子序列相比，最长公共子序列有以下不同点：
 
 - 针对的是两个序列，求它们的最长公共子序列。
-- 在最长递增子序列中，dp[i] 表示以 S<sub>i</sub> 为结尾的最长递增子序列长度，子序列必须包含 S<sub>i</sub> ；在最长公共子序列中，dp[i][j] 表示 S1 中前 i 个字符与 S2 中前 j 个字符的最长公共子序列长度，不一定包含 S1<sub>i</sub> 和 S2<sub>j</sub> 。
+- 在最长递增子序列中，dp[i] 表示以 S<sub>i</sub> 为结尾的最长递增子序列长度，子序列必须包含 S<sub>i</sub> ；在最长公共子序列中，dp[i][j] 表示 S1 中前 i 个字符与 S2 中前 j 个字符的最长公共子序列长度，不一定包含 S1<sub>i</sub> 和 S2<sub>j</sub>。
 - 在求最终解时，最长公共子序列中 dp[N][M] 就是最终解，而最长递增子序列中 dp[N] 不是最终解，因为以 S<sub>N</sub> 为结尾的最长递增子序列不一定是整个序列最长递增子序列，需要遍历一遍 dp 数组找到最大者。
 
 ```java
@@ -2956,9 +2951,7 @@ public int lengthOfLCS(int[] nums1, int[] nums2) {
 - 第 i 件物品没添加到背包，总体积不超过 j 的前 i 件物品的最大价值就是总体积不超过 j 的前 i-1 件物品的最大价值，dp[i][j] = dp[i-1][j]。
 - 第 i 件物品添加到背包中，dp[i][j] = dp[i-1][j-w] + v。
 
-第 i 件物品可添加也可以不添加，取决于哪种情况下最大价值更大。
-
-综上，0-1 背包的状态转移方程为：
+第 i 件物品可添加也可以不添加，取决于哪种情况下最大价值更大。因此，0-1 背包的状态转移方程为：
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=max(dp[i-1][j],dp[i-1][j-w]+v)"/></div> <br>
 
@@ -2981,11 +2974,11 @@ public int knapsack(int W, int N, int[] weights, int[] values) {
 
 **空间优化** 
 
-在程序实现时可以对 0-1 背包做优化。观察状态转移方程可以知道，前 i 件物品的状态仅由前 i-1 件物品的状态有关，因此可以将 dp 定义为一维数组，其中 dp[j] 既可以表示 dp[i-1][j] 也可以表示 dp[i][j]。此时，
+在程序实现时可以对 0-1 背包做优化。观察状态转移方程可以知道，前 i 件物品的状态仅与前 i-1 件物品的状态有关，因此可以将 dp 定义为一维数组，其中 dp[j] 既可以表示 dp[i-1][j] 也可以表示 dp[i][j]。此时，
 
 <div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[j]=max(dp[j],dp[j-w]+v)"/></div> <br>
 
-因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，以防止将 dp[i-1][j-w] 覆盖。也就是说要先计算 dp[i][j] 再计算 dp[i][j-w]，在程序实现时需要按倒序来循环求解。
+因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，以防将 dp[i-1][j-w] 覆盖。也就是说要先计算 dp[i][j] 再计算 dp[i][j-w]，在程序实现时需要按倒序来循环求解。
 
 ```java
 public int knapsack(int W, int N, int[] weights, int[] values) {
@@ -3205,7 +3198,7 @@ public int findMaxForm(String[] strs, int m, int n) {
 }
 ```
 
-**找零钱的方法数** 
+**找零钱的最少硬币数** 
 
 [322. Coin Change (Medium)](https://leetcode.com/problems/coin-change/description/)
 
@@ -3286,60 +3279,6 @@ public int combinationSum4(int[] nums, int target) {
 }
 ```
 
-**只能进行 k 次的股票交易** 
-
-[188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)
-
-```java
-public int maxProfit(int k, int[] prices) {
-    int n = prices.length;
-    if (k >= n / 2) {   // 这种情况下该问题退化为普通的股票交易问题
-        int maxProfit = 0;
-        for (int i = 1; i < n; i++) {
-            if (prices[i] > prices[i - 1]) {
-                maxProfit += prices[i] - prices[i - 1];
-            }
-        }
-        return maxProfit;
-    }
-    int[][] maxProfit = new int[k + 1][n];
-    for (int i = 1; i <= k; i++) {
-        int localMax = maxProfit[i - 1][0] - prices[0];
-        for (int j = 1; j < n; j++) {
-            maxProfit[i][j] = Math.max(maxProfit[i][j - 1], prices[j] + localMax);
-            localMax = Math.max(localMax, maxProfit[i - 1][j] - prices[j]);
-        }
-    }
-    return maxProfit[k][n - 1];
-}
-```
-
-**只能进行两次的股票交易** 
-
-[123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/)
-
-```java
-public int maxProfit(int[] prices) {
-    int firstBuy = Integer.MIN_VALUE, firstSell = 0;
-    int secondBuy = Integer.MIN_VALUE, secondSell = 0;
-    for (int curPrice : prices) {
-        if (firstBuy < -curPrice) {
-            firstBuy = -curPrice;
-        }
-        if (firstSell < firstBuy + curPrice) {
-            firstSell = firstBuy + curPrice;
-        }
-        if (secondBuy < firstSell - curPrice) {
-            secondBuy = firstSell - curPrice;
-        }
-        if (secondSell < secondBuy + curPrice) {
-            secondSell = secondBuy + curPrice;
-        }
-    }
-    return secondSell;
-}
-```
-
 ### 股票交易
 
 **需要冷却期的股票交易** 
@@ -3432,6 +3371,60 @@ public int maxProfit(int[] prices) {
 }
 ```
 
+**只能进行两次的股票交易** 
+
+[123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/)
+
+```java
+public int maxProfit(int[] prices) {
+    int firstBuy = Integer.MIN_VALUE, firstSell = 0;
+    int secondBuy = Integer.MIN_VALUE, secondSell = 0;
+    for (int curPrice : prices) {
+        if (firstBuy < -curPrice) {
+            firstBuy = -curPrice;
+        }
+        if (firstSell < firstBuy + curPrice) {
+            firstSell = firstBuy + curPrice;
+        }
+        if (secondBuy < firstSell - curPrice) {
+            secondBuy = firstSell - curPrice;
+        }
+        if (secondSell < secondBuy + curPrice) {
+            secondSell = secondBuy + curPrice;
+        }
+    }
+    return secondSell;
+}
+```
+
+**只能进行 k 次的股票交易** 
+
+[188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/)
+
+```java
+public int maxProfit(int k, int[] prices) {
+    int n = prices.length;
+    if (k >= n / 2) {   // 这种情况下该问题退化为普通的股票交易问题
+        int maxProfit = 0;
+        for (int i = 1; i < n; i++) {
+            if (prices[i] > prices[i - 1]) {
+                maxProfit += prices[i] - prices[i - 1];
+            }
+        }
+        return maxProfit;
+    }
+    int[][] maxProfit = new int[k + 1][n];
+    for (int i = 1; i <= k; i++) {
+        int localMax = maxProfit[i - 1][0] - prices[0];
+        for (int j = 1; j < n; j++) {
+            maxProfit[i][j] = Math.max(maxProfit[i][j - 1], prices[j] + localMax);
+            localMax = Math.max(localMax, maxProfit[i - 1][j] - prices[j]);
+        }
+    }
+    return maxProfit[k][n - 1];
+}
+```
+
 ### 字符串编辑
 
 **删除两个字符串的字符使它们相等** 
@@ -3450,11 +3443,8 @@ Explanation: You need one step to make "sea" to "ea" and another step to make "e
 public int minDistance(String word1, String word2) {
     int m = word1.length(), n = word2.length();
     int[][] dp = new int[m + 1][n + 1];
-    for (int i = 0; i <= m; i++) {
-        for (int j = 0; j <= n; j++) {
-            if (i == 0 || j == 0) {
-                continue;
-            }
+    for (int i = 1; i <= m; i++) {
+        for (int j = 1; j <= n; j++) {
             if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                 dp[i][j] = dp[i - 1][j - 1] + 1;
             } else {
@@ -3612,7 +3602,7 @@ public int countPrimes(int n) {
 
 ```java
 int gcd(int a, int b) {
-    return b == 0 ? a : gcd(b, a% b);
+    return b == 0 ? a : gcd(b, a % b);
 }
 ```
 
@@ -3683,7 +3673,7 @@ public String convertToBase7(int num) {
 }
 ```
 
-Java 中 static String toString(int num, int radix) 可以将一个整数转换为 redix 进制表示的字符串。
+Java 中 static String toString(int num, int radix) 可以将一个整数转换为 radix 进制表示的字符串。
 
 ```java
 public String convertToBase7(int num) {
@@ -6413,7 +6403,6 @@ public int maxChunksToSorted(int[] arr) {
 }
 ```
 
-
 ## 图
 
 ### 二分图
@@ -6630,6 +6619,7 @@ public int[] findRedundantConnection(int[][] edges) {
 }
 
 private class UF {
+
     private int[] id;
 
     UF(int N) {
@@ -6675,7 +6665,7 @@ x ^ x = 0       x & x = x       x | x = x
 ```
 
 - 利用 x ^ 1s = \~x 的特点，可以将位级表示翻转；利用 x ^ x = 0 的特点，可以将三个数中重复的两个数去除，只留下另一个数。
-- 利用 x & 0s = 0 和 x & 1s = x 的特点，可以实现掩码操作。一个数 num 与 mask ：00111100 进行位与操作，只保留 num 中与 mask 的 1 部分相对应的位。
+- 利用 x & 0s = 0 和 x & 1s = x 的特点，可以实现掩码操作。一个数 num 与 mask：00111100 进行位与操作，只保留 num 中与 mask 的 1 部分相对应的位。
 - 利用 x | 0s = x 和 x | 1s = 1s 的特点，可以实现设值操作。一个数 num 与 mask：00111100 进行位或操作，将 num 中与 mask 的 1 部分相对应的位都设置为 1。
 
 位与运算技巧：
@@ -6920,7 +6910,6 @@ public boolean isPowerOfFour(int num) {
 }
 ```
 
-
 **判断一个数的位级表示是否不会出现连续的 0 和 1** 
 
 [693. Binary Number with Alternating Bits (Easy)](https://leetcode.com/problems/binary-number-with-alternating-bits/description/)
diff --git a/notes/Linux.md b/notes/Linux.md
index 14d3b317..73a2e9ec 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -129,15 +129,16 @@ info 与 man 类似，但是 info 将文档分成一个个页面，每个页面
 /usr/local/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/home/dmtsai/.local/bin:/home/dmtsai/bin
 ```
 
-env 命令可以获取当前终端的环境变量。
-
 ## sudo
 
 sudo 允许一般用户使用 root 可执行的命令，不过只有在 /etc/sudoers 配置文件中添加的用户才能使用该指令。
 
 ## 包管理工具
 
-RPM 和 DPKG 为最常见的两类软件包管理工具。RPM 全称为 Redhat Package Manager，最早由 Red Hat 公司制定实施，随后被 GNU 开源操作系统接受并成为很多 Linux 系统 (RHEL) 的既定软件标准。与 RPM 进行竞争的是基于 Debian 操作系统 (UBUNTU) 的 DEB 软件包管理工具 DPKG，全称为 Debian Package，功能方面与 RPM 相似。
+RPM 和 DPKG 为最常见的两类软件包管理工具：
+
+- RPM 全称为 Redhat Package Manager，最早由 Red Hat 公司制定实施，随后被 GNU 开源操作系统接受并成为很多 Linux 系统 (RHEL) 的既定软件标准。
+- 与 RPM 进行竞争的是基于 Debian 操作系统 (Ubuntu) 的 DEB 软件包管理工具 DPKG，全称为 Debian Package，功能方面与 RPM 相似。
 
 YUM 基于 RPM，具有依赖管理功能，并具有软件升级的功能。
 
@@ -194,13 +195,13 @@ IDE（ATA）全称 Advanced Technology Attachment，接口速度最大为 133MB/
 
 ### 2. SATA
 
-SATA 全称 Serial ATA，也就是使用串口的 ATA 接口，因抗干扰性强，且对数据线的长度要求比 ATA 低很多，支持热插拔等功能，SATA-II 的接口速度为 300MiB/s，而新的 SATA-III 标准可达到 600MiB/s 的传输速度。SATA 的数据线也比 ATA 的细得多，有利于机箱内的空气流通，整理线材也比较方便。
+SATA 全称 Serial ATA，也就是使用串口的 ATA 接口，抗干扰性强，且对数据线的长度要求比 ATA 低很多，支持热插拔等功能。SATA-II 的接口速度为 300MiB/s，而新的 SATA-III 标准可达到 600MiB/s 的传输速度。SATA 的数据线也比 ATA 的细得多，有利于机箱内的空气流通，整理线材也比较方便。
 
 <div align="center"> <img src="../pics//f9f2a16b-4843-44d1-9759-c745772e9bcf.jpg" width=""/> </div><br>
 
 ### 3. SCSI
 
-SCSI 全称是 Small Computer System Interface（小型机系统接口），经历多代的发展，从早期的 SCSI-II，到目前的 Ultra320 SCSI 以及 Fiber-Channel（光纤通道），接口型式也多种多样。SCSI 硬盘广为工作站级个人电脑以及服务器所使用，因此会使用较为先进的技术，如碟片转速 15000rpm 的高转速，且资料传输时 CPU 占用率较低，但是单价也比相同容量的 ATA 及 SATA 硬盘更加昂贵。
+SCSI 全称是 Small Computer System Interface（小型机系统接口），经历多代的发展，从早期的 SCSI-II 到目前的 Ultra320 SCSI 以及 Fiber-Channel（光纤通道），接口型式也多种多样。SCSI 硬盘广为工作站级个人电脑以及服务器所使用，因此会使用较为先进的技术，如碟片转速 15000rpm 的高转速，且传输时 CPU 占用率较低，但是单价也比相同容量的 ATA 及 SATA 硬盘更加昂贵。
 
 <div align="center"> <img src="../pics//f0574025-c514-49f5-a591-6d6a71f271f7.jpg" width=""/> </div><br>
 
@@ -229,7 +230,7 @@ Linux 中每个硬件都被当做一个文件，包括磁盘。磁盘以磁盘
 
 MBR 中，第一个扇区最重要，里面有主要开机记录（Master boot record, MBR）及分区表（partition table），其中主要开机记录占 446 bytes，分区表占 64 bytes。
 
-分区表只有 64 bytes，最多只能存储 4 个分区，这 4 个分区为主分区（Primary）和扩展分区（Extended）。其中扩展分区只有一个，它将其它扇区用来记录分区表，因此通过扩展分区可以分出更多分区，这些分区称为逻辑分区。
+分区表只有 64 bytes，最多只能存储 4 个分区，这 4 个分区为主分区（Primary）和扩展分区（Extended）。其中扩展分区只有一个，它使用其它扇区用记录额外的分区表，因此通过扩展分区可以分出更多分区，这些分区称为逻辑分区。
 
 Linux 也把分区当成文件，分区文件的命名方式为：磁盘文件名 + 编号，例如 /dev/sda1。注意，逻辑分区的编号从 5 开始。
 
@@ -251,10 +252,10 @@ MBR 不支持 2.2 TB 以上的硬盘，GPT 则最多支持到 2<sup>33</sup> TB
 
 BIOS（Basic Input/Output System，基本输入输出系统），它是一个固件（嵌入在硬件中的软件），BIOS 程序存放在断电后内容不会丢失的只读内存中。
 
-BIOS 是开机的时候计算机执行的第一个程序，这个程序知道可以开机的磁盘，并读取磁盘第一个扇区的主要开机记录（MBR），由主要开机记录（MBR）执行其中的开机管理程序，这个开机管理程序会加载操作系统的核心文件。
-
 <div align="center"> <img src="../pics//50831a6f-2777-46ea-a571-29f23c85cc21.jpg"/> </div><br>
 
+BIOS 是开机的时候计算机执行的第一个程序，这个程序知道可以开机的磁盘，并读取磁盘第一个扇区的主要开机记录（MBR），由主要开机记录（MBR）执行其中的开机管理程序，这个开机管理程序会加载操作系统的核心文件。
+
 主要开机记录（MBR）中的开机管理程序提供以下功能：选单、载入核心文件以及转交其它开机管理程序。转交这个功能可以用来实现了多重引导，只需要将另一个操作系统的开机管理程序安装在其它分区的启动扇区上，在启动开机管理程序时，就可以通过选单选择启动当前的操作系统或者转交给其它开机管理程序从而启动另一个操作系统。
 
 下图中，第一扇区的主要开机记录（MBR）中的开机管理程序提供了两个选单：M1、M2，M1 指向了 Windows 操作系统，而 M2 指向其它分区的启动扇区，里面包含了另外一个开机管理程序，提供了一个指向 Linux 的选单。
@@ -283,11 +284,10 @@ BIOS 不可以读取 GPT 分区表，而 UEFI 可以。
 除此之外还包括：
 
 - superblock：记录文件系统的整体信息，包括 inode 和 block 的总量、使用量、剩余量，以及文件系统的格式与相关信息等；
-- block bitmap：记录 block 是否被使用的位域；
+- block bitmap：记录 block 是否被使用的位域。
 
 <div align="center"> <img src="../pics//BSD_disk.png" width="800"/> </div><br>
 
-
 ## 文件读取
 
 对于 Ext2 文件系统，当要读取一个文件的内容时，先在 inode 中去查找文件内容所在的所有 block，然后把所有 block 的内容读出来。
@@ -1169,7 +1169,7 @@ dmtsai lines: 5 columns: 9
 | S | interruptible sleep (waiting for an event to complete) |
 | Z | zombie (terminated but not reaped by its parent) |
 | T | stopped (either by a job control signal or because it is being traced) |
-
+<br>
 <div align="center"> <img src="../pics//76a49594323247f21c9b3a69945445ee.png" width=""/> </div><br>
 
 ## SIGCHLD
@@ -1179,12 +1179,12 @@ dmtsai lines: 5 columns: 9
 - 得到 SIGCHLD 信号；
 - waitpid() 或者 wait() 调用会返回。
 
-<div align="center"> <img src="../pics//flow.png" width=""/> </div><br>
-
 其中子进程发送的 SIGCHLD 信号包含了子进程的信息，包含了进程 ID、进程状态、进程使用 CPU 的时间等。
 
 在子进程退出时，它的进程描述符不会立即释放，这是为了让父进程得到子进程信息，父进程通过 wait() 和 waitpid() 来获得一个已经退出的子进程的信息。
 
+<div align="center"> <img src="../pics//flow.png" width=""/> </div><br>
+
 ## wait()
 
 ```c
diff --git a/notes/Redis.md b/notes/Redis.md
index 8f231382..629fcdd8 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -549,7 +549,7 @@ def main():
 
 # 十二、Sentinel
 
-Sentinel（哨兵）可以监听主服务器，并在主服务器进入下线状态时，自动从从服务器中选举出新的主服务器。
+Sentinel（哨兵）可以监听集群中的服务器，并在主服务器进入下线状态时，自动从从服务器中选举出新的主服务器。
 
 # 十三、分片
 
diff --git a/notes/分布式.md b/notes/分布式.md
index c048b43e..e98f2860 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -19,10 +19,11 @@
 * [五、Paxos](#五paxos)
     * [执行过程](#执行过程)
     * [约束条件](#约束条件)
-* [五、Raft](#五raft)
+* [六、Raft](#六raft)
     * [单个 Candidate 的竞选](#单个-candidate-的竞选)
     * [多个 Candidate 竞选](#多个-candidate-竞选)
-    * [日志复制](#日志复制)
+    * [数据同步](#数据同步)
+* [参考](#参考)
 <!-- GFM-TOC -->
 
 
@@ -35,7 +36,7 @@
 - 互斥量为 1 表示有其它进程在使用锁，此时处于锁定状态；
 - 互斥量为 0 表示未锁定状态。
 
-1 和 0 可以用一个整型值表示，也可以用某个数据存在或者不存在表示，存在表示互斥量为 1。
+1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示，存在表示互斥量为 1。
 
 ## 数据库的唯一索引
 
@@ -43,7 +44,7 @@
 
 存在以下几个问题：
 
-- 锁没有失效时间，解锁失败的话其它进程无法再获得锁。
+- 锁没有失效时间，解锁失败的话其它进程无法再获得该锁。
 - 只能是非阻塞锁，插入失败直接就报错了，无法重试。
 - 不可重入，已经获得锁的进程也必须重新获取锁。
 
@@ -59,7 +60,7 @@ EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了
 
 使用了多个 Redis 实例来实现分布式锁，这是为了保证在发生单点故障时仍然可用。
 
-- 尝试从 N 个相互独立 Redis 实例获取锁，如果一个实例不可用，应该尽快尝试下一个；
+- 尝试从 N 个相互独立 Redis 实例获取锁；
 - 计算获取锁消耗的时间，只有当这个时间小于锁的过期时间，并且从大多数（N / 2 + 1）实例上获取了锁，那么就认为锁获取成功了；
 - 如果锁获取失败，就到每个实例上释放锁。
 
@@ -67,7 +68,7 @@ EXPIRE 指令可以为一个键值对设置一个过期时间，从而避免了
 
 ### 1. Zookeeper 抽象模型
 
-Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示它的父节点为 /app1。
+Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父节点为 /app1。
 
 <div align="center"> <img src="../pics//31d99967-1171-448e-8531-bccf5c14cffe.jpg" width="400"/> </div><br>
 
@@ -96,21 +97,15 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 一个节点未获得锁，需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
 
-参考：
-
-- [Distributed locks with Redis](https://redis.io/topics/distlock)
-- [浅谈分布式锁](http://www.linkedkeeper.com/detail/blog.action?bid=1023)
-- [基于 Zookeeper 的分布式锁](http://www.dengshenyu.com/java/%E5%88%86%E5%B8%83%E5%BC%8F%E7%B3%BB%E7%BB%9F/2017/10/23/zookeeper-distributed-lock.html)
-
 # 二、分布式事务
 
-指事务的操作位于不同的节点上，需要保证事务的 AICD 特性。例如在下单场景下，库存和订单如果不在同一个节点上，就涉及分布式事务。
+指事务的操作位于不同的节点上，需要保证事务的 AICD 特性。
+
+例如在下单场景下，库存和订单如果不在同一个节点上，就涉及分布式事务。
 
 ## 本地消息表
 
-### 1. 原理
-
-本地消息表与业务数据表处于同一个数据库中，这样就能利用本地事务来保证在对这两个表的操作满足事务特性。
+本地消息表与业务数据表处于同一个数据库中，这样就能利用本地事务来保证在对这两个表的操作满足事务特性，并且使用了消息队列来保证最终一致性。
 
 1. 在分布式事务操作的一方完成写业务数据的操作之后向本地消息表发送一个消息，本地事务能保证这个消息一定会被写入本地消息表中。
 2. 之后将本地消息表中的消息转发到 Kafka 等消息队列中，如果转发成功则将消息从本地消息表中删除，否则继续重新转发。
@@ -118,23 +113,19 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 <div align="center"> <img src="../pics//e3bf5de4-ab1e-4a9b-896d-4b0ad7e9220a.jpg"/> </div><br>
 
-### 2. 分析
-
-本地消息表利用了本地事务来实现分布式事务，并且使用了消息队列来保证最终一致性。
-
 ## 2PC
 
 两阶段提交（Two-phase Commit，2PC），通过引入协调者（Coordinator）来协调参与者的行为，并最终决定这些参与者是否要真正执行事务。
 
 ### 1. 运行过程
 
-（一）准备阶段
+#### 1.1 准备阶段
 
 协调者询问参与者事务是否执行成功，参与者发回事务执行结果。
 
 <div align="center"> <img src="../pics//04f41228-375d-4b7d-bfef-738c5a7c8f07.jpg"/> </div><br>
 
-（二）提交阶段
+#### 1.2 提交阶段
 
 如果事务在每个参与者上都执行成功，事务协调者发送通知让参与者提交事务；否则，协调者发送通知让参与者回滚事务。
 
@@ -144,28 +135,22 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 ### 2. 存在的问题
 
-（一）同步阻塞
+#### 2.1 同步阻塞
 
 所有事务参与者在等待其它参与者响应的时候都处于同步阻塞状态，无法进行其它操作。
 
-（二）单点问题
+#### 2.2 单点问题
 
 协调者在 2PC 中起到非常大的作用，发生故障将会造成很大影响。特别是在阶段二发生故障，所有参与者会一直等待状态，无法完成其它操作。
 
-（三）数据不一致
+#### 2.3 数据不一致
 
 在阶段二，如果协调者只发送了部分 Commit 消息，此时网络发生异常，那么只有部分参与者接收到 Commit 消息，也就是说只有部分参与者提交了事务，使得系统数据不一致。
 
-（四）太过保守
+#### 2.4 太过保守
 
 任意一个节点失败就会导致整个事务失败，没有完善的容错机制。
 
-参考：
-
-- [聊聊分布式事务，再说说解决方案](https://www.cnblogs.com/savorboard/p/distributed-system-transaction-consistency.html)
-- [分布式系统的事务处理](https://coolshell.cn/articles/10910.html)
-- [深入理解分布式事务](https://juejin.im/entry/577c6f220a2b5800573492be)
-
 # 三、CAP
 
 分布式系统不可能同时满足一致性（C：Consistency）、可用性（A：Availability）和分区容忍性（P：Partition Tolerance），最多只能同时满足其中两项。
@@ -184,7 +169,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 可用性指分布式系统在面对各种异常时可以提供正常服务的能力，可以用系统可用时间占总时间的比值来衡量，4 个 9 的可用性表示系统 99.99% 的时间是可用的。
 
-在可用性条件下，系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
+在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
 
 ## 分区容忍性
 
@@ -198,16 +183,11 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点表示
 
 可用性和一致性往往是冲突的，很难都使它们同时满足。在多个节点之间进行数据同步时，
 
-*   为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
-*   为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
+- 为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
+- 为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
 
 <div align="center"> <img src="../pics//0b587744-c0a8-46f2-8d72-e8f070d67b4b.jpg"/> </div><br>
 
-参考：
-
-- 倪超. 从 Paxos 到 ZooKeeper : 分布式一致性原理与实践 [M]. 电子工业出版社, 2015.
-- [What is CAP theorem in distributed database system?](http://www.colooshiki.com/index.php/2017/04/20/what-is-cap-theorem-in-distributed-database-system/)
-
 # 四、BASE
 
 BASE 是基本可用（Basically Available）、软状态（Soft State）和最终一致性（Eventually Consistent）三个短语的缩写。
@@ -292,66 +272,69 @@ Acceptor 接收到接受请求时，如果序号大于等于该 Acceptor 承诺
 
 Paxos 协议能够让 Proposer 发送的提议朝着能被大多数 Acceptor 接受的那个提议靠拢，因此能够保证可终止性。
 
-参考：
+# 六、Raft
 
-- [NEAT ALGORITHMS - PAXOS](http://harry.me/blog/2014/12/27/neat-algorithms-paxos/)
-- [Paxos By Example](https://angus.nyc/2012/paxos-by-example/)
-
-# 五、Raft
-
-Raft 和 Paxos 类似，但是更容易理解，也更容易实现。
-
-Raft 主要是用来竞选主节点。
+Raft 也是分布式一致性协议，主要是用来竞选主节点。
 
 ## 单个 Candidate 的竞选
 
 有三种节点：Follower、Candidate 和 Leader。Leader 会周期性的发送心跳包给 Follower。每个 Follower 都设置了一个随机的竞选超时时间，一般为 150ms\~300ms，如果在这个时间内没有收到 Leader 的心跳包，就会变成 Candidate，进入竞选阶段。
 
-*   下图表示一个分布式系统的最初阶段，此时只有 Follower，没有 Leader。Follower A 等待一个随机的竞选超时时间之后，没收到 Leader 发来的心跳包，因此进入竞选阶段。
+- 下图展示一个分布式系统的最初阶段，此时只有 Follower 没有 Leader。Node A 等待一个随机的竞选超时时间之后，没收到 Leader 发来的心跳包，因此进入竞选阶段。
 
 <div align="center"> <img src="../pics//111521118015898.gif"/> </div><br>
 
-*   此时 A 发送投票请求给其它所有节点。
+- 此时 Node A 发送投票请求给其它所有节点。
 
 <div align="center"> <img src="../pics//111521118445538.gif"/> </div><br>
 
-*   其它节点会对请求进行回复，如果超过一半的节点回复了，那么该 Candidate 就会变成 Leader。
+- 其它节点会对请求进行回复，如果超过一半的节点回复了，那么该 Candidate 就会变成 Leader。
 
 <div align="center"> <img src="../pics//111521118483039.gif"/> </div><br>
 
-*   之后 Leader 会周期性地发送心跳包给 Follower，Follower 接收到心跳包，会重新开始计时。
+- 之后 Leader 会周期性地发送心跳包给 Follower，Follower 接收到心跳包，会重新开始计时。
 
 <div align="center"> <img src="../pics//111521118640738.gif"/> </div><br>
 
 ## 多个 Candidate 竞选
 
-*   如果有多个 Follower 成为 Candidate，并且所获得票数相同，那么就需要重新开始投票，例如下图中 Candidate B 和 Candidate D 都获得两票，因此需要重新开始投票。
+- 如果有多个 Follower 成为 Candidate，并且所获得票数相同，那么就需要重新开始投票。例如下图中 Node B 和 Node D 都获得两票，需要重新开始投票。
 
 <div align="center"> <img src="../pics//111521119203347.gif"/> </div><br>
 
-*   当重新开始投票时，由于每个节点设置的随机竞选超时时间不同，因此能下一次再次出现多个 Candidate 并获得同样票数的概率很低。
+- 由于每个节点设置的随机竞选超时时间不同，因此下一次再次出现多个 Candidate 并获得同样票数的概率很低。
 
 <div align="center"> <img src="../pics//111521119368714.gif"/> </div><br>
 
-## 日志复制
+## 数据同步
 
-*   来自客户端的修改都会被传入 Leader。注意该修改还未被提交，只是写入日志中。
+- 来自客户端的修改都会被传入 Leader。注意该修改还未被提交，只是写入日志中。
 
 <div align="center"> <img src="../pics//7.gif"/> </div><br>
 
-*   Leader 会把修改复制到所有 Follower。
+- Leader 会把修改复制到所有 Follower。
 
 <div align="center"> <img src="../pics//9.gif"/> </div><br>
 
-*   Leader 会等待大多数的 Follower 也进行了修改，然后才将修改提交。
+- Leader 会等待大多数的 Follower 也进行了修改，然后才将修改提交。
 
 <div align="center"> <img src="../pics//10.gif"/> </div><br>
 
-*   此时 Leader 会通知的所有 Follower 让它们也提交修改，此时所有节点的值达成一致。
+- 此时 Leader 会通知的所有 Follower 让它们也提交修改，此时所有节点的值达成一致。
 
 <div align="center"> <img src="../pics//11.gif"/> </div><br>
 
-参考：
+# 参考
 
+- 倪超. 从 Paxos 到 ZooKeeper : 分布式一致性原理与实践 [M]. 电子工业出版社, 2015.
+- [Distributed locks with Redis](https://redis.io/topics/distlock)
+- [浅谈分布式锁](http://www.linkedkeeper.com/detail/blog.action?bid=1023)
+- [基于 Zookeeper 的分布式锁](http://www.dengshenyu.com/java/%E5%88%86%E5%B8%83%E5%BC%8F%E7%B3%BB%E7%BB%9F/2017/10/23/zookeeper-distributed-lock.html)
 - [Raft: Understandable Distributed Consensus](http://thesecretlivesofdata.com/raft)
+- [聊聊分布式事务，再说说解决方案](https://www.cnblogs.com/savorboard/p/distributed-system-transaction-consistency.html)
+- [分布式系统的事务处理](https://coolshell.cn/articles/10910.html)
+- [深入理解分布式事务](https://juejin.im/entry/577c6f220a2b5800573492be)
+- [What is CAP theorem in distributed database system?](http://www.colooshiki.com/index.php/2017/04/20/what-is-cap-theorem-in-distributed-database-system/)
+- [NEAT ALGORITHMS - PAXOS](http://harry.me/blog/2014/12/27/neat-algorithms-paxos/)
+- [Paxos By Example](https://angus.nyc/2012/paxos-by-example/)
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 1bac6bbb..d88b6168 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -1082,7 +1082,7 @@ false
 
 ```java
 public boolean isNumeric(char[] str) {
-    if (str == null)
+    if (str == null || str.length == 0)
         return false;
     return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
 }
@@ -2779,19 +2779,25 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
 
 ```java
 public boolean isContinuous(int[] nums) {
+
     if (nums.length < 5)
         return false;
+
     Arrays.sort(nums);
+
+    // 统计癞子数量
     int cnt = 0;
-    for (int num : nums)                   /* 统计癞子数量 */
+    for (int num : nums)
         if (num == 0)
             cnt++;
 
+    // 使用癞子去补全不连续的顺子
     for (int i = cnt; i < nums.length - 1; i++) {
         if (nums[i + 1] == nums[i])
             return false;
-        cnt -= nums[i + 1] - nums[i] - 1;  /* 使用癞子去补全不连续的顺子 */
+        cnt -= nums[i + 1] - nums[i] - 1;
     }
+
     return cnt >= 0;
 }
 ```
@@ -2824,7 +2830,7 @@ public int LastRemaining_Solution(int n, int m) {
 
 ## 题目描述
 
-可以有一次买入和一次卖出，买入必须在前。求最大收益。
+可以有一次买入和一次卖出，那么买入必须在前。求最大收益。
 
 ## 解题思路
 
diff --git a/notes/攻击技术.md b/notes/攻击技术.md
index d5a735ba..46dd4616 100644
--- a/notes/攻击技术.md
+++ b/notes/攻击技术.md
@@ -207,7 +207,7 @@ ResultSet rs = stmt.executeQuery();
 
 拒绝服务攻击（denial-of-service attack，DoS），亦称洪水攻击，其目的在于使目标电脑的网络或系统资源耗尽，使服务暂时中断或停止，导致其正常用户无法访问。
 
-分布式拒绝服务攻击（distributed denial-of-service attack，DDoS），指攻击者使用网络上两个或以上被攻陷的电脑作为“僵尸”向特定的目标发动“拒绝服务”式攻击。
+分布式拒绝服务攻击（distributed denial-of-service attack，DDoS），指攻击者使用两个或以上被攻陷的电脑作为“僵尸”向特定的目标发动“拒绝服务”式攻击。
 
 # 参考资料
 
diff --git a/notes/构建工具.md b/notes/构建工具.md
index c0e4acd5..ce157934 100644
--- a/notes/构建工具.md
+++ b/notes/构建工具.md
@@ -15,7 +15,7 @@
 
 ## 运行单元测试
 
-不再需要在项目代码中添加测试代码，从而污染项目代码。
+不再需要在项目代码中添加测试代码，从而避免了污染项目代码。
 
 ## 将源代码转化为可执行文件
 
diff --git a/notes/消息队列.md b/notes/消息队列.md
index c5687e8a..209d962a 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -9,6 +9,7 @@
 * [三、可靠性](#三可靠性)
     * [发送端的可靠性](#发送端的可靠性)
     * [接收端的可靠性](#接收端的可靠性)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -29,15 +30,10 @@
 发布与订阅模式和观察者模式有以下不同：
 
 - 观察者模式中，观察者和主题都知道对方的存在；而在发布与订阅模式中，发布者与订阅者不知道对方的存在，它们之间通过频道进行通信。
-- 观察者模式是同步的，当事件触发时，主题会去调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息。
+- 观察者模式是同步的，当事件触发时，主题会调用观察者的方法，然后等待方法返回；而发布与订阅模式是异步的，发布者向频道发送一个消息之后，就不需要关心订阅者何时去订阅这个消息，可以立即返回。
 
 <div align="center"> <img src="../pics//bee1ff1d-c80f-4b3c-b58c-7073a8896ab2.jpg"/> </div><br>
 
-参考：
-
-- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
-- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
-
 # 二、使用场景
 
 ## 异步处理
@@ -78,3 +74,8 @@
 
 - 保证接收端处理消息的业务逻辑具有幂等性：只要具有幂等性，那么消费多少次消息，最后处理的结果都是一样的。
 - 保证消息具有唯一编号，并使用一张日志表来记录已经消费的消息编号。
+
+# 参考资料
+
+- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
+- [消息队列中点对点与发布订阅区别](https://blog.csdn.net/lizhitao/article/details/47723105)
diff --git a/notes/算法.md b/notes/算法.md
index 30467b42..a9f9e689 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -23,7 +23,7 @@
 * [五、栈和队列](#五栈和队列)
     * [栈](#栈)
     * [队列](#队列)
-* [六、查找](#六查找)
+* [六、符号表](#六符号表)
     * [初级实现](#初级实现)
     * [二叉查找树](#二叉查找树)
     * [2-3 查找树](#2-3-查找树)
@@ -507,7 +507,7 @@ public class QuickSort<T extends Comparable<T>> extends Sort<T> {
 
 取 a[l] 作为切分元素，然后从数组的左端向右扫描直到找到第一个大于等于它的元素，再从数组的右端向左扫描找到第一个小于等于它的元素，交换这两个元素。不断进行这个过程，就可以保证左指针 i 的左侧元素都不大于切分元素，右指针 j 的右侧元素都不小于切分元素。当两个指针相遇时，将切分元素 a[l] 和 a[j] 交换位置。
 
-<div align="center"> <img src="../pics//864bfa7d-1149-420c-a752-f9b3d4e782ec.png" width="400"/> </div><br>
+<div align="center"> <img src="../pics//766aedd0-1b00-4065-aa2b-7d31138df84f.png" width="400"/> </div><br>
 
 ```java
 private int partition(T[] nums, int l, int h) {
@@ -770,7 +770,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 | 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
-| 选择排序 | √| N<sup>2</sup> | 1 | |
+| 选择排序 | × | N<sup>2</sup> | 1 | |
 | 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
 | 插入排序 | √ |  N \~ N<sup>2</sup> | 1 | 时间复杂度和初始顺序有关 |
 | 希尔排序 | ×  |  N 的若干倍乘于递增序列的长度 | 1 | |
@@ -1290,7 +1290,7 @@ public class ListQueue<Item> implements MyQueue<Item> {
 
 
 
-# 六、查找
+# 六、符号表
 
 符号表（Symbol Table）是一种存储键值对的数据结构，可以支持快速查找操作。
 
@@ -1411,9 +1411,9 @@ public class ListUnorderedST<Key, Value> implements UnorderedST<Key, Value> {
 
 使用一对平行数组，一个存储键一个存储值。
 
-rank() 方法至关重要，当键在表中时，它能够知道该键的位置；当键不在表中时，它也能知道在何处插入新键。
+二分查找的 rank() 方法至关重要，当键在表中时，它能够知道该键的位置；当键不在表中时，它也能知道在何处插入新键。
 
-复杂度：二分查找最多需要 logN+1 次比较，使用二分查找实现的符号表的查找操作所需要的时间最多是对数级别的。但是插入操作需要移动数组元素，是线性级别的。
+二分查找最多需要 logN+1 次比较，使用二分查找实现的符号表的查找操作所需要的时间最多是对数级别的。但是插入操作需要移动数组元素，是线性级别的。
 
 ```java
 public class BinarySearchOrderedST<Key extends Comparable<Key>, Value> implements OrderedST<Key, Value> {
@@ -2254,7 +2254,7 @@ from H1 to H3
 - c : 40
 - d : 80
 
-可以将每种字符转换成二进制编码，例如将 a 转换为 00，b 转换为 01，c 转换为 10，d 转换为 11。这是最简单的一种编码方式，没有考虑各个字符的权值（出现频率）。而哈夫曼编码能让出现频率最高的字符的编码最短，从而保证整体的编码长度最短。
+可以将每种字符转换成二进制编码，例如将 a 转换为 00，b 转换为 01，c 转换为 10，d 转换为 11。这是最简单的一种编码方式，没有考虑各个字符的权值（出现频率）。而哈夫曼编码采用了贪心策略，使出现频率最高的字符的编码最短，从而保证整体的编码长度最短。
 
 首先生成一颗哈夫曼树，每次生成过程中选取频率最少的两个节点，生成一个新节点作为它们的父节点，并且新节点的频率为两个节点的和。选取频率最少的原因是，生成过程使得先选取的节点在树的最底层，那么需要的编码长度更长，频率更少可以使得总编码长度更少。
 
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index 805764a9..b6a3ddf5 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -68,7 +68,7 @@
 
 应用服务器只要不具有状态，那么就可以很容易地通过负载均衡器向集群中添加新的服务器。
 
-关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器存储空间限制。
+关系型数据库的伸缩性通过 Sharding 来实现，将数据按一定的规则分布到不同的节点上，从而解决单台存储服务器的存储空间限制。
 
 对于非关系型数据库，它们天生就是为海量数据而诞生，对伸缩性的支持特别好。
 
@@ -78,7 +78,7 @@
 
 实现可扩展主要有两种方式：
 
-- 使用消息队列进行解耦，应用之间通过消息传递的方式进行通信；
+- 使用消息队列进行解耦，应用之间通过消息传递进行通信；
 - 使用分布式服务将业务和可复用的服务分离开来，业务使用分布式服务框架调用可复用的服务。新增的产品可以通过调用可复用的服务来实现业务逻辑，对其它产品没有影响。
 
 # 四、可用性
@@ -87,7 +87,7 @@
 
 保证高可用的主要手段是使用冗余，当某个服务器故障时就请求其它服务器。
 
-应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上不会对用户有任何影响。
+应用服务器的冗余比较容易实现，只要保证应用服务器不具有状态，那么某个应用服务器故障时，负载均衡器将该应用服务器原先的用户请求转发到另一个应用服务器上，不会对用户有任何影响。
 
 存储服务器的冗余需要使用主从复制来实现，当主服务器故障时，需要提升从服务器为主服务器，这个过程称为切换。
 
@@ -97,7 +97,7 @@
 
 ## 服务降级
 
-服务器降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
+服务降级是系统为了应对大量的请求，主动关闭部分功能，从而保证核心功能可用。
 
 # 五、安全性
 
diff --git a/notes/缓存.md b/notes/缓存.md
index 9a9f2810..ed834253 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -6,6 +6,7 @@
 * [五、缓存问题](#五缓存问题)
 * [六、数据分布](#六数据分布)
 * [七、一致性哈希](#七一致性哈希)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -29,10 +30,6 @@
 
 - LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最远的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
 
-参考资料：
-
-- [缓存那些事](https://tech.meituan.com/cache_about.html)
-
 # 二、LRU
 
 以下是一个基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
@@ -143,10 +140,6 @@ public class LRU<K, V> implements Iterable<K> {
 }
 ```
 
-源代码：
-
-- [CyC2018/Algorithm](https://github.com/CyC2018/Algorithm/tree/master/Caching)
-
 # 三、缓存位置
 
 ## 浏览器
@@ -169,7 +162,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 使用 Redis、Memcache 等分布式缓存将数据缓存在分布式缓存系统中。
 
-相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存。而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
+相对于本地缓存来说，分布式缓存单独部署，可以根据需求分配硬件资源。不仅如此，服务器集群都可以访问分布式缓存，而本地缓存需要在服务器集群之间进行同步，实现和性能开销上都非常大。
 
 ## 数据库缓存
 
@@ -177,7 +170,7 @@ MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查
 
 # 四、CDN
 
-内容分发网络（Content distribution network，CDN）是一种通过互连的网络系统，利用更靠近用户的服务器更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
+内容分发网络（Content distribution network，CDN）是一种互连的网络系统，它利用更靠近用户的服务器从而更快更可靠地将 HTML、CSS、JavaScript、音乐、图片、视频等静态资源分发给用户。
 
 CDN 主要有以下优点：
 
@@ -187,11 +180,6 @@ CDN 主要有以下优点：
 
 <div align="center"> <img src="../pics//15313ed8-a520-4799-a300-2b6b36be314f.jpg"/> </div><br>
 
-参考资料：
-
-- [内容分发网络](https://zh.wikipedia.org/wiki/%E5%85%A7%E5%AE%B9%E5%82%B3%E9%81%9E%E7%B6%B2%E8%B7%AF)
-- [How Aspiration CDN helps to improve your website loading speed?](https://www.aspirationhosting.com/aspiration-cdn/)
-
 # 五、缓存问题
 
 ## 缓存穿透
@@ -213,7 +201,7 @@ CDN 主要有以下优点：
 
 - 为了防止缓存在同一时间大面积过期导致的缓存雪崩，可以通过观察用户行为，合理设置缓存过期时间来实现；
 - 为了防止缓存服务器宕机出现的缓存雪崩，可以使用分布式缓存，分布式缓存中每一个节点只缓存部分的数据，当某个节点宕机时可以保证其它节点的缓存仍然可用。
-- 也可以在进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
+- 也可以进行缓存预热，避免在系统刚启动不久由于还未将大量数据进行缓存而导致缓存雪崩。
 
 ## 缓存一致性
 
@@ -243,10 +231,6 @@ CDN 主要有以下优点：
 - 能保持数据原有的顺序；
 - 并且能够准确控制每台服务器存储的数据量，从而使得存储空间的利用率最大。
 
-参考资料：
-
-- 大规模分布式存储系统
-
 # 七、一致性哈希
 
 Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据失效的问题。
@@ -267,6 +251,10 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
 
-参考资料：
+# 参考资料
 
+- 大规模分布式存储系统
+- [缓存那些事](https://tech.meituan.com/cache_about.html)
 - [一致性哈希算法](https://my.oschina.net/jayhu/blog/732849)
+- [内容分发网络](https://zh.wikipedia.org/wiki/%E5%85%A7%E5%AE%B9%E5%82%B3%E9%81%9E%E7%B6%B2%E8%B7%AF)
+- [How Aspiration CDN helps to improve your website loading speed?](https://www.aspirationhosting.com/aspiration-cdn/)
diff --git a/notes/设计模式.md b/notes/设计模式.md
index b8f4dc9b..531595e3 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -60,9 +60,9 @@
 
 （一）懒汉式-线程不安全
 
-以下实现中，私有静态变量 uniqueInstance 被延迟化实例化，这样做的好处是，如果没有用到该类，那么就不会实例化 uniqueInstance，从而节约资源。
+以下实现中，私有静态变量 uniqueInstance 被延迟实例化，这样做的好处是，如果没有用到该类，那么就不会实例化 uniqueInstance，从而节约资源。
 
-这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么多个线程会执行 `uniqueInstance = new Singleton();` 语句，这将导致多次实例化 uniqueInstance。
+这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么会有多个线程执行 `uniqueInstance = new Singleton();` 语句，这将导致多次实例化 uniqueInstance。
 
 ```java
 public class Singleton {
@@ -81,11 +81,21 @@ public class Singleton {
 }
 ```
 
-（二）懒汉式-线程安全
+（二）饿汉式-线程安全
 
-只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了对 uniqueInstance 进行多次实例化的问题。
+线程不安全问题主要是由于 uniqueInstance 被多次实例化，采取直接实例化 uniqueInstance 的方式就不会产生线程不安全问题。
 
-但是这样有一个问题，就是当一个线程进入该方法之后，其它线程试图进入该方法都必须等待，因此性能上有一定的损耗。
+但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
+
+```java
+private static Singleton uniqueInstance = new Singleton();
+```
+
+（三）懒汉式-线程安全
+
+只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了多次实例化 uniqueInstance 的问题。
+
+但是当一个线程进入该方法之后，其它试图进入该方法的线程都必须等待，因此性能上有一定的损耗。
 
 ```java
 public static synchronized Singleton getUniqueInstance() {
@@ -96,16 +106,6 @@ public static synchronized Singleton getUniqueInstance() {
 }
 ```
 
-（三）饿汉式-线程安全
-
-线程不安全问题主要是由于 uniqueInstance 被实例化了多次，如果 uniqueInstance 采用直接实例化的话，就不会被实例化多次，也就不会产生线程不安全问题。
-
-但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
-
-```java
-private static Singleton uniqueInstance = new Singleton();
-```
-
 （四）双重校验锁-线程安全
 
 uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
@@ -175,7 +175,7 @@ public class Singleton {
 }
 ```
 
-（五）枚举实现
+（六）枚举实现
 
 这是单例模式的最佳实践，它实现简单，并且在面对复杂的序列化或者反射攻击的时候，能够防止实例化多次。
 
@@ -231,27 +231,9 @@ public class Singleton implements Serializable {
 
 简单工厂不是设计模式，更像是一种编程习惯。它把实例化的操作单独放到一个类中，这个类就成为简单工厂类，让简单工厂类来决定应该用哪个具体子类来实例化。
 
-<div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
-
 这样做能把客户类和具体子类的实现解耦，客户类不再需要知道有哪些子类以及应当实例化哪个子类。因为客户类往往有多个，如果不使用简单工厂，所有的客户类都要知道所有子类的细节。而且一旦子类发生改变，例如增加子类，那么所有的客户类都要进行修改。
 
-如果存在下面这种代码，就需要使用简单工厂将对象实例化的部分放到简单工厂中。
-
-```java
-public class Client {
-    public static void main(String[] args) {
-        int type = 1;
-        Product product;
-        if (type == 1) {
-            product = new ConcreteProduct1();
-        } else if (type == 2) {
-            product = new ConcreteProduct2();
-        } else {
-            product = new ConcreteProduct();
-        }
-    }
-}
-```
+<div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
 
 ### 实现
 
@@ -275,6 +257,27 @@ public class ConcreteProduct2 implements Product {
 }
 ```
 
+以下的 Client 类中包含了实例化的代码，这是一种错误的实现，如果在客户类中存在实例化代码，就需要将代码放到简单工厂中。
+
+```java
+public class Client {
+    public static void main(String[] args) {
+        int type = 1;
+        Product product;
+        if (type == 1) {
+            product = new ConcreteProduct1();
+        } else if (type == 2) {
+            product = new ConcreteProduct2();
+        } else {
+            product = new ConcreteProduct();
+        }
+        // do something with the product
+    }
+}
+```
+
+以下的 SimpleFactory 是简单工厂实现，它被所有需要进行实例化的客户类调用。
+
 ```java
 public class SimpleFactory {
     public Product createProduct(int type) {
@@ -293,6 +296,7 @@ public class Client {
     public static void main(String[] args) {
         SimpleFactory simpleFactory = new SimpleFactory();
         Product product = simpleFactory.createProduct(1);
+        // do something with the product
     }
 }
 ```
@@ -2906,7 +2910,7 @@ Java 利用缓存来加速大量小对象的访问时间。
 - 远程代理（Remote Proxy）：控制对远程对象（不同地址空间）的访问，它负责将请求及其参数进行编码，并向不同地址空间中的对象发送已经编码的请求。
 - 虚拟代理（Virtual Proxy）：根据需要创建开销很大的对象，它可以缓存实体的附加信息，以便延迟对它的访问，例如在网站加载一个很大图片时，不能马上完成，可以用虚拟代理缓存图片的大小信息，然后生成一张临时图片代替原始图片。
 - 保护代理（Protection Proxy）：按权限控制对象的访问，它负责检查调用者是否具有实现一个请求所必须的访问权限。
-- 智能代理（Smart Reference）：取代了简单的指针，它在访问对象时执行一些附加操作：记录对象的引用次数，比如智能智能；当第一次引用一个持久化对象时，将它装入内存；在访问一个实际对象前，检查是否已经锁定了它，以确保其它对象不能改变它。
+- 智能代理（Smart Reference）：取代了简单的指针，它在访问对象时执行一些附加操作：记录对象的引用次数；当第一次引用一个持久化对象时，将它装入内存；在访问一个实际对象前，检查是否已经锁定了它，以确保其它对象不能改变它。
 
 <div align="center"> <img src="../pics//a6c20f60-5eba-427d-9413-352ada4b40fe.png"/> </div><br>
 
diff --git a/notes/面向对象思想.md b/notes/面向对象思想.md
index 9b59e43d..203c7ebf 100644
--- a/notes/面向对象思想.md
+++ b/notes/面向对象思想.md
@@ -37,6 +37,7 @@
 
 ```java
 public class Person {
+
     private String name;
     private int gender;
     private int age;
@@ -63,17 +64,20 @@ public class Person {
 
 继承实现了  **IS-A**  关系，例如 Cat 和 Animal 就是一种 IS-A 关系，因此 Cat 可以继承自 Animal，从而获得 Animal 非 private 的属性和方法。
 
+继承应该遵循里氏替换原则，子类对象必须能够替换掉所有父类对象。
+
 Cat 可以当做 Animal 来使用，也就是说可以使用 Animal 引用 Cat 对象。父类引用指向子类对象称为  **向上转型** 。
 
 ```java
 Animal animal = new Cat();
 ```
 
-继承应该遵循里氏替换原则，子类对象必须能够替换掉所有父类对象。
-
 ## 多态
 
-多态分为编译时多态和运行时多态。编译时多态主要指方法的重载，运行时多态指程序中定义的对象引用所指向的具体类型在运行期间才确定。
+多态分为编译时多态和运行时多态：
+
+- 编译时多态主要指方法的重载
+- 运行时多态指程序中定义的对象引用所指向的具体类型在运行期间才确定
 
 运行时多态有三个条件：
 
@@ -116,7 +120,7 @@ public class Music {
 
 # 二、类图
 
-以下类图使用 [PlantUML](https://www.planttext.com/) 绘制，更多语法及使用请参考：http://plantuml.com/
+以下类图使用 [PlantUML](https://www.planttext.com/) 绘制，更多语法及使用请参考：http://plantuml.com/ 。
 
 ## 泛化关系 (Generalization)
 
@@ -327,7 +331,7 @@ Vihicle .. N
 
 ### 2. 合成复用原则
 
-尽量使用对象组合，而不是继承来达到复用的目的。
+尽量使用对象组合，而不是通过继承来达到复用的目的。
 
 ### 3. 共同封闭原则
 
@@ -349,3 +353,4 @@ Vihicle .. N
 - [看懂 UML 类图和时序图](http://design-patterns.readthedocs.io/zh_CN/latest/read_uml.html#generalization)
 - [UML 系列——时序图（顺序图）sequence diagram](http://www.cnblogs.com/wolf-sun/p/UML-Sequence-diagram.html)
 - [面向对象编程三大特性 ------ 封装、继承、多态](http://blog.csdn.net/jianyuerensheng/article/details/51602015)
+

From a8104d3f5e7df44fe56b55e16b07639d5811e381 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Sat, 25 Aug 2018 23:30:03 +0800
Subject: [PATCH 142/172] Update README.md

---
 README.md | 36 ++++++++++++++++++------------------
 1 file changed, 18 insertions(+), 18 deletions(-)

diff --git a/README.md b/README.md
index d511fa7c..27edaf24 100644
--- a/README.md
+++ b/README.md
@@ -11,7 +11,7 @@
 
 <!-- [![](https://img.shields.io/badge/>-gitter-blue.svg)](https://gitter.im/CyC2018-Interview-Notebook/Lobby?utm_source=share-link&utm_medium=link&utm_campaign=share-link) [![](https://img.shields.io/badge/_-gitbook-4ab8a1.svg)](https://legacy.gitbook.com/book/cyc2018/interview-notebook/details)  -->
 
-## :pencil2: 算法
+### :pencil2: 算法
 
 - [剑指 Offer 题解](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/剑指%20offer%20题解.md)
 
@@ -25,7 +25,7 @@
 
    排序、并查集、栈和队列、红黑树、散列表。
 
-## :computer: 操作系统
+### :computer: 操作系统
 
 - [计算机操作系统](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机操作系统.md)
 
@@ -35,7 +35,7 @@
 
   基本实现原理以及基本操作。
 
-## :cloud: 网络 
+### :cloud: 网络 
 
 - [计算机网络](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/计算机网络.md)
 
@@ -49,7 +49,7 @@
 
   I/O 模型、I/O 多路复用。
 
-## :couple: 面向对象
+### :couple: 面向对象
 
 - [设计模式](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/设计模式.md)
 
@@ -59,7 +59,7 @@
 
   三大原则（继承、封装、多态）、类图、设计原则。
 
-## :floppy_disk: 数据库 
+### :floppy_disk: 数据库 
 
 - [数据库系统原理](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/数据库系统原理.md)
 
@@ -81,7 +81,7 @@
 
   五种数据类型、字典和跳跃表数据结构、使用场景、和 Memcache 的比较、淘汰策略、持久化、文件事件的 Reactor 模式、复制。
 
-## :coffee: Java
+### :coffee: Java
 
 - [Java 基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Java%20基础.md)
 
@@ -103,7 +103,7 @@
 
   NIO 的原理以及实例。
 
-## :bulb: 系统设计 
+### :bulb: 系统设计 
 
 - [系统设计基础](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/系统设计基础.md)
 
@@ -129,7 +129,7 @@
 
   消息处理模型、使用场景、可靠性
 
-## :hammer: 工具 
+### :hammer: 工具 
 
 - [Git](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/Git.md)
 
@@ -147,7 +147,7 @@
 
   构建工具的基本概念、主流构建工具介绍。
 
-## :speak_no_evil: 编码实践 
+### :speak_no_evil: 编码实践 
 
 - [重构](https://github.com/CyC2018/InnterviewNotes/blob/master/notes/重构.md)
 
@@ -161,9 +161,9 @@
 
   Google 开源项目的代码风格规范。
 
-## :memo: 后记 
+### :memo: 后记 
 
-### About
+#### About
 
 这个仓库是笔者的一个学习笔记，主要总结一些比较重要的知识点，希望对大家有所帮助。
 
@@ -171,7 +171,7 @@
 
 [BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)，这个书单是笔者至今看的一些比较好的技术书籍，虽然没有全都看完，但每本书多多少少都看了一部分。
 
-### How To Contribute
+#### How To Contribute
 
 笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
 
@@ -179,7 +179,7 @@
 
 欢迎在 Issue 中提交对本仓库的改进建议~
 
-### Typesetting
+#### Typesetting
 
 笔记内容按照 [中文文案排版指北](http://mazhuang.org/wiki/chinese-copywriting-guidelines/) 进行排版，以保证内容的可读性。
 
@@ -187,7 +187,7 @@
 
 笔者将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Markdown-Typesetting)。
 
-### Uploading
+#### Uploading
 
 笔者在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。
 
@@ -195,15 +195,15 @@
 
 笔者将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
 
-### Statement
+#### Statement
 
 本仓库不参与商业行为，不向读者收取任何费用。(This repository is not engaging in business activities, and does not charge readers any fee.)
 
-### Logo
+#### Logo
 
 Power by [logomakr](https://logomakr.com/).
 
-### Acknowledgements
+#### Acknowledgements
 
 感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与笔者联系。
 
@@ -229,7 +229,7 @@ Power by [logomakr](https://logomakr.com/).
     <img src="other/10072416.jpg" width="50px">
 </a>
 
-### License
+#### License
 
 在对本作品进行演绎时，请署名并以相同方式共享。
 

From b75f85c61731f3e128504bcbe0c0ce55294836b1 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 13:12:11 +0800
Subject: [PATCH 143/172] auto commit

---
 notes/MySQL.md | 46 ++++++++++++++++++++++++++--------------------
 1 file changed, 26 insertions(+), 20 deletions(-)

diff --git a/notes/MySQL.md b/notes/MySQL.md
index bb465ff3..04b6ea61 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -4,7 +4,7 @@
     * [MySQL 索引](#mysql-索引)
     * [索引优化](#索引优化)
     * [索引的优点](#索引的优点)
-    * [索引的使用场景](#索引的使用场景)
+    * [索引的使用条件](#索引的使用条件)
 * [二、查询性能优化](#二查询性能优化)
     * [使用 Explain 进行分析](#使用-explain-进行分析)
     * [优化数据访问](#优化数据访问)
@@ -58,13 +58,13 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 平衡树查找操作的时间复杂度等于树高 h，而树高大致为 O(h)=O(log<sub>d</sub>N)，其中 d 为每个节点的出度。
 
-红黑树的出度为 2，而 B+ Tree 的出度一般都非常大，所以红黑树的树高 h 很明显比 B+ Tree 大非常多，检索的次数也就更多。
+红黑树的出度为 2，而 B+ Tree 的出度一般都非常大，所以红黑树的树高 h 很明显比 B+ Tree 大非常多，查找的次数也就更多。
 
-（二）利用计算机预读特性
+（二）利用磁盘预读特性
 
-为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，因此速度会非常快。
+为了减少磁盘 I/O，磁盘往往不是严格按需读取，而是每次都会预读。预读过程中，磁盘进行顺序读取，顺序读取不需要进行磁盘寻道，并且只需要很短的旋转时间，速度会非常快。
 
-操作系统一般将内存和磁盘分割成固态大小的块，每一块称为一页，内存与磁盘以页为单位交换数据。数据库系统将索引的一个节点的大小设置为页的大小，使得一次 I/O 就能完全载入一个节点，并且可以利用预读特性，相邻的节点也能够被预先载入。
+操作系统一般将内存和磁盘分割成固态大小的块，每一块称为一页，内存与磁盘以页为单位交换数据。数据库系统将索引的一个节点的大小设置为页的大小，使得一次 I/O 就能完全载入一个节点。并且可以利用预读特性，相邻的节点也能够被预先载入。
 
 ## MySQL 索引
 
@@ -74,15 +74,15 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 是大多数 MySQL 存储引擎的默认索引类型。
 
-因为不再需要进行全表扫描，只需要对树进行搜索即可，因此查找速度快很多。除了用于查找，还可以用于排序和分组。
+因为不再需要进行全表扫描，只需要对树进行搜索即可，所以查找速度快很多。
+
+除了用于查找，还可以用于排序和分组。
 
 可以指定多个列作为索引列，多个索引列共同组成键。
 
 适用于全键值、键值范围和键前缀查找，其中键前缀查找只适用于最左前缀查找。如果不是按照索引列的顺序进行查找，则无法使用索引。
 
-InnoDB 的 B+Tree 索引分为主索引和辅助索引。
-
-主索引的叶子节点 data 域记录着完整的数据记录，这种索引方式被称为聚簇索引。因为无法把数据行存放在两个不同的地方，所以一个表只能有一个聚簇索引。
+InnoDB 的 B+Tree 索引分为主索引和辅助索引。主索引的叶子节点 data 域记录着完整的数据记录，这种索引方式被称为聚簇索引。因为无法把数据行存放在两个不同的地方，所以一个表只能有一个聚簇索引。
 
 <div align="center"> <img src="../pics//c28c6fbc-2bc1-47d9-9b2e-cf3d4034f877.jpg"/> </div><br>
 
@@ -92,7 +92,7 @@ InnoDB 的 B+Tree 索引分为主索引和辅助索引。
 
 ### 2. 哈希索引
 
-哈希索引能以 O(1) 时间进行查找，但是失去了有序性，它具有以下限制：
+哈希索引能以 O(1) 时间进行查找，但是失去了有序性：
 
 - 无法用于排序与分组；
 - 只支持精确查找，无法用于部分查找和范围查找。
@@ -101,9 +101,11 @@ InnoDB 存储引擎有一个特殊的功能叫“自适应哈希索引”，当
 
 ### 3. 全文索引
 
-MyISAM 存储引擎支持全文索引，用于查找文本中的关键词，而不是直接比较是否相等。查找条件使用 MATCH AGAINST，而不是普通的 WHERE。
+MyISAM 存储引擎支持全文索引，用于查找文本中的关键词，而不是直接比较是否相等。
 
-全文索引一般使用倒排索引实现，它记录着关键词到其所在文档的映射。
+查找条件使用 MATCH AGAINST，而不是普通的 WHERE。
+
+全文索引使用倒排索引实现，它记录着关键词到其所在文档的映射。
 
 InnoDB 存储引擎在 MySQL 5.6.4 版本中也开始支持全文索引。
 
@@ -136,7 +138,9 @@ WHERE actor_id = 1 AND film_id = 1;
 
 ### 3. 索引列的顺序
 
-让选择性最强的索引列放在前面，索引的选择性是指：不重复的索引值和记录总数的比值。最大值为 1，此时每个记录都有唯一的索引与其对应。选择性越高，查询效率也越高。
+让选择性最强的索引列放在前面。
+
+索引的选择性是指：不重复的索引值和记录总数的比值。最大值为 1，此时每个记录都有唯一的索引与其对应。选择性越高，查询效率也越高。
 
 例如下面显示的结果中 customer_id 的选择性比 staff_id 更高，因此最好把 customer_id 列放在多列索引的前面。
 
@@ -173,14 +177,16 @@ customer_id_selectivity: 0.0373
 
 - 大大减少了服务器需要扫描的数据行数。
 
-- 帮助服务器避免进行排序和分组，也就不需要创建临时表（B+Tree 索引是有序的，可以用于 ORDER BY 和 GROUP BY 操作。临时表主要是在排序和分组过程中创建，因为不需要排序和分组，也就不需要创建临时表）。
+- 帮助服务器避免进行排序和分组，以及避免创建临时表（B+Tree 索引是有序的，可以用于 ORDER BY 和 GROUP BY 操作。临时表主要是在排序和分组过程中创建，因为不需要排序和分组，也就不需要创建临时表）。
 
-- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，也就将相邻的数据都存储在一起）。
+- 将随机 I/O 变为顺序 I/O（B+Tree 索引是有序的，会将相邻的数据都存储在一起）。
 
-## 索引的使用场景
+## 索引的使用条件
+
+- 对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效；
+
+- 对于中到大型的表，索引就非常有效；
 
-- 对于非常小的表、大部分情况下简单的全表扫描比建立索引更高效。
-- 对于中到大型的表，索引就非常有效。
 - 但是对于特大型的表，建立和维护索引的代价将会随之增长。这种情况下，需要用到一种技术可以直接区分出需要查询的一组数据，而不是一条记录一条记录地匹配，例如可以使用分区技术。
 
 # 二、查询性能优化
@@ -214,7 +220,7 @@ Explain 用来分析 SELECT 查询语句，开发人员可以通过分析 Explai
 一个大查询如果一次性执行的话，可能一次锁住很多数据、占满整个事务日志、耗尽系统资源、阻塞很多小的但重要的查询。
 
 ```sql
-DELETE FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
+DELEFT FROM messages WHERE create < DATE_SUB(NOW(), INTERVAL 3 MONTH);
 ```
 
 ```sql
@@ -342,7 +348,7 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 水平切分又称为 Sharding，它是将同一个表中的记录拆分到多个结构相同的表中。
 
-当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓解单个数据库的压力。
+当一个表的数据不断增多时，Sharding 是必然的选择，它可以将数据分布到集群的不同节点上，从而缓存单个数据库的压力。
 
 <div align="center"> <img src="../pics//63c2909f-0c5f-496f-9fe5-ee9176b31aba.jpg"/> </div><br>
 

From e57025b873e8f330c8a36f9601965e9997e0e4e8 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 15:41:50 +0800
Subject: [PATCH 144/172] auto commit

---
 notes/Java 基础.md   |  2 +-
 notes/Java 虚拟机.md | 16 +++++-----------
 notes/MySQL.md       | 23 ++++++++++-------------
 3 files changed, 16 insertions(+), 25 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 44ca873a..1f172f30 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -215,7 +215,7 @@ String s5 = "bbb";
 System.out.println(s4 == s5);  // true
 ```
 
-在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到 Native Method 中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
+在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到堆中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
 
 - [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning)
 - [深入解析 String#intern](https://tech.meituan.com/in_depth_understanding_string_intern.html)
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 1bba557b..edc43e8f 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -38,7 +38,7 @@
 
 ## Java 虚拟机栈
 
-每个 Java 方法在执行的同时会创建一个栈帧用于存储局部变量表、操作数栈、常量池引用等信息，从调用直至执行完成的过程，就对应着一个栈帧在 Java 虚拟机栈中入栈和出栈的过程。
+每个 Java 方法在执行的同时会创建一个栈帧用于存储局部变量表、操作数栈、常量池引用等信息。从方法调用直至执行完成的过程，就对应着一个栈帧在 Java 虚拟机栈中入栈和出栈的过程。
 
 <div align="center"> <img src="../pics//926c7438-c5e1-4b94-840a-dcb24ff1dafe.png" width="450"/> </div><br>
 
@@ -55,27 +55,21 @@ java -Xss512M HackTheJava
 
 ## 本地方法栈
 
-本地方法一般是用其它语言（C、C++ 或汇编语言等）编写的，并且被编译为基于本机硬件和操作系统的程序，对待这些方法需要特别处理。
-
 本地方法栈与 Java 虚拟机栈类似，它们之间的区别只不过是本地方法栈为本地方法服务。
 
+本地方法一般是用其它语言（C、C++ 或汇编语言等）编写的，并且被编译为基于本机硬件和操作系统的程序，对待这些方法需要特别处理。
+
 <div align="center"> <img src="../pics//JNI-Java-Native-Interface.jpg" width="350"/> </div><br>
 
 ## 堆
 
 所有对象都在这里分配内存，是垃圾收集的主要区域（"GC 堆"）。
 
-现代的垃圾收集器基本都是采用分代收集算法，针对不同类型的对象采取不同的垃圾回收算法，可以将堆分成两块：
+现代的垃圾收集器基本都是采用分代收集算法，其主要的思想是针对不同类型的对象采取不同的垃圾回收算法，可以将堆分成两块：
 
 - 新生代（Young Generation）
 - 老年代（Old Generation）
 
-新生代可以继续划分成以下三个空间：
-
-- Eden（伊甸园）
-- From Survivor（幸存者）
-- To Survivor
-
 堆不需要连续内存，并且可以动态增加其内存，增加失败会抛出 OutOfMemoryError 异常。
 
 可以通过 -Xms 和 -Xmx 两个虚拟机参数来指定一个程序的堆内存大小，第一个参数设置初始值，第二个参数设置最大值。
@@ -104,7 +98,7 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 ## 直接内存
 
-在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存，然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。
+在 JDK 1.4 中新加入了 NIO 类，它可以使用 Native 函数库直接分配堆外内存（Native 堆），然后通过一个存储在 Java 堆里的 DirectByteBuffer 对象作为这块内存的引用进行操作。
 
 这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index 04b6ea61..bf795d34 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -354,17 +354,17 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 ## 垂直切分
 
-<div align="center"> <img src="../pics//e130e5b8-b19a-4f1e-b860-223040525cf6.jpg"/> </div><br>
-
 垂直切分是将一张表按列切分成多个表，通常是按照列的关系密集程度进行切分，也可以利用垂直切分将经常被使用的列和不经常被使用的列切分到不同的表中。
 
 在数据库的层面使用垂直切分将按数据库中表的密集程度部署到不同的库中，例如将原来的电商数据库垂直切分成商品数据库、用户数据库等。
 
+<div align="center"> <img src="../pics//e130e5b8-b19a-4f1e-b860-223040525cf6.jpg"/> </div><br>
+
 ## Sharding 策略
 
-- 哈希取模：hash(key) % NUM_DB
-- 范围：可以是 ID 范围也可以是时间范围
-- 映射表：使用单独的一个数据库来存储映射关系
+- 哈希取模：hash(key) % N；
+- 范围：可以是 ID 范围也可以是时间范围；
+- 映射表：使用单独的一个数据库来存储映射关系。
 
 ## Sharding 存在的问题及解决方案
 
@@ -382,20 +382,15 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 - 为每个分片指定一个 ID 范围
 - 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)
 
-更多内容请参考：
-
-- [How Sharding Works](https://medium.com/@jeeyoungk/how-sharding-works-b4dec46b3f6)
-- [大众点评订单系统分库分表实践](https://tech.meituan.com/dianping_order_db_sharding.html)
-
 # 六、复制
 
 ## 主从复制
 
 主要涉及三个线程：binlog 线程、I/O 线程和 SQL 线程。
 
--  **binlog 线程** ：负责将主服务器上的数据更改写入二进制日志中。
--  **I/O 线程** ：负责从主服务器上读取二进制日志，并写入从服务器的中继日志中。
--  **SQL 线程** ：负责读取中继日志并重放其中的 SQL 语句。
+-  **binlog 线程** ：负责将主服务器上的数据更改写入二进制日志（Binary log）中。
+-  **I/O 线程** ：负责从主服务器上读取二进制日志，并写入从服务器的重放日志（Replay log）中。
+-  **SQL 线程** ：负责读取重放日志并重放其中的 SQL 语句。
 
 <div align="center"> <img src="../pics//master-slave.png"/> </div><br>
 
@@ -423,3 +418,5 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 - [SQL Azure Federation – Introduction](http://geekswithblogs.net/shaunxu/archive/2012/01/07/sql-azure-federation-ndash-introduction.aspx "Title of this entry.")
 - [MySQL 索引背后的数据结构及算法原理](http://blog.codinglabs.org/articles/theory-of-mysql-index.html)
 - [MySQL 性能优化神器 Explain 使用分析](https://segmentfault.com/a/1190000008131735)
+- [How Sharding Works](https://medium.com/@jeeyoungk/how-sharding-works-b4dec46b3f6)
+- [大众点评订单系统分库分表实践](https://tech.meituan.com/dianping_order_db_sharding.html)

From 4a6a3583f0adf864a81dafa51546893dbad71d72 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Sun, 26 Aug 2018 15:44:44 +0800
Subject: [PATCH 145/172] Update BOOKLIST.md

---
 BOOKLIST.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/BOOKLIST.md b/BOOKLIST.md
index 4805a7dc..e88ba706 100644
--- a/BOOKLIST.md
+++ b/BOOKLIST.md
@@ -54,7 +54,7 @@
 - [STL 源码剖析](https://book.douban.com/subject/1110934/)
 - [深度探索 C++ 对象模型](https://book.douban.com/subject/1091086/)
 
-# 网站架构/分布式
+# 系统设计
 
 - [大规模分布式存储系统](https://book.douban.com/subject/25723658/)
 - [从 Paxos 到 Zookeeper](https://book.douban.com/subject/26292004/)

From 1c85575ed1cc2894bae203cb305f6e8c2de3bba9 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 16:17:09 +0800
Subject: [PATCH 146/172] auto commit

---
 notes/Java 虚拟机.md | 17 ++++++++---------
 1 file changed, 8 insertions(+), 9 deletions(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index edc43e8f..c3ae5889 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -136,8 +136,6 @@ public class ReferenceCountingGC {
 
 通过 GC Roots 作为起始点进行搜索，能够到达到的对象都是存活的，不可达的对象可被回收。
 
-<div align="center"> <img src="../pics//0635cbe8.png" width=""/> </div><br>
-
 Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC Roots 一般包含以下内容：
 
 - 虚拟机栈中引用的对象
@@ -145,6 +143,8 @@ Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC
 - 方法区中类静态属性引用的对象
 - 方法区中的常量引用的对象
 
+<div align="center"> <img src="../pics//0635cbe8.png" width=""/> </div><br>
+
 ### 3. 方法区的回收
 
 因为方法区主要存放永久代对象，而永久代对象的回收率比新生代低很多，因此在方法区上进行回收性价比不高。
@@ -165,13 +165,13 @@ Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC
 
 finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。但是 try-finally 等方式可以做的更好，并且该方法运行代价高昂，不确定性大，无法保证各个对象的调用顺序，因此最好不要使用。
 
-当一个对象可被回收时，如果需要执行该对象的 finalize() 方法，那么就有可能通过在该方法中让对象重新被引用，从而实现自救。自救只能进行一次，如果回收的对象之前调用了 finalize() 方法自救，后面回收时不会调用 finalize() 方法。
+当一个对象可被回收时，如果需要执行该对象的 finalize() 方法，那么就有可能在该方法中让对象重新被引用，从而实现自救。自救只能进行一次，如果回收的对象之前调用了 finalize() 方法自救，后面回收时不会调用 finalize() 方法。
 
 ## 引用类型
 
 无论是通过引用计算算法判断对象的引用数量，还是通过可达性分析算法判断对象是否可达，判定对象是否可被回收都与引用有关。
 
-Java 具有四种强度不同的引用类型。
+Java 提供了四种强度不同的引用类型。
 
 ### 1. 强引用
 
@@ -280,7 +280,7 @@ Serial 翻译为串行，也就是说它以串行的方式执行。
 
 它的优点是简单高效，对于单个 CPU 环境来说，由于没有线程交互的开销，因此拥有最高的单线程收集效率。
 
-它是 Client 模式下的默认新生代收集器，因为在用户的桌面应用场景下，分配给虚拟机管理的内存一般来说不会很大。Serial 收集器收集几十兆甚至一两百兆的新生代停顿时间可以控制在一百多毫秒以内，只要不是太频繁，这点停顿是可以接受的。
+它是 Client 模式下的默认新生代收集器，因为在该应用场景下，分配给虚拟机管理的内存一般来说不会很大。Serial 收集器收集几十兆甚至一两百兆的新生代停顿时间可以控制在一百多毫秒以内，只要不是太频繁，这点停顿是可以接受的。
 
 ### 2. ParNew 收集器
 
@@ -298,11 +298,11 @@ Serial 翻译为串行，也就是说它以串行的方式执行。
 
 其它收集器关注点是尽可能缩短垃圾收集时用户线程的停顿时间，而它的目标是达到一个可控制的吞吐量，它被称为“吞吐量优先”收集器。这里的吞吐量指 CPU 用于运行用户代码的时间占总时间的比值。
 
-停顿时间越短就越适合需要与用户交互的程序，良好的响应速度能提升用户体验。而高吞吐量则可以高效率地利用 CPU 时间，尽快完成程序的运算任务，主要适合在后台运算而不需要太多交互的任务。
+停顿时间越短就越适合需要与用户交互的程序，良好的响应速度能提升用户体验。而高吞吐量则可以高效率地利用 CPU 时间，尽快完成程序的运算任务，适合在后台运算而不需要太多交互的任务。
 
 缩短停顿时间是以牺牲吞吐量和新生代空间来换取的：新生代空间变小，垃圾回收变得频繁，导致吞吐量下降。
 
-可以通过一个开关参数打卡 GC 自适应的调节策略（GC Ergonomics），就不需要手工指定新生代的大小（-Xmn）、Eden 和 Survivor 区的比例、晋升老年代对象年龄等细节参数了。虚拟机会根据当前系统的运行情况收集性能监控信息，动态调整这些参数以提供最合适的停顿时间或者最大的吞吐量，这种方式称为 。
+可以通过一个开关参数打开 GC 自适应的调节策略（GC Ergonomics），就不需要手工指定新生代的大小（-Xmn）、Eden 和 Survivor 区的比例、晋升老年代对象年龄等细节参数了。虚拟机会根据当前系统的运行情况收集性能监控信息，动态调整这些参数以提供最合适的停顿时间或者最大的吞吐量。
 
 ### 4. Serial Old 收集器
 
@@ -372,8 +372,6 @@ G1 把堆划分成多个大小相等的独立区域（Region），新生代和
 - 空间整合：整体来看是基于“标记 - 整理”算法实现的收集器，从局部（两个 Region 之间）上来看是基于“复制”算法实现的，这意味着运行期间不会产生内存空间碎片。
 - 可预测的停顿：能让使用者明确指定在一个长度为 M 毫秒的时间片段内，消耗在 GC 上的时间不得超过 N 毫秒。
 
-更详细内容请参考：[Getting Started with the G1 Garbage Collector](http://www.oracle.com/webfolder/technetwork/tutorials/obe/java/G1GettingStarted/index.html)
-
 # 三、内存分配与回收策略
 
 ## Minor GC 和 Full GC
@@ -738,6 +736,7 @@ public class FileSystemClassLoader extends ClassLoader {
 - 周志明. 深入理解 Java 虚拟机 [M]. 机械工业出版社, 2011.
 - [Chapter 2. The Structure of the Java Virtual Machine](https://docs.oracle.com/javase/specs/jvms/se8/html/jvms-2.html#jvms-2.5.4)
 - [Jvm memory](https://www.slideshare.net/benewu/jvm-memory)
+[Getting Started with the G1 Garbage Collector](http://www.oracle.com/webfolder/technetwork/tutorials/obe/java/G1GettingStarted/index.html)
 - [JNI Part1: Java Native Interface Introduction and “Hello World” application](http://electrofriends.com/articles/jni/jni-part1-java-native-interface/)
 - [Memory Architecture Of JVM(Runtime Data Areas)](https://hackthejava.wordpress.com/2015/01/09/memory-architecture-by-jvmruntime-data-areas/)
 - [JVM Run-Time Data Areas](https://www.programcreek.com/2013/04/jvm-run-time-data-areas/)

From dce6d0ee3f0605e5553db6d46b15ea0ff58540c6 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 19:57:38 +0800
Subject: [PATCH 147/172] auto commit

---
 notes/Java 容器.md   | 187 ++++++++++++++++++++-----------------------
 notes/Java 虚拟机.md |   2 +-
 2 files changed, 87 insertions(+), 102 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 001f864c..57e5f475 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -126,7 +126,91 @@ public class ArrayList<E> extends AbstractList<E>
 private static final int DEFAULT_CAPACITY = 10;
 ```
 
-### 2. 序列化
+### 2. 扩容
+
+添加元素时使用 ensureCapacityInternal() 方法来保证容量足够，如果不够时，需要使用 grow() 方法进行扩容，新容量的大小为 `oldCapacity + (oldCapacity >> 1)`，也就是旧容量的 1.5 倍。
+
+扩容操作需要调用 `Arrays.copyOf()` 把原数组整个复制到新数组中，这个操作代价很高，因此最好在创建 ArrayList 对象时就指定大概的容量大小，减少扩容操作的次数。
+
+```java
+public boolean add(E e) {
+    ensureCapacityInternal(size + 1);  // Increments modCount!!
+    elementData[size++] = e;
+    return true;
+}
+
+private void ensureCapacityInternal(int minCapacity) {
+    if (elementData == DEFAULTCAPACITY_EMPTY_ELEMENTDATA) {
+        minCapacity = Math.max(DEFAULT_CAPACITY, minCapacity);
+    }
+    ensureExplicitCapacity(minCapacity);
+}
+
+private void ensureExplicitCapacity(int minCapacity) {
+    modCount++;
+    // overflow-conscious code
+    if (minCapacity - elementData.length > 0)
+        grow(minCapacity);
+}
+
+private void grow(int minCapacity) {
+    // overflow-conscious code
+    int oldCapacity = elementData.length;
+    int newCapacity = oldCapacity + (oldCapacity >> 1);
+    if (newCapacity - minCapacity < 0)
+        newCapacity = minCapacity;
+    if (newCapacity - MAX_ARRAY_SIZE > 0)
+        newCapacity = hugeCapacity(minCapacity);
+    // minCapacity is usually close to size, so this is a win:
+    elementData = Arrays.copyOf(elementData, newCapacity);
+}
+```
+
+### 3. 删除元素
+
+需要调用 System.arraycopy() 将 index+1 后面的元素都复制到 index 位置上，该操作的时间复杂度为 O(N)，可以看出 ArrayList 删除元素的代价是非常高的。
+
+```java
+public E remove(int index) {
+    rangeCheck(index);
+    modCount++;
+    E oldValue = elementData(index);
+    int numMoved = size - index - 1;
+    if (numMoved > 0)
+        System.arraycopy(elementData, index+1, elementData, index, numMoved);
+    elementData[--size] = null; // clear to let GC do its work
+    return oldValue;
+}
+```
+
+### 4. Fail-Fast
+
+modCount 用来记录 ArrayList 结构发生变化的次数。结构发生变化是指添加或者删除至少一个元素的所有操作，或者是调整内部数组的大小，仅仅只是设置元素的值不算结构发生变化。
+
+在进行序列化或者迭代等操作时，需要比较操作前后 modCount 是否改变，如果改变了需要抛出 ConcurrentModificationException。
+
+```java
+private void writeObject(java.io.ObjectOutputStream s)
+    throws java.io.IOException{
+    // Write out element count, and any hidden stuff
+    int expectedModCount = modCount;
+    s.defaultWriteObject();
+
+    // Write out size as capacity for behavioural compatibility with clone()
+    s.writeInt(size);
+
+    // Write out all elements in the proper order.
+    for (int i=0; i<size; i++) {
+        s.writeObject(elementData[i]);
+    }
+
+    if (modCount != expectedModCount) {
+        throw new ConcurrentModificationException();
+    }
+}
+```
+
+### 5. 序列化
 
 ArrayList 基于数组实现，并且具有动态扩容特性，因此保存元素的数组不一定都会被使用，那么就没必要全部进行序列化。
 
@@ -191,90 +275,6 @@ ObjectOutputStream oos = new ObjectOutputStream(new FileOutputStream(file));
 oos.writeObject(list);
 ```
 
-### 3. 扩容
-
-添加元素时使用 ensureCapacityInternal() 方法来保证容量足够，如果不够时，需要使用 grow() 方法进行扩容，新容量的大小为 `oldCapacity + (oldCapacity >> 1)`，也就是旧容量的 1.5 倍。
-
-扩容操作需要调用 `Arrays.copyOf()` 把原数组整个复制到新数组中，这个操作代价很高，因此最好在创建 ArrayList 对象时就指定大概的容量大小，减少扩容操作的次数。
-
-```java
-public boolean add(E e) {
-    ensureCapacityInternal(size + 1);  // Increments modCount!!
-    elementData[size++] = e;
-    return true;
-}
-
-private void ensureCapacityInternal(int minCapacity) {
-    if (elementData == DEFAULTCAPACITY_EMPTY_ELEMENTDATA) {
-        minCapacity = Math.max(DEFAULT_CAPACITY, minCapacity);
-    }
-    ensureExplicitCapacity(minCapacity);
-}
-
-private void ensureExplicitCapacity(int minCapacity) {
-    modCount++;
-    // overflow-conscious code
-    if (minCapacity - elementData.length > 0)
-        grow(minCapacity);
-}
-
-private void grow(int minCapacity) {
-    // overflow-conscious code
-    int oldCapacity = elementData.length;
-    int newCapacity = oldCapacity + (oldCapacity >> 1);
-    if (newCapacity - minCapacity < 0)
-        newCapacity = minCapacity;
-    if (newCapacity - MAX_ARRAY_SIZE > 0)
-        newCapacity = hugeCapacity(minCapacity);
-    // minCapacity is usually close to size, so this is a win:
-    elementData = Arrays.copyOf(elementData, newCapacity);
-}
-```
-
-### 4. 删除元素
-
-需要调用 System.arraycopy() 将 index+1 后面的元素都复制到 index 位置上，该操作的时间复杂度为 O(N)，可以看出 ArrayList 删除元素的代价是非常高的。
-
-```java
-public E remove(int index) {
-    rangeCheck(index);
-    modCount++;
-    E oldValue = elementData(index);
-    int numMoved = size - index - 1;
-    if (numMoved > 0)
-        System.arraycopy(elementData, index+1, elementData, index, numMoved);
-    elementData[--size] = null; // clear to let GC do its work
-    return oldValue;
-}
-```
-
-### 5. Fail-Fast
-
-modCount 用来记录 ArrayList 结构发生变化的次数。结构发生变化是指添加或者删除至少一个元素的所有操作，或者是调整内部数组的大小，仅仅只是设置元素的值不算结构发生变化。
-
-在进行序列化或者迭代等操作时，需要比较操作前后 modCount 是否改变，如果改变了需要抛出 ConcurrentModificationException。
-
-```java
-private void writeObject(java.io.ObjectOutputStream s)
-    throws java.io.IOException{
-    // Write out element count, and any hidden stuff
-    int expectedModCount = modCount;
-    s.defaultWriteObject();
-
-    // Write out size as capacity for behavioural compatibility with clone()
-    s.writeInt(size);
-
-    // Write out all elements in the proper order.
-    for (int i=0; i<size; i++) {
-        s.writeObject(elementData[i]);
-    }
-
-    if (modCount != expectedModCount) {
-        throw new ConcurrentModificationException();
-    }
-}
-```
-
 ## Vector
 
 ### 1. 同步
@@ -461,21 +461,6 @@ static class Entry<K,V> implements Map.Entry<K,V> {
     public final String toString() {
         return getKey() + "=" + getValue();
     }
-
-    /**
-     * This method is invoked whenever the value in an entry is
-     * overwritten by an invocation of put(k,v) for a key k that's already
-     * in the HashMap.
-     */
-    void recordAccess(HashMap<K,V> m) {
-    }
-
-    /**
-     * This method is invoked whenever the entry is
-     * removed from the table.
-     */
-    void recordRemoval(HashMap<K,V> m) {
-    }
 }
 ```
 
@@ -929,7 +914,7 @@ JDK 1.8 使用了 CAS 操作来支持更高的并发度，在 CAS 操作失败
 public class LinkedHashMap<K,V> extends HashMap<K,V> implements Map<K,V>
 ```
 
-内存维护了一个双向链表，用来维护插入顺序或者 LRU 顺序。
+内部维护了一个双向链表，用来维护插入顺序或者 LRU 顺序。
 
 ```java
 /**
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index c3ae5889..2b565cb5 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -531,7 +531,7 @@ public class Test {
 
 - 与类的构造函数（或者说实例构造器 &lt;init>()）不同，不需要显式的调用父类的构造器。虚拟机会自动保证在子类的 &lt;clinit>() 方法运行之前，父类的 &lt;clinit>() 方法已经执行结束。因此虚拟机中第一个执行 &lt;clinit>() 方法的类肯定为 java.lang.Object。
 
-- 由于父类的 &lt;clinit>() 方法先执行，也就意味着父类中定义的静态语句块要优先于子类的变量赋值操作。例如以下代码：
+- 由于父类的 &lt;clinit>() 方法先执行，也就意味着父类中定义的静态语句块的执行要优先于子类。例如以下代码：
 
 ```java
 static class Parent {

From 322a6c58bfd857dfcd78c6ed291ee001c9dc8ae6 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 20:53:07 +0800
Subject: [PATCH 148/172] auto commit

---
 notes/Java 基础.md | 76 ++++++++++++++++++++++++++++++++++++++++------
 1 file changed, 67 insertions(+), 9 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 1f172f30..95249103 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -6,7 +6,8 @@
     * [概览](#概览)
     * [不可变的好处](#不可变的好处)
     * [String, StringBuffer and StringBuilder](#string,-stringbuffer-and-stringbuilder)
-    * [String.intern()](#stringintern)
+    * [String Pool](#string-pool)
+    * [new String("abc")](#new-string"abc")
 * [三、运算](#三运算)
     * [参数传递](#参数传递)
     * [float 与 double](#float-与-double)
@@ -64,7 +65,7 @@ int y = x;         // 拆箱
 
 new Integer(123) 与 Integer.valueOf(123) 的区别在于：
 
-- new Integer(123) 每次都会新建一个对象
+- new Integer(123) 每次都会新建一个对象；
 - Integer.valueOf(123) 会使用缓存池中的对象，多次调用会取得同一个对象的引用。
 
 ```java
@@ -193,33 +194,90 @@ String 不可变性天生具备线程安全，可以在多个线程中安全地
 
 [StackOverflow : String, StringBuffer, and StringBuilder](https://stackoverflow.com/questions/2971315/string-stringbuffer-and-stringbuilder)
 
-## String.intern()
+## String Pool
 
-使用 String.intern() 可以保证相同内容的字符串变量引用同一的内存对象。
+字符串常量池（String Poll）保存着所有字符串字面量（literal strings），这些字面量在编译时期就确定。不仅如此，还可以使用 String 的 intern() 方法在运行过程中将字符串添加到 String Poll 中。
 
-下面示例中，s1 和 s2 采用 new String() 的方式新建了两个不同对象，而 s3 是通过 s1.intern() 方法取得一个对象引用。intern() 首先把 s1 引用的对象放到 String Pool（字符串常量池）中，然后返回这个对象引用。因此 s3 和 s1 引用的是同一个字符串常量池的对象。
+当一个字符串调用 intern() 方法时，如果 String Poll 中已经存在一个字符串和该字符串值相等（使用 equals() 方法进行确定），那么就会返回 String Poll 中字符串的引用；否则，就会在 String Poll 中添加一个新的字符串，并返回这个新字符串的引用。
+
+下面示例中，s1 和 s2 采用 new String() 的方式新建了两个不同字符串，而 s3 和 s4 是通过 s1.intern() 方法取得一个字符串引用。intern() 首先把 s1 引用的字符串放到 String Pool 中，然后返回这个字符串引用。因此 s3 和 s4 引用的是同一个字符串。
 
 ```java
 String s1 = new String("aaa");
 String s2 = new String("aaa");
 System.out.println(s1 == s2);           // false
 String s3 = s1.intern();
-System.out.println(s1.intern() == s3);  // true
+String s4 = s1.intern();
+System.out.println(s3 == s4);           // true
 ```
 
-如果是采用 "bbb" 这种使用双引号的形式创建字符串实例，会自动地将新建的对象放入 String Pool 中。
+如果是采用 "bbb" 这种字面量的形式创建字符串，会自动地将字符串放入 String Pool 中。
 
 ```java
-String s4 = "bbb";
 String s5 = "bbb";
+String s6 = "bbb";
 System.out.println(s4 == s5);  // true
 ```
 
-在 Java 7 之前，字符串常量池被放在运行时常量池中，它属于永久代。而在 Java 7，字符串常量池被移到堆中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
+在 Java 7 之前，String Poll 被放在运行时常量池中，它属于永久代。而在 Java 7，String Poll 被移到堆中。这是因为永久代的空间有限，在大量使用字符串的场景下会导致 OutOfMemoryError 错误。
 
 - [StackOverflow : What is String interning?](https://stackoverflow.com/questions/10578984/what-is-string-interning)
 - [深入解析 String#intern](https://tech.meituan.com/in_depth_understanding_string_intern.html)
 
+## new String("abc")
+
+使用这种方式一共会创建两个字符串对象（前提是 String Poll 中还没有 "abc" 字符串对象）。
+
+- "abc" 属于字符串字面量，因此编译时期会在 String Poll 中创建一个字符串对象，指向这个 "abc" 字符串字面量；
+- 而使用 new 的方式会在堆中创建一个字符串对象。
+
+创建一个测试类，其 main 方法中使用这种方式来创建字符串对象。
+
+```java
+public class NewStringTest {
+    public static void main(String[] args) {
+        String s = new String("abc");
+    }
+}
+```
+
+使用 javap -verbose 进行反编译，得到以下内容：
+
+```java
+// ...
+Constant pool:
+// ...
+   #2 = Class              #18            // java/lang/String
+   #3 = String             #19            // abc
+// ...
+  #18 = Utf8               java/lang/String
+  #19 = Utf8               abc
+// ...
+
+  public static void main(java.lang.String[]);
+    descriptor: ([Ljava/lang/String;)V
+    flags: ACC_PUBLIC, ACC_STATIC
+    Code:
+      stack=3, locals=2, args_size=1
+         0: new           #2                  // class java/lang/String
+         3: dup
+         4: ldc           #3                  // String abc
+         6: invokespecial #4                  // Method java/lang/String."<init>":(Ljava/lang/String;)V
+         9: astore_1
+// ...
+```
+
+在 Constant Poll 中，#19 存储这字符串字面量 "abc"，#3 是 String Poll 的字符串对象，它指向 #19 这个字符串字面量。在 main 方法中，0: 行使用 new #2 在堆中创建一个字符串对象，并且使用 ldc #3 将 String Poll 中的字符串对象作为 String 构造函数的参数。
+
+以下是 String 构造函数的源码，可以看到，在将一个字符串对象作为另一个字符串对象的构造函数参数时，并不会完全复制 value 数组内容，而是都会指向同一个 value 数组。
+
+```java
+public String(String original) {
+    this.value = original.value;
+    this.hash = original.hash;
+}
+```
+
 # 三、运算
 
 ## 参数传递

From 9e987425855fa2f150754ba15d242157f720c3a4 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 26 Aug 2018 23:17:04 +0800
Subject: [PATCH 149/172] auto commit

---
 notes/Java 基础.md | 19 ++++++++++++-------
 1 file changed, 12 insertions(+), 7 deletions(-)

diff --git a/notes/Java 基础.md b/notes/Java 基础.md
index 95249103..5b689cf8 100644
--- a/notes/Java 基础.md	
+++ b/notes/Java 基础.md	
@@ -461,6 +461,7 @@ public class AccessExample {
 
 ```java
 public class AccessWithInnerClassExample {
+
     private class InnerClass {
         int x;
     }
@@ -526,6 +527,7 @@ ac2.func1();
 
 ```java
 public interface InterfaceExample {
+
     void func1();
 
     default void func2(){
@@ -577,7 +579,7 @@ System.out.println(InterfaceExample.x);
 - 需要能控制继承来的成员的访问权限，而不是都为 public。
 - 需要继承非静态和非常量字段。
 
-在很多情况下，接口优先于抽象类，因为接口没有抽象类严格的类层次结构要求，可以灵活地为一个类添加行为。并且从 Java 8 开始，接口也可以有默认的方法实现，使得修改接口的成本也变的很低。
+在很多情况下，接口优先于抽象类。因为接口没有抽象类严格的类层次结构要求，可以灵活地为一个类添加行为。并且从 Java 8 开始，接口也可以有默认的方法实现，使得修改接口的成本也变的很低。
 
 - [深入理解 abstract class 和 interface](https://www.ibm.com/developerworks/cn/java/l-javainterface-abstract/)
 - [When to Use Abstract Class and Interface](https://dzone.com/articles/when-to-use-abstract-class-and-intreface)
@@ -718,7 +720,7 @@ x.equals(y) == x.equals(y); // true
 x.equals(null); // false;
 ```
 
-**2. equals() 与 ==** 
+**2. 等价与相等** 
 
 - 对于基本类型，== 判断两个值是否相等，基本类型没有 equals() 方法。
 - 对于引用类型，== 判断两个变量是否引用同一个对象，而 equals() 判断引用的对象是否等价。
@@ -781,7 +783,7 @@ set.add(e2);
 System.out.println(set.size());   // 2
 ```
 
-理想的散列函数应当具有均匀性，即不相等的对象应当均匀分布到所有可能的散列值上。这就要求了散列函数要把所有域的值都考虑进来，可以将每个域都当成 R 进制的某一位，然后组成一个 R 进制的整数。R 一般取 31，因为它是一个奇素数，如果是偶数的话，当出现乘法溢出，信息就会丢失，因为与 2 相乘相当于向左移一位。
+理想的散列函数应当具有均匀性，即不相等的对象应当均匀分布到所有可能的散列值上。这就要求了散列函数要把所有域的值都考虑进来。可以将每个域都当成 R 进制的某一位，然后组成一个 R 进制的整数。R 一般取 31，因为它是一个奇素数，如果是偶数的话，当出现乘法溢出，信息就会丢失，因为与 2 相乘相当于向左移一位。
 
 一个数与 31 相乘可以转换成移位和减法：`31*x == (x<<5)-x`，编译器会自动进行这个优化。
 
@@ -886,6 +888,7 @@ public class CloneExample implements Cloneable {
 
 ```java
 public class ShallowCloneExample implements Cloneable {
+
     private int[] arr;
 
     public ShallowCloneExample() {
@@ -928,6 +931,7 @@ System.out.println(e2.get(2)); // 222
 
 ```java
 public class DeepCloneExample implements Cloneable {
+
     private int[] arr;
 
     public DeepCloneExample() {
@@ -975,6 +979,7 @@ System.out.println(e2.get(2)); // 2
 
 ```java
 public class CloneConstructorExample {
+
     private int[] arr;
 
     public CloneConstructorExample() {
@@ -1040,7 +1045,7 @@ private 方法隐式地被指定为 final，如果在子类中定义的方法和
 
 **1. 静态变量** 
 
-- 静态变量：又称为类变量，也就是说这个变量属于类的，类所有的实例都共享静态变量，可以直接通过类名来访问它；静态变量在内存中只存在一份。
+- 静态变量：又称为类变量，也就是说这个变量属于类的，类所有的实例都共享静态变量，可以直接通过类名来访问它。静态变量在内存中只存在一份。
 - 实例变量：每创建一个实例就会产生一个实例变量，它与该实例同生共死。
 
 ```java
@@ -1059,7 +1064,7 @@ public class A {
 
 **2. 静态方法** 
 
-静态方法在类加载的时候就存在了，它不依赖于任何实例。所以静态方法必须有实现，也就是说它不能是抽象方法（abstract）。
+静态方法在类加载的时候就存在了，它不依赖于任何实例。所以静态方法必须有实现，也就是说它不能是抽象方法。
 
 ```java
 public abstract class A {
@@ -1182,7 +1187,7 @@ public InitialOrderTest() {
 
 每个类都有一个  **Class**  对象，包含了与类有关的信息。当编译一个新类时，会产生一个同名的 .class 文件，该文件内容保存着 Class 对象。
 
-类加载相当于 Class 对象的加载。类在第一次使用时才动态加载到 JVM 中，可以使用 `Class.forName("com.mysql.jdbc.Driver")` 这种方式来控制类的加载，该方法会返回一个 Class 对象。
+类加载相当于 Class 对象的加载，类在第一次使用时才动态加载到 JVM 中。也可以使用 `Class.forName("com.mysql.jdbc.Driver")` 这种方式来控制类的加载，该方法会返回一个 Class 对象。
 
 反射可以提供运行时的类信息，并且这个类可以在运行时才加载进来，甚至在编译时期该类的 .class 不存在也可以加载进来。
 
@@ -1279,7 +1284,7 @@ Java 注解是附加在代码中的一些元信息，用于一些工具在编译
 - Java 没有指针，它的引用可以理解为安全指针，而 C++ 具有和 C 一样的指针。
 - Java 支持自动垃圾回收，而 C++ 需要手动回收。
 - Java 不支持多重继承，只能通过实现多个接口来达到相同目的，而 C++ 支持多重继承。
-- Java 不支持操作符重载，虽然可以对两个 String 对象支持加法运算，但是这是语言内置支持的操作，不属于操作符重载，而 C++ 可以。
+- Java 不支持操作符重载，虽然可以对两个 String 对象执行加法运算，但是这是语言内置支持的操作，不属于操作符重载，而 C++ 可以。
 - Java 的 goto 是保留字，但是不可用，C++ 可以使用 goto。
 - Java 不支持条件编译，C++ 通过 #ifdef #ifndef 等预处理命令从而实现条件编译。
 

From 60e2aa5551b685ba0f9493cf92bbf46229298240 Mon Sep 17 00:00:00 2001
From: ziyitony <602232939@qq.com>
Date: Sun, 26 Aug 2018 23:52:20 +0800
Subject: [PATCH 150/172] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E7=A9=BA?=
 =?UTF-8?q?=E5=88=86=E5=A4=8D=E7=94=A8=E6=8A=80=E6=9C=AF=E7=9A=84=E4=BE=8B?=
 =?UTF-8?q?=E5=AD=90?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

1.3 小节增加了虚拟内存作为空分复用的例子
---
 notes/计算机操作系统.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 6bfb559c..5c6596b2 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -58,7 +58,7 @@
 
 虚拟技术把一个物理实体转换为多个逻辑实体。
 
-主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。
+主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个线程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个线程工作时使用物理内存，线程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
 
 ### 4. 异步
 

From 3dfc0c7cd8f72b44b238a8a91c7e5dfc92a32622 Mon Sep 17 00:00:00 2001
From: ziyitony <602232939@qq.com>
Date: Mon, 27 Aug 2018 00:00:01 +0800
Subject: [PATCH 151/172] =?UTF-8?q?Update=20=E8=AE=A1=E7=AE=97=E6=9C=BA?=
 =?UTF-8?q?=E6=93=8D=E4=BD=9C=E7=B3=BB=E7=BB=9F.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/计算机操作系统.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 5c6596b2..0be09349 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -58,7 +58,7 @@
 
 虚拟技术把一个物理实体转换为多个逻辑实体。
 
-主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个线程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个线程工作时使用物理内存，线程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
+主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个进程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个进程工作时使用物理内存，进程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
 
 ### 4. 异步
 

From 4e4c0c2da14680f69d4b28aa07899606980cd580 Mon Sep 17 00:00:00 2001
From: ziyitony <602232939@qq.com>
Date: Mon, 27 Aug 2018 00:00:14 +0800
Subject: [PATCH 152/172] =?UTF-8?q?Update=20=E8=AE=A1=E7=AE=97=E6=9C=BA?=
 =?UTF-8?q?=E6=93=8D=E4=BD=9C=E7=B3=BB=E7=BB=9F.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/计算机操作系统.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 5c6596b2..0be09349 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -58,7 +58,7 @@
 
 虚拟技术把一个物理实体转换为多个逻辑实体。
 
-主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个线程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个线程工作时使用物理内存，线程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
+主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个进程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个进程工作时使用物理内存，进程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
 
 ### 4. 异步
 

From 785f48ba52e2427456a8c419c9aeeaa5b3699aca Mon Sep 17 00:00:00 2001
From: YiliaZhang <532104074@qq.com>
Date: Mon, 27 Aug 2018 23:21:40 +0800
Subject: [PATCH 153/172] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87=20offer=20?=
 =?UTF-8?q?=E9=A2=98=E8=A7=A3.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

5. 替换空格   在for (int i = 0; i < str.length( ); i++)中，str还在一直append，for循环出不去，最终会报OOM
---
 notes/剑指 offer 题解.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index d88b6168..1162fa71 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -230,7 +230,7 @@ Output:
 ```java
 public String replaceSpace(StringBuffer str) {
     int P1 = str.length() - 1;
-    for (int i = 0; i < str.length(); i++)
+    for (int i = 0; i < P1+1; i++)
         if (str.charAt(i) == ' ')
             str.append("  ");
 

From 0fd224b3ab2b241cb3d4795a162cb99eac1242e9 Mon Sep 17 00:00:00 2001
From: MasterSJ <452857582@qq.com>
Date: Tue, 28 Aug 2018 13:31:55 +0800
Subject: [PATCH 154/172] =?UTF-8?q?=E5=BE=AA=E7=8E=AF=E4=B8=AD=E6=9F=A5?=
 =?UTF-8?q?=E6=89=BE=E7=9A=84=E7=A9=BA=E6=A0=BC=EF=BC=8C=E5=A6=82=E6=9E=9C?=
 =?UTF-8?q?=E5=A1=AB=E5=85=85=E7=9A=84=E4=B9=9F=E6=98=AF=E7=A9=BA=E6=A0=BC?=
 =?UTF-8?q?=E5=B0=B1=E6=AD=BB=E5=BE=AA=E7=8E=AF=E4=BA=86?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/剑指 offer 题解.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index d88b6168..43990090 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -232,7 +232,7 @@ public String replaceSpace(StringBuffer str) {
     int P1 = str.length() - 1;
     for (int i = 0; i < str.length(); i++)
         if (str.charAt(i) == ' ')
-            str.append("  ");
+            str.append("aa");
 
     int P2 = str.length() - 1;
     while (P1 >= 0 && P2 > P1) {

From 46f19d913c5e9c54e98a80e0ed805c30f7eb9513 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 28 Aug 2018 21:52:21 +0800
Subject: [PATCH 155/172] auto commit

---
 notes/Java IO.md         | 14 +++++++-------
 notes/MySQL.md           | 10 +++++-----
 notes/Redis.md           |  9 ++++-----
 notes/Socket.md          | 14 +++++++-------
 notes/剑指 offer 题解.md |  4 ++--
 notes/计算机操作系统.md  |  3 +--
 notes/计算机网络.md      | 26 ++++++++++++--------------
 7 files changed, 38 insertions(+), 42 deletions(-)

diff --git a/notes/Java IO.md b/notes/Java IO.md
index a14ea902..ee23a4bb 100644
--- a/notes/Java IO.md	
+++ b/notes/Java IO.md	
@@ -6,7 +6,7 @@
     * [装饰者模式](#装饰者模式)
 * [四、字符操作](#四字符操作)
     * [编码与解码](#编码与解码)
-    * [String](#string)
+    * [String 的编码方式](#string-的编码方式)
     * [Reader 与 Writer](#reader-与-writer)
     * [实现逐行输出文本文件的内容](#实现逐行输出文本文件的内容)
 * [五、对象操作](#五对象操作)
@@ -121,7 +121,7 @@ UTF-16be 中的 be 指的是 Big Endian，也就是大端。相应地也有 UTF-
 
 Java 使用双字节编码 UTF-16be，这不是指 Java 只支持这一种编码方式，而是说 char 这种类型使用 UTF-16be 进行编码。char 类型占 16 位，也就是两个字节，Java 使用这种双字节编码是为了让一个中文或者一个英文都能使用一个 char 来存储。
 
-## String
+## String 的编码方式
 
 String 可以看成一个字符序列，可以指定一个编码方式将它编码为字节序列，也可以指定一个编码方式将一个字节序列解码为 String。
 
@@ -371,7 +371,7 @@ public static void fastCopy(String src, String dist) throws IOException {
     /* 获取目标文件的输出字节流 */
     FileOutputStream fout = new FileOutputStream(dist);
 
-    /* 获取输出字节流的通道 */
+    /* 获取输出字节流的文件通道 */
     FileChannel fcout = fout.getChannel();
 
     /* 为缓冲区分配 1024 个字节 */
@@ -392,7 +392,7 @@ public static void fastCopy(String src, String dist) throws IOException {
 
         /* 把缓冲区的内容写入输出文件中 */
         fcout.write(buffer);
-        
+
         /* 清空缓冲区 */
         buffer.clear();
     }
@@ -407,7 +407,7 @@ NIO 实现了 IO 多路复用中的 Reactor 模型，一个线程 Thread 使用
 
 通过配置监听的通道 Channel 为非阻塞，那么当 Channel 上的 IO 事件还未到达时，就不会进入阻塞状态一直等待，而是继续轮询其它 Channel，找到 IO 事件已经到达的 Channel 执行。
 
-因为创建和切换线程的开销很大，因此使用一个线程来处理多个事件而不是一个线程处理一个事件具有更好的性能。
+因为创建和切换线程的开销很大，因此使用一个线程来处理多个事件而不是一个线程处理一个事件，对于 IO 密集型的应用具有很好地性能。
 
 应该注意的是，只有套接字 Channel 才能配置为非阻塞，而 FileChannel 不能，为 FileChannel 配置非阻塞也没有意义。
 
@@ -601,8 +601,8 @@ MappedByteBuffer mbb = fc.map(FileChannel.MapMode.READ_WRITE, 0, 1024);
 
 NIO 与普通 I/O 的区别主要有以下两点：
 
-- NIO 是非阻塞的
-- NIO 面向块，I/O 面向流
+- NIO 是非阻塞的；
+- NIO 面向块，I/O 面向流。
 
 # 八、参考资料
 
diff --git a/notes/MySQL.md b/notes/MySQL.md
index bf795d34..f115ce92 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -48,7 +48,7 @@ B+ Tree 是基于 B Tree 和叶子节点顺序访问指针进行实现，它具
 
 进行查找操作时，首先在根节点进行二分查找，找到一个 key 所在的指针，然后递归地在指针所指向的节点进行查找。直到查找到叶子节点，然后在叶子节点上进行二分查找，找出 key 所对应的 data。
 
-插入删除操作记录会破坏平衡树的平衡性，因此在插入删除操作之后，需要对树进行一个分裂、合并、旋转等操作来维护平衡性。
+插入删除操作会破坏平衡树的平衡性，因此在插入删除操作之后，需要对树进行一个分裂、合并、旋转等操作来维护平衡性。
 
 ### 3. 与红黑树的比较
 
@@ -233,7 +233,7 @@ do {
 
 ### 2. 分解大连接查询
 
-将一个大连接查询分解成对每一个表进行一次单表查询，然后将结果在应用程序中进行关联，这样做的好处有：
+将一个大连接查询分解成对每一个表进行一次单表查询，然后在应用程序中进行关联，这样做的好处有：
 
 - 让缓存更高效。对于连接查询，如果其中一个表发生变化，那么整个查询缓存就无法使用。而分解后的多个查询，即使其中一个表发生变化，对其它表的查询缓存依然可以使用。
 - 分解成多个单表查询，这些单表查询的缓存结果更可能被其它查询使用到，从而减少冗余记录的查询。
@@ -332,7 +332,7 @@ MySQL 提供了两种相似的日期时间类型：DATETIME 和 TIMESTAMP。
 
 ### 2. TIMESTAMP
 
-和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年 到 2038 年。
+和 UNIX 时间戳相同，保存从 1970 年 1 月 1 日午夜（格林威治时间）以来的秒数，使用 4 个字节，只能表示从 1970 年到 2038 年。
 
 它和时区有关，也就是说一个时间戳在不同的时区所代表的具体时间是不同的。
 
@@ -372,9 +372,9 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 
 使用分布式事务来解决，比如 XA 接口。
 
-### 2. 链接
+### 2. JOIN
 
-可以将原来的 JOIN 分解成多个单表查询，然后在用户程序中进行 JOIN。
+可以将原来的 JOIN 分解成多个单表 JOIN 查询，然后在用户程序中进行 JOIN。
 
 ### 3. ID 唯一性
 
diff --git a/notes/Redis.md b/notes/Redis.md
index 629fcdd8..58320dfa 100644
--- a/notes/Redis.md
+++ b/notes/Redis.md
@@ -49,7 +49,7 @@
 
 Redis 是速度非常快的非关系型（NoSQL）内存键值数据库，可以存储键和五种不同类型的值之间的映射。
 
-键的类型只能为字符串，值支持的五种类型数据类型为：字符串、列表、集合、散列表、有序集合。
+键的类型只能为字符串，值支持五种数据类型：字符串、列表、集合、散列表、有序集合。
 
 Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，使用复制来扩展读性能，使用分片来扩展写性能。
 
@@ -58,7 +58,7 @@ Redis 支持很多特性，例如将内存中的数据持久化到硬盘中，
 | 数据类型 | 可以存储的值 | 操作 |
 | :--: | :--: | :--: |
 | STRING | 字符串、整数或者浮点数 | 对整个字符串或者字符串的其中一部分执行操作</br> 对整数和浮点数执行自增或者自减操作 |
-| LIST | 列表 | 从两端压入或者弹出元素</br> 对单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
+| LIST | 列表 | 从两端压入或者弹出元素 </br> 对单个或者多个元素</br> 进行修剪，只保留一个范围内的元素 |
 | SET | 无序集合 | 添加、获取、移除单个元素</br> 检查一个元素是否存在于集合中</br> 计算交集、并集、差集</br> 从集合里面随机获取元素 |
 | HASH | 包含键值对的无序散列表 | 添加、获取、移除单个键值对</br> 获取所有键值对</br> 检查某个键是否存在|
 | ZSET | 有序集合 | 添加、获取、删除元素</br> 根据分值范围或者成员来获取元素</br> 计算一个键的排名 |
@@ -357,7 +357,7 @@ List 是一个双向链表，可以通过 lpop 和 lpush 写入和读取消息
 
 在分布式场景下具有多个应用服务器，可以使用 Redis 来统一存储这些应用服务器的会话信息。
 
-当应用服务器不再存储用户的会话信息，也就不再具有状态，一个用户可以请求任意一个应用服务器。
+当应用服务器不再存储用户的会话信息，也就不再具有状态，一个用户可以请求任意一个应用服务器，从而更容易实现高可用性以及可伸缩性。
 
 ## 分布式锁实现
 
@@ -393,7 +393,7 @@ Redis Cluster 实现了分布式的支持。
 
 - 在 Redis 中，并不是所有数据都一直存储在内存中，可以将一些很久没用的 value 交换到磁盘，而 Memcached 的数据则会一直在内存中。
 
-- Memcached 将内存分割成特定长度的块来存储数据，以完全解决内存碎片的问题，但是这种方式会使得内存的利用率不高，例如块的大小为 128 bytes，只存储 100 bytes 的数据，那么剩下的 28 bytes 就浪费掉了。
+- Memcached 将内存分割成特定长度的块来存储数据，以完全解决内存碎片的问题。但是这种方式会使得内存的利用率不高，例如块的大小为 128 bytes，只存储 100 bytes 的数据，那么剩下的 28 bytes 就浪费掉了。
 
 # 六、键的过期时间
 
@@ -605,5 +605,4 @@ Redis 没有关系型数据库中的表这一概念来将同种类型的数据
 - [论述 Redis 和 Memcached 的差异](http://www.cnblogs.com/loveincode/p/7411911.html)
 - [Redis 3.0 中文版- 分片](http://wiki.jikexueyuan.com/project/redis-guide)
 - [Redis 应用场景](http://www.scienjus.com/redis-use-case/)
-- [Observer vs Pub-Sub](http://developers-club.com/posts/270339/)
 - [Using Redis as an LRU cache](https://redis.io/topics/lru-cache)
diff --git a/notes/Socket.md b/notes/Socket.md
index 8e7c0b89..482bafe6 100644
--- a/notes/Socket.md
+++ b/notes/Socket.md
@@ -25,9 +25,9 @@
 - 等待数据准备好
 - 从内核向进程复制数据
 
-对于一个套接字上的输入操作，第一步通常涉及等待数据从网络中到达。当所等待分组到达时，它被复制到内核中的某个缓冲区。第二步就是把数据从内核缓冲区复制到应用进程缓冲区。
+对于一个套接字上的输入操作，第一步通常涉及等待数据从网络中到达。当所等待数据到达时，它被复制到内核中的某个缓冲区。第二步就是把数据从内核缓冲区复制到应用进程缓冲区。
 
-Unix 下有五种 I/O 模型：
+Unix 有五种 I/O 模型：
 
 - 阻塞式 I/O
 - 非阻塞式 I/O
@@ -39,7 +39,7 @@ Unix 下有五种 I/O 模型：
 
 应用进程被阻塞，直到数据复制到应用进程缓冲区中才返回。
 
-应该注意到，在阻塞的过程中，其它程序还可以执行，因此阻塞不意味着整个操作系统都被阻塞。因为其他程序还可以执行，因此不消耗 CPU 时间，这种模型的执行效率会比较高。
+应该注意到，在阻塞的过程中，其它程序还可以执行，因此阻塞不意味着整个操作系统都被阻塞。因为其他程序还可以执行，因此不消耗 CPU 时间，这种模型的 CPU 利用率效率会比较高。
 
 下图中，recvfrom 用于接收 Socket 传来的数据，并复制到应用进程的缓冲区 buf 中。这里把 recvfrom() 当成系统调用。
 
@@ -53,13 +53,13 @@ ssize_t recvfrom(int sockfd, void *buf, size_t len, int flags, struct sockaddr *
 
 应用进程执行系统调用之后，内核返回一个错误码。应用进程可以继续执行，但是需要不断的执行系统调用来获知 I/O 是否完成，这种方式称为轮询（polling）。
 
-由于 CPU 要处理更多的系统调用，因此这种模型是比较低效的。
+由于 CPU 要处理更多的系统调用，因此这种模型的 CPU 利用率是比较低的。
 
 <div align="center"> <img src="../pics//1492929000361_5.png"/> </div><br>
 
 ## I/O 复用
 
-使用 select 或者 poll 等待数据，并且可以等待多个套接字中的任何一个变为可读，这一过程会被阻塞，当某一个套接字可读时返回。之后再使用 recvfrom 把数据从内核复制到进程中。
+使用 select 或者 poll 等待数据，并且可以等待多个套接字中的任何一个变为可读。这一过程会被阻塞，当某一个套接字可读时返回，之后再使用 recvfrom 把数据从内核复制到进程中。
 
 它可以让单个进程具有处理多个 I/O 事件的能力。又被称为 Event Driven I/O，即事件驱动 I/O。
 
@@ -77,7 +77,7 @@ ssize_t recvfrom(int sockfd, void *buf, size_t len, int flags, struct sockaddr *
 
 ## 异步 I/O
 
-进行 aio_read 系统调用会立即返回，应用进程继续执行，不会被阻塞，内核会在所有操作完成之后向应用进程发送信号。
+应用进程执行 aio_read 系统调用会立即返回，应用进程可以继续执行，不会被阻塞，内核会在所有操作完成之后向应用进程发送信号。
 
 异步 I/O 与信号驱动 I/O 的区别在于，异步 I/O 的信号是通知应用进程 I/O 完成，而信号驱动 I/O 的信号是通知应用进程可以开始 I/O。
 
@@ -198,7 +198,7 @@ else
 select 和 poll 的功能基本相同，不过在一些实现细节上有所不同。
 
 - select 会修改描述符，而 poll 不会；
-- select 的描述符类型使用数组实现，FD_SETSIZE 大小默认为 1024，因此默认只能监听 1024 个描述符。如果要监听更多描述符的话，需要修改 FD_SETSIZE 之后重新编译；而 poll 的描述符类型使用链表实现，没有描述符的数量的限制；
+- select 的描述符类型使用数组实现，FD_SETSIZE 大小默认为 1024，因此默认只能监听 1024 个描述符。如果要监听更多描述符的话，需要修改 FD_SETSIZE 之后重新编译；而 poll 的描述符类型使用链表实现，没有描述符数量的限制；
 - poll 提供了更多的事件类型，并且对描述符的重复利用上比 select 高。
 - 如果一个线程对某个描述符调用了 select 或者 poll，另一个线程关闭了该描述符，会导致调用结果不确定。
 
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 83f03d8e..63cf170e 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -230,9 +230,9 @@ Output:
 ```java
 public String replaceSpace(StringBuffer str) {
     int P1 = str.length() - 1;
-    for (int i = 0; i < P1+1; i++)
+    for (int i = 0; i < P1 + 1; i++)
         if (str.charAt(i) == ' ')
-            str.append("aa");
+            str.append("  ");
 
     int P2 = str.length() - 1;
     while (P1 >= 0 && P2 > P1) {
diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 6bfb559c..b3b0a41b 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -791,12 +791,11 @@ FIFO 算法可能会把经常使用的页面置换出去，为了避免这一问
 
 <div align="center"> <img src="../pics//ecf8ad5d-5403-48b9-b6e7-f2e20ffe8fca.png"/> </div><br>
 
-
 ### 6. 时钟
 
 > Clock
 
-第二次机会算法需要在链表中移动页面，降低了效率。时钟算法使用环形链表将页面链接起来，再使用一个指针指向最老的页面。
+第二次机会算法需要在链表中移动页面，降低了效率。时钟算法使用环形链表将页面连接起来，再使用一个指针指向最老的页面。
 
 <div align="center"> <img src="../pics//5f5ef0b6-98ea-497c-a007-f6c55288eab1.png"/> </div><br>
 
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index e80ba7f8..99a6fb43 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -59,7 +59,7 @@
 
 网络把主机连接起来，而互联网是把多种不同的网络连接起来，因此互联网是网络的网络。
 
-<div align="center"> <img src="../pics//network-of-networks.gif" width="500"/> </div><br>
+<div align="center"> <img src="../pics//network-of-networks.gif" width="400"/> </div><br>
 
 ## ISP
 
@@ -67,9 +67,7 @@
 
 <div align="center"> <img src="../pics//46cec213-3048-4a80-aded-fdd577542801.jpg" width="500"/> </div><br>
 
-目前的互联网是一种多层次 ISP 结构，ISP 根据覆盖面积的大小分为第一层 ISP、区域 ISP 和接入 ISP。
-
-互联网交换点 IXP 允许两个 ISP 直接相连而不用经过第三个 ISP。
+目前的互联网是一种多层次 ISP 结构，ISP 根据覆盖面积的大小分为第一层 ISP、区域 ISP 和接入 ISP。互联网交换点 IXP 允许两个 ISP 直接相连而不用经过第三个 ISP。
 
 <div align="center"> <img src="../pics//168e893c-e4a0-4ba4-b81f-9d993483abd0.jpg" width="500"/> </div><br>
 
@@ -125,7 +123,7 @@
 
 ## 计算机网络体系结构*
 
-<div align="center"> <img src="../pics//426df589-6f97-4622-b74d-4a81fcb1da8e.png" width="800"/> </div><br>
+<div align="center"> <img src="../pics//426df589-6f97-4622-b74d-4a81fcb1da8e.png" width="600"/> </div><br>
 
 ### 1. 五层协议
 
@@ -143,7 +141,7 @@
 
 其中表示层和会话层用途如下：
 
--  **表示层** ：数据压缩、加密以及数据描述，这使得应用程序不必担心在各台主机中数据内部格式不同的问题。
+-  **表示层** ：数据压缩、加密以及数据描述，这使得应用程序不必关心在各台主机中数据内部格式不同的问题。
 
 -  **会话层** ：建立及管理会话。
 
@@ -700,7 +698,7 @@ TCP 使用超时重传来实现可靠传输：如果一个已经发送的报文
 
 <div align="center"> <img src="../pics//51e2ed95-65b8-4ae9-8af3-65602d452a25.jpg" width="500"/> </div><br>
 
-TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、快重传、快恢复。
+TCP 主要通过四个算法来进行拥塞控制：慢开始、拥塞避免、快重传、快恢复。
 
 发送方需要维护一个叫做拥塞窗口（cwnd）的状态变量，注意拥塞窗口与发送方窗口的区别：拥塞窗口只是一个状态变量，实际决定发送方能发送多少数据的是发送方窗口。
 
@@ -713,11 +711,11 @@ TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、
 
 ### 1. 慢开始与拥塞避免
 
-发送的最初执行慢开始，令 cwnd=1，发送方只能发送 1 个报文段；当收到确认后，将 cwnd 加倍，因此之后发送方能够发送的报文段数量为：2、4、8 ...
+发送的最初执行慢开始，令 cwnd = 1，发送方只能发送 1 个报文段；当收到确认后，将 cwnd 加倍，因此之后发送方能够发送的报文段数量为：2、4、8 ...
 
-注意到慢开始每个轮次都将 cwnd 加倍，这样会让 cwnd 增长速度非常快，从而使得发送方发送的速度增长速度过快，网络拥塞的可能也就更高。设置一个慢开始门限 ssthresh，当 cwnd >= ssthresh 时，进入拥塞避免，每个轮次只将 cwnd 加 1。
+注意到慢开始每个轮次都将 cwnd 加倍，这样会让 cwnd 增长速度非常快，从而使得发送方发送的速度增长速度过快，网络拥塞的可能性也就更高。设置一个慢开始门限 ssthresh，当 cwnd >= ssthresh 时，进入拥塞避免，每个轮次只将 cwnd 加 1。
 
-如果出现了超时，则令 ssthresh = cwnd/2，然后重新执行慢开始。
+如果出现了超时，则令 ssthresh = cwnd / 2，然后重新执行慢开始。
 
 ### 2. 快重传与快恢复
 
@@ -725,7 +723,7 @@ TCP 主要通过四种算法来进行拥塞控制：慢开始、拥塞避免、
 
 在发送方，如果收到三个重复确认，那么可以知道下一个报文段丢失，此时执行快重传，立即重传下一个报文段。例如收到三个 M<sub>2</sub>，则 M<sub>3</sub> 丢失，立即重传 M<sub>3</sub>。
 
-在这种情况下，只是丢失个别报文段，而不是网络拥塞。因此执行快恢复，令 ssthresh = cwnd/2 ，cwnd = ssthresh，注意到此时直接进入拥塞避免。
+在这种情况下，只是丢失个别报文段，而不是网络拥塞。因此执行快恢复，令 ssthresh = cwnd / 2 ，cwnd = ssthresh，注意到此时直接进入拥塞避免。
 
 慢开始和快恢复的快慢指的是 cwnd 的设定值，而不是 cwnd 的增长速率。慢开始 cwnd 设定为 1，而快恢复 cwnd 设定为 ssthresh。
 
@@ -743,8 +741,8 @@ DNS 是一个分布式数据库，提供了主机名和 IP 地址之间相互转
 
 DNS 可以使用 UDP 或者 TCP 进行传输，使用的端口号都为 53。大多数情况下 DNS 使用 UDP 进行传输，这就要求域名解析器和域名服务器都必须自己处理超时和重传来保证可靠性。在两种情况下会使用 TCP 进行传输：
 
-- 如果返回的响应超过的 512 字节就改用 TCP 进行传输（UDP 最大只支持 512 字节的数据）。
-- 区域传送需要使用 TCP 进行传输（区域传送是主域名服务器向辅助域名服务器传送变化的那部分数据）。
+- 如果返回的响应超过的 512 字节（UDP 最大只支持 512 字节的数据）。
+- 区域传送（区域传送是主域名服务器向辅助域名服务器传送变化的那部分数据）。
 
 ## 文件传送协议
 
@@ -806,7 +804,7 @@ POP3 的特点是只要用户从服务器上读取了邮件，就把该邮件删
 
 ### 3. IMAP
 
-IMAP 协议中客户端和服务器上的邮件保持同步，如果不去手动删除邮件，那么服务器上的邮件也不会被删除。IMAP 这种做法可以让用户随时随地去访问服务器上的邮件。
+IMAP 协议中客户端和服务器上的邮件保持同步，如果不手动删除邮件，那么服务器上的邮件也不会被删除。IMAP 这种做法可以让用户随时随地去访问服务器上的邮件。
 
 ## 常用端口
 

From 03bb846a22ab22acd608af2e27d0063c44ae1a7a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 28 Aug 2018 22:02:30 +0800
Subject: [PATCH 156/172] auto commit

---
 notes/Java 容器.md      | 2 +-
 notes/计算机操作系统.md | 6 +++++-
 2 files changed, 6 insertions(+), 2 deletions(-)

diff --git a/notes/Java 容器.md b/notes/Java 容器.md
index 57e5f475..fd1ee97e 100644
--- a/notes/Java 容器.md	
+++ b/notes/Java 容器.md	
@@ -1182,7 +1182,7 @@ ListIterator <-- List
 - Eckel B. Java 编程思想 [M]. 机械工业出版社, 2002.
 - [Java Collection Framework](https://www.w3resource.com/java-tutorial/java-collections.php)
 - [Iterator 模式](https://openhome.cc/Gossip/DesignPattern/IteratorPattern.htm)
-- [Java 8 系列之重新认识 HashMap](https://tech.meituan.com/java-hashmap.html)
+- [Java 8 系列之重新认识 HashMap](https://tech.meituan.com/java_hashmap.html)
 - [What is difference between HashMap and Hashtable in Java?](http://javarevisited.blogspot.hk/2010/10/difference-between-hashmap-and.html)
 - [Java 集合之 HashMap](http://www.zhangchangle.com/2018/02/07/Java%E9%9B%86%E5%90%88%E4%B9%8BHashMap/)
 - [The principle of ConcurrentHashMap analysis](http://www.programering.com/a/MDO3QDNwATM.html)
diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 4154cf66..73f67c62 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -58,7 +58,11 @@
 
 虚拟技术把一个物理实体转换为多个逻辑实体。
 
-主要有两种虚拟技术：时分复用技术和空分复用技术。例如多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。例如多个进程占用的总内存可以超过电脑实际安装的物理内存，但是采用虚拟内存 swap 和页面调度等技术，让每个进程工作时使用物理内存，进程切换的时候将内存内容写入到虚拟内存保存，通过工作在物理内存保存在虚拟内存的方式实现空分复用技术。
+主要有两种虚拟技术：时分复用技术和空分复用技术。
+
+多个进程能在同一个处理器上并发执行使用了时分复用技术，让每个进程轮流占有处理器，每次只执行一小个时间片并快速切换。
+
+虚拟内存使用了空分复用技术，它将物理内存抽象为地址空间，每个进程都有各自的地址空间。地址空间和物理内存使用页进行交换，地址空间的页并不需要全部在物理内存中，当使用到一个没有在物理内存的页时，执行页面置换算法，将该页置换到内存中。
 
 ### 4. 异步
 

From 06190f2f1947f4421234ea7912c51fb07d5c79a9 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Tue, 28 Aug 2018 22:55:26 +0800
Subject: [PATCH 157/172] auto commit

---
 notes/算法.md | 78 +++++++++++++++++++++++++++++++++++++--------------
 1 file changed, 57 insertions(+), 21 deletions(-)

diff --git a/notes/算法.md b/notes/算法.md
index a9f9e689..53edc1c1 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -104,17 +104,21 @@ public class ThreeSumSlow implements ThreeSum {
     public int count(int[] nums) {
         int N = nums.length;
         int cnt = 0;
-        for (int i = 0; i < N; i++)
-            for (int j = i + 1; j < N; j++)
-                for (int k = j + 1; k < N; k++)
-                    if (nums[i] + nums[j] + nums[k] == 0)
+        for (int i = 0; i < N; i++) {
+            for (int j = i + 1; j < N; j++) {
+                for (int k = j + 1; k < N; k++) {
+                    if (nums[i] + nums[j] + nums[k] == 0) {
                         cnt++;
+                    }
+                }
+            }
+        }
         return cnt;
     }
 }
 ```
 
-### 2. ThreeSumFast
+### 2. ThreeSumBinarySearch
 
 通过将数组先排序，对两个元素求和，并用二分查找方法查找是否存在该和的相反数，如果存在，就说明存在三元组的和为 0。
 
@@ -123,19 +127,23 @@ public class ThreeSumSlow implements ThreeSum {
 该方法可以将 ThreeSum 算法增长数量级降低为 O(N<sup>2</sup>logN)。
 
 ```java
-public class ThreeSumFast {
-    public static int count(int[] nums) {
+public class ThreeSumBinarySearch implements ThreeSum {
+
+    @Override
+    public int count(int[] nums) {
         Arrays.sort(nums);
         int N = nums.length;
         int cnt = 0;
-        for (int i = 0; i < N; i++)
+        for (int i = 0; i < N; i++) {
             for (int j = i + 1; j < N; j++) {
                 int target = -nums[i] - nums[j];
                 int index = BinarySearch.search(nums, target);
                 // 应该注意这里的下标必须大于 j，否则会重复统计。
-                if (index > j)
+                if (index > j) {
                     cnt++;
+                }
             }
+        }
         return cnt;
     }
 }
@@ -143,22 +151,59 @@ public class ThreeSumFast {
 
 ```java
 public class BinarySearch {
+
     public static int search(int[] nums, int target) {
         int l = 0, h = nums.length - 1;
         while (l <= h) {
             int m = l + (h - l) / 2;
-            if (target == nums[m])
+            if (target == nums[m]) {
                 return m;
-            else if (target > nums[m])
+            } else if (target > nums[m]) {
                 l = m + 1;
-            else
+            } else {
                 h = m - 1;
+            }
         }
         return -1;
     }
 }
 ```
 
+### 3. ThreeSumTwoPointer
+
+更有效的方法是先将数组排序，然后使用双指针进行查找，时间复杂度为 O(N<sup>2</sup>)。
+
+```java
+public class ThreeSumTwoPointer implements ThreeSum {
+
+    @Override
+    public int count(int[] nums) {
+        int N = nums.length;
+        int cnt = 0;
+        Arrays.sort(nums);
+        for (int i = 0; i < N - 2; i++) {
+            int l = i + 1, h = N - 1, target = -nums[i];
+            if (i > 0 && nums[i] == nums[i - 1]) continue;
+            while (l < h) {
+                int sum = nums[l] + nums[h];
+                if (sum == target) {
+                    cnt++;
+                    while (l < h && nums[l] == nums[l + 1]) l++;
+                    while (l < h && nums[h] == nums[h - 1]) h--;
+                    l++;
+                    h--;
+                } else if (sum < target) {
+                    l++;
+                } else {
+                    h--;
+                }
+            }
+        }
+        return cnt;
+    }
+}
+```
+
 ## 倍率实验
 
 如果 T(N) \~ aN<sup>b</sup>logN，那么 T(2N)/T(N) \~ 2<sup>b</sup>。
@@ -180,29 +225,20 @@ public class BinarySearch {
 public class RatioTest {
 
     public static void main(String[] args) {
-
         int N = 500;
         int loopTimes = 7;
         double preTime = -1;
-
         while (loopTimes-- > 0) {
-
             int[] nums = new int[N];
-
             StopWatch.start();
-
             ThreeSum threeSum = new ThreeSumSlow();
-
             int cnt = threeSum.count(nums);
             System.out.println(cnt);
-
             double elapsedTime = StopWatch.elapsedTime();
             double ratio = preTime == -1 ? 0 : elapsedTime / preTime;
             System.out.println(N + "  " + elapsedTime + "  " + ratio);
-
             preTime = elapsedTime;
             N *= 2;
-
         }
     }
 }

From 5426af758eff2c61373154c7b888eab5d70556cd Mon Sep 17 00:00:00 2001
From: Zhiwei Tian <hebeitianzhiwei@outlook.com>
Date: Tue, 28 Aug 2018 23:22:13 +0800
Subject: [PATCH 158/172] mistakes

---
 notes/Java 虚拟机.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 2b565cb5..1a1adc5a 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -268,7 +268,7 @@ HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了
 以上是 HotSpot 虚拟机中的 7 个垃圾收集器，连线表示垃圾收集器可以配合使用。
 
 - 单线程与多线程：单线程指的是垃圾收集器只使用一个线程进行收集，而多线程使用多个线程；
-- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并形指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
+- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并行指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
 
 ### 1. Serial 收集器
 

From 58f11f78a85d80a28aac5b6e769027c86b51ea89 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <zhengyc101@163.com>
Date: Wed, 29 Aug 2018 10:31:07 +0800
Subject: [PATCH 159/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 27edaf24..b9f43ec8 100644
--- a/README.md
+++ b/README.md
@@ -4,7 +4,7 @@
 
 <br>
 <div align="center">
-    <img src="other/LogoMakr_0zpEzN.png" width="150px">
+    <img src="other/LogoMakr_0zpEzN.png" width="200px">
     <br>
     <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>

From 90e46ab379964e38a9e16b8ee08493ff9ede383c Mon Sep 17 00:00:00 2001
From: showCodes <22638002+showCodes@users.noreply.github.com>
Date: Thu, 30 Aug 2018 11:02:33 +0800
Subject: [PATCH 160/172] =?UTF-8?q?Update=20Java=20=E8=99=9A=E6=8B=9F?=
 =?UTF-8?q?=E6=9C=BA.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 notes/Java 虚拟机.md | 26 +++++++++++++++++++++++++-
 1 file changed, 25 insertions(+), 1 deletion(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 1a1adc5a..40972bfd 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -7,6 +7,8 @@
     * [方法区](#方法区)
     * [运行时常量池](#运行时常量池)
     * [直接内存](#直接内存)
+    * [类的创建过程](#类的创建过程)
+    * [对象的结构](#对象的结构)
 * [二、垃圾收集](#二垃圾收集)
     * [判断一个对象是否可被回收](#判断一个对象是否可被回收)
     * [引用类型](#引用类型)
@@ -102,6 +104,28 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
 
+## 对象的创建过程
+
+[图解JAVA对象的创建过程](https://www.cnblogs.com/chenyangyao/p/5296807.html)
+
+![111](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235423381-1926278401.png)
+
+## 对象的结构
+
+对象包含 3 个部分：
+1. Header(对象头)：如下图所示；
+2. InstanceData(实例数据)：将等宽的类型放在一起；
+3. Padding(对齐填充)：Hotspot 虚拟机的内存管理系统要求对象的起始地址要是 8 个字节的整数倍，而对象头就是 8 个字节的整数倍。padding 是用来填充实例数据不足 8 个字节整数倍的部分，可以理解为占位符。
+
+![112](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235633553-1910724119.png)
+
+## 对象的地址访问
+
+1. 直接指针：对象的引用直接指向堆中的内存地址 (Hotspot 采用的方式)；
+2. 使用句柄：Java 堆中划分出一块内存作为句柄池，对象的引用指向句柄池。
+
+![113](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235555303-769658219.jpg)
+
 # 二、垃圾收集
 
 垃圾收集主要是针对堆和方法区进行。
@@ -138,7 +162,7 @@ public class ReferenceCountingGC {
 
 Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC Roots 一般包含以下内容：
 
-- 虚拟机栈中引用的对象
+- 虚拟机栈 (局部变量表) 中引用的对象
 - 本地方法栈中引用的对象
 - 方法区中类静态属性引用的对象
 - 方法区中的常量引用的对象

From 3edd3984f48e76a22a95add26f9f29f8b232926b Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 30 Aug 2018 20:25:57 +0800
Subject: [PATCH 161/172] auto commit

---
 notes/Docker.md       | 32 +++++++-------------
 notes/Git.md          | 32 ++++++++++----------
 notes/HTTP.md         | 46 ++++++++++++++--------------
 notes/Java 并发.md    | 33 ++++++++++----------
 notes/Java 虚拟机.md  | 30 ++-----------------
 notes/Linux.md        | 10 +++----
 notes/分布式.md       | 26 ++++++++--------
 notes/构建工具.md     | 29 +++++++-----------
 notes/消息队列.md     |  2 +-
 notes/算法.md         | 70 +++++++++++++++++++++----------------------
 notes/缓存.md         | 26 ++++++++--------
 notes/计算机网络.md   |  6 ++--
 notes/设计模式.md     | 13 +++++---
 notes/重构.md         |  6 ++--
 notes/集群.md         |  6 ++--
 notes/面向对象思想.md |  5 +++-
 16 files changed, 169 insertions(+), 203 deletions(-)

diff --git a/notes/Docker.md b/notes/Docker.md
index 80108896..1b61a7fe 100644
--- a/notes/Docker.md
+++ b/notes/Docker.md
@@ -4,6 +4,7 @@
 * [三、优势](#三优势)
 * [四、使用场景](#四使用场景)
 * [五、镜像与容器](#五镜像与容器)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -15,11 +16,6 @@
 
 Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程进行隔离，被隔离的进程独立于宿主操作系统和其它隔离的进程。使用 Docker 可以不修改应用程序代码，不需要开发人员学习特定环境下的技术，就能够将现有的应用程序部署在其他机器中。
 
-参考资料：
-
-- [DOCKER 101: INTRODUCTION TO DOCKER WEBINAR RECAP](https://blog.docker.com/2017/08/docker-101-introduction-docker-webinar-recap/)
-- [Docker 入门教程](http://www.ruanyifeng.com/blog/2018/02/docker-tutorial.html)
-
 # 二、与虚拟机的比较
 
 虚拟机也是一种虚拟化技术，它与 Docker 最大的区别在于它是通过模拟硬件，并在硬件上安装操作系统来实现。
@@ -40,30 +36,22 @@ Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程
 
 而 Docker 只是一个进程，只需要将应用以及相关的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
 
-参考资料：
-
-- [Docker container vs Virtual machine](http://www.bogotobogo.com/DevOps/Docker/Docker_Container_vs_Virtual_Machine.php)
-
 # 三、优势
 
 除了启动速度快以及占用资源少之外，Docker 具有以下优势：
 
 ## 更容易迁移
 
-Docker 可以提供一致性的运行环境，可以在不同的机器上进行迁移，而不用担心环境变化导致无法运行。
+提供一致性的运行环境，可以在不同的机器上进行迁移，而不用担心环境变化导致无法运行。
 
 ## 更容易维护
 
-Docker 使用分层技术和镜像，使得应用可以更容易复用重复部分。复用程度越高，维护工作也越容易。
+使用分层技术和镜像，使得应用可以更容易复用重复部分。复用程度越高，维护工作也越容易。
 
 ## 更容易扩展
 
 可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像可以非常容易得到我们想要的镜像。
 
-参考资料：
-
-- [为什么要使用 Docker？](https://yeasy.gitbooks.io/docker_practice/introduction/why.html)
-
 # 四、使用场景
 
 ## 持续集成
@@ -80,11 +68,6 @@ Docker 具有轻量级以及隔离性的特点，在将代码集成到一个 Doc
 
 Docker 轻量级的特点使得它很适合用于部署、维护、组合微服务。
 
-参考资料：
-
-- [What is Docker](https://www.docker.com/what-docker)
-- [持续集成是什么？](http://www.ruanyifeng.com/blog/2015/09/continuous-integration.html)
-
 # 五、镜像与容器
 
 镜像是一种静态的结构，可以看成面向对象里面的类，而容器是镜像的一个实例。
@@ -95,9 +78,14 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 
 <div align="center"> <img src="../pics//docker-filesystems-busyboxrw.png"/> </div><br>
 
-参考资料：
+# 参考资料
 
+- [DOCKER 101: INTRODUCTION TO DOCKER WEBINAR RECAP](https://blog.docker.com/2017/08/docker-101-introduction-docker-webinar-recap/)
+- [Docker 入门教程](http://www.ruanyifeng.com/blog/2018/02/docker-tutorial.html)
+- [Docker container vs Virtual machine](http://www.bogotobogo.com/DevOps/Docker/Docker_Container_vs_Virtual_Machine.php)
 - [How to Create Docker Container using Dockerfile](https://linoxide.com/linux-how-to/dockerfile-create-docker-container/)
 - [理解 Docker（2）：Docker 镜像](http://www.cnblogs.com/sammyliu/p/5877964.html)
-
+- [为什么要使用 Docker？](https://yeasy.gitbooks.io/docker_practice/introduction/why.html)
+- [What is Docker](https://www.docker.com/what-docker)
+- [持续集成是什么？](http://www.ruanyifeng.com/blog/2015/09/continuous-integration.html)
 
diff --git a/notes/Git.md b/notes/Git.md
index 4bb0a543..71ff502c 100644
--- a/notes/Git.md
+++ b/notes/Git.md
@@ -1,8 +1,7 @@
 <!-- GFM-TOC -->
-* [学习资料](#学习资料)
 * [集中式与分布式](#集中式与分布式)
-* [Git 的中心服务器](#git-的中心服务器)
-* [Git 工作流](#git-工作流)
+* [中心服务器](#中心服务器)
+* [工作流](#工作流)
 * [分支实现](#分支实现)
 * [冲突](#冲突)
 * [Fast forward](#fast-forward)
@@ -11,16 +10,10 @@
 * [SSH 传输设置](#ssh-传输设置)
 * [.gitignore 文件](#gitignore-文件)
 * [Git 命令一览](#git-命令一览)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 学习资料
-
-- [Git - 简明指南](http://rogerdudler.github.io/git-guide/index.zh.html)
-- [图解 Git](http://marklodato.github.io/visual-git-guide/index-zh-cn.html)
-- [廖雪峰 : Git 教程](https://www.liaoxuefeng.com/wiki/0013739516305929606dd18361248578c67b8067c8c017b000)
-- [Learn Git Branching](https://learngitbranching.js.org/)
-
 # 集中式与分布式
 
 Git 属于分布式版本控制系统，而 SVN 属于集中式。
@@ -33,11 +26,13 @@ Git 属于分布式版本控制系统，而 SVN 属于集中式。
 
 分布式版本控制新建分支、合并分支操作速度非常快，而集中式版本控制新建一个分支相当于复制一份完整代码。
 
-# Git 的中心服务器
+# 中心服务器
 
-Git 的中心服务器用来交换每个用户的修改。没有中心服务器也能工作，但是中心服务器能够 24 小时保持开机状态，这样就能更方便的交换修改。Github 就是一种 Git 中心服务器。
+中心服务器用来交换每个用户的修改，没有中心服务器也能工作，但是中心服务器能够 24 小时保持开机状态，这样就能更方便的交换修改。
 
-# Git 工作流
+Github 就是一个中心服务器。
+
+# 工作流
 
 <div align="center"> <img src="../pics//a1198642-9159-4d88-8aec-c3b04e7a2563.jpg"/> </div><br>
 
@@ -54,14 +49,14 @@ Git 版本库有一个称为 stage 的暂存区，还有自动创建的 master 
 
 <div align="center"> <img src="../pics//17976404-95f5-480e-9cb4-250e6aa1d55f.png"/> </div><br>
 
-可以跳过暂存区域直接从分支中取出修改或者直接提交修改到分支中
+可以跳过暂存区域直接从分支中取出修改，或者直接提交修改到分支中。
 
 - git commit -a 直接把所有文件的修改添加到暂缓区然后执行提交
 - git checkout HEAD -- files 取出最后一次修改，可以用来进行回滚操作
 
 # 分支实现
 
-Git 把每次提交都连成一条时间线。分支使用指针来实现，例如 master 分支指针指向时间线的最后一个节点，也就是最后一次提交。HEAD 指针指向的是当前分支。
+使用指针将每个提交连接成一条时间线，HEAD 指针指向当前分支指针。
 
 <div align="center"> <img src="../pics//fb546e12-e1fb-4b72-a1fb-8a7f5000dce6.jpg"/> </div><br>
 
@@ -69,7 +64,7 @@ Git 把每次提交都连成一条时间线。分支使用指针来实现，例
 
 <div align="center"> <img src="../pics//bc775758-89ab-4805-9f9c-78b8739cf780.jpg"/> </div><br>
 
-每次提交只会让当前分支向前移动，而其它分支不会移动。
+每次提交只会让当前分支指针向前移动，而其它分支指针不会移动。
 
 <div align="center"> <img src="../pics//5292faa6-0141-4638-bf0f-bb95b081dcba.jpg"/> </div><br>
 
@@ -155,4 +150,9 @@ $ ssh-keygen -t rsa -C "youremail@example.com"
 
 比较详细的地址：http://www.cheat-sheets.org/saved-copy/git-cheat-sheet.pdf
 
+# 参考资料
 
+- [Git - 简明指南](http://rogerdudler.github.io/git-guide/index.zh.html)
+- [图解 Git](http://marklodato.github.io/visual-git-guide/index-zh-cn.html)
+- [廖雪峰 : Git 教程](https://www.liaoxuefeng.com/wiki/0013739516305929606dd18361248578c67b8067c8c017b000)
+- [Learn Git Branching](https://learngitbranching.js.org/)
diff --git a/notes/HTTP.md b/notes/HTTP.md
index 537b24fe..aae807e6 100644
--- a/notes/HTTP.md
+++ b/notes/HTTP.md
@@ -25,6 +25,7 @@
     * [实体首部字段](#实体首部字段)
 * [五、具体应用](#五具体应用)
     * [Cookie](#cookie)
+    * [6. Secure](#6-secure)
     * [缓存](#缓存)
     * [连接管理](#连接管理)
     * [内容协商](#内容协商)
@@ -310,7 +311,7 @@ HTTP 协议是无状态的，主要是为了让 HTTP 协议尽可能简单，使
 
 Cookie 是服务器发送到用户浏览器并保存在本地的一小块数据，它会在浏览器之后向同一服务器再次发起请求时被携带上，用于告知服务端两个请求是否来自同一浏览器。由于之后每次请求都会需要携带 Cookie 数据，因此会带来额外的性能开销（尤其是在移动环境下）。
 
-Cookie 曾一度用于客户端数据的存储，因为当时并没有其它合适的存储办法而作为唯一的存储手段，但现在随着现代浏览器开始支持各种各样的存储方式，Cookie 渐渐被淘汰。新的浏览器 API 已经允许开发者直接将数据存储到本地，如使用 Web storage API （本地存储和会话存储）或 IndexedDB。
+Cookie 曾一度用于客户端数据的存储，因为当时并没有其它合适的存储办法而作为唯一的存储手段，但现在随着现代浏览器开始支持各种各样的存储方式，Cookie 渐渐被淘汰。新的浏览器 API 已经允许开发者直接将数据存储到本地，如使用 Web storage API（本地存储和会话存储）或 IndexedDB。
 
 ### 1. 用途
 
@@ -348,7 +349,17 @@ Cookie: yummy_cookie=choco; tasty_cookie=strawberry
 Set-Cookie: id=a3fWa; Expires=Wed, 21 Oct 2015 07:28:00 GMT;
 ```
 
-### 4. JavaScript 获取 Cookie
+### 4. 作用域
+
+Domain 标识指定了哪些主机可以接受 Cookie。如果不指定，默认为当前文档的主机（不包含子域名）。如果指定了 Domain，则一般包含子域名。例如，如果设置 Domain=mozilla.org，则 Cookie 也包含在子域名中（如 developer.mozilla.org）。
+
+Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径必须存在于请求 URL 中）。以字符 %x2F ("/") 作为路径分隔符，子路径也会被匹配。例如，设置 Path=/docs，则以下地址都会匹配：
+
+- /docs
+- /docs/Web/
+- /docs/Web/HTTP
+
+### 5. JavaScript
 
 通过 `Document.cookie` 属性可创建新的 Cookie，也可通过该属性访问非 HttpOnly 标记的 Cookie。
 
@@ -358,9 +369,7 @@ document.cookie = "tasty_cookie=strawberry";
 console.log(document.cookie);
 ```
 
-### 5. Secure 和 HttpOnly
-
-标记为 Secure 的 Cookie 只能通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
+### 6.   HttpOnly
 
 标记为 HttpOnly 的 Cookie 不能被 JavaScript 脚本调用。跨站脚本攻击 (XSS) 常常使用 JavaScript 的 `Document.cookie` API 窃取用户的 Cookie 信息，因此使用 HttpOnly 标记可以在一定程度上避免 XSS 攻击。
 
@@ -368,15 +377,9 @@ console.log(document.cookie);
 Set-Cookie: id=a3fWa; Expires=Wed, 21 Oct 2015 07:28:00 GMT; Secure; HttpOnly
 ```
 
-### 6. 作用域
+## 6. Secure
 
-Domain 标识指定了哪些主机可以接受 Cookie。如果不指定，默认为当前文档的主机（不包含子域名）。如果指定了 Domain，则一般包含子域名。例如，如果设置 Domain=mozilla.org，则 Cookie 也包含在子域名中（如 developer.mozilla.org）。
-
-Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径必须存在于请求 URL 中）。以字符 %x2F ("/") 作为路径分隔符，子路径也会被匹配。例如，设置 Path=/docs，则以下地址都会匹配：
-
-- /docs
-- /docs/Web/
-- /docs/Web/HTTP
+标记为 Secure 的 Cookie 只能通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
 
 ### 7. Session
 
@@ -387,8 +390,7 @@ Session 可以存储在服务器上的文件、数据库或者内存中。也可
 使用 Session 维护用户登录状态的过程如下：
 
 - 用户进行登录时，用户提交包含用户名和密码的表单，放入 HTTP 请求报文中；
-- 服务器验证该用户名和密码；
-- 如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 Key 称为 Session ID；
+- 服务器验证该用户名和密码，如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 Key 称为 Session ID；
 - 服务器返回的响应报文的 Set-Cookie 首部字段包含了这个 Session ID，客户端收到响应报文之后将该 Cookie 值存入浏览器中；
 - 客户端之后对同一个服务器进行请求时会包含该 Cookie 值，服务器收到之后提取出 Session ID，从 Redis 中取出用户信息，继续之前的业务操作。
 
@@ -462,13 +464,13 @@ Cache-Control: max-age=31536000
 
 Expires 首部字段也可以用于告知缓存服务器该资源什么时候会过期。
 
-- 在 HTTP/1.1 中，会优先处理 max-age 指令；
-- 在 HTTP/1.0 中，max-age 指令会被忽略掉。
-
 ```html
 Expires: Wed, 04 Jul 2012 08:26:05 GMT
 ```
 
+- 在 HTTP/1.1 中，会优先处理 max-age 指令；
+- 在 HTTP/1.0 中，max-age 指令会被忽略掉。
+
 ### 4. 缓存验证
 
 需要先了解 ETag 首部字段的含义，它是资源的唯一标识。URL 不能唯一表示资源，例如 `http://www.google.com/` 有中文和英文两个资源，只有 ETag 才能对这两个资源进行唯一标识。
@@ -727,7 +729,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 
 ## HTTP/1.x 缺陷
 
- HTTP/1.x 实现简单是以牺牲性能为代价的：
+HTTP/1.x 实现简单是以牺牲性能为代价的：
 
 - 客户端需要使用多个连接才能实现并发和缩短延迟；
 - 不会压缩请求和响应首部，从而导致不必要的网络流量；
@@ -741,9 +743,9 @@ HTTP/2.0 将报文分成 HEADERS 帧和 DATA 帧，它们都是二进制格式
 
 在通信过程中，只会有一个 TCP 连接存在，它承载了任意数量的双向数据流（Stream）。
 
-- 一个数据流都有一个唯一标识符和可选的优先级信息，用于承载双向信息。
-- 消息（Message）是与逻辑请求或响应消息对应的完整的一系列帧。
-- 帧（Fram）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
+- 一个数据流（Stream）都有一个唯一标识符和可选的优先级信息，用于承载双向信息。
+- 消息（Message）是与逻辑请求或响应对应的完整的一系列帧。
+- 帧（Frame）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
 
 <div align="center"> <img src="../pics//af198da1-2480-4043-b07f-a3b91a88b815.png" width="600"/> </div><br>
 
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 1403f764..03aaa9fd 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -637,6 +637,7 @@ B
 
 ```java
 public class WaitNotifyExample {
+
     public synchronized void before() {
         System.out.println("before");
         notifyAll();
@@ -674,12 +675,15 @@ after
 
 ## await() signal() signalAll()
 
-java.util.concurrent 类库中提供了 Condition 类来实现线程之间的协调，可以在 Condition 上调用 await() 方法使线程等待，其它线程调用 signal() 或 signalAll() 方法唤醒等待的线程。相比于 wait() 这种等待方式，await() 可以指定等待的条件，因此更加灵活。
+java.util.concurrent 类库中提供了 Condition 类来实现线程之间的协调，可以在 Condition 上调用 await() 方法使线程等待，其它线程调用 signal() 或 signalAll() 方法唤醒等待的线程。
+
+相比于 wait() 这种等待方式，await() 可以指定等待的条件，因此更加灵活。
 
 使用 Lock 来获取一个 Condition 对象。
 
 ```java
 public class AwaitSignalExample {
+
     private Lock lock = new ReentrantLock();
     private Condition condition = lock.newCondition();
 
@@ -809,7 +813,7 @@ before..before..before..before..before..before..before..before..before..before..
 
 ## Semaphore
 
-Semaphore 就是操作系统中的信号量，可以控制对互斥资源的访问线程数。
+Semaphore 类似于操作系统中的信号量，可以控制对互斥资源的访问线程数。
 
 <div align="center"> <img src="../pics//Semaphore.png" width=""/> </div><br>
 
@@ -1098,11 +1102,11 @@ Java 内存模型定义了 8 个操作来完成主内存和工作内存的交互
 
 Java 内存模型保证了 read、load、use、assign、store、write、lock 和 unlock 操作具有原子性，例如对一个 int 类型的变量执行 assign 赋值操作，这个操作就是原子性的。但是 Java 内存模型允许虚拟机将没有被 volatile 修饰的 64 位数据（long，double）的读写操作划分为两次 32 位的操作来进行，即 load、store、read 和 write 操作可以不具备原子性。
 
-有一个错误认识就是，int 等原子性的变量在多线程环境中不会出现线程安全问题。前面的线程不安全示例代码中，cnt 变量属于 int 类型变量，1000 个线程对它进行自增操作之后，得到的值为 997 而不是 1000。
+有一个错误认识就是，int 等原子性的类型在多线程环境中不会出现线程安全问题。前面的线程不安全示例代码中，cnt 属于 int 类型变量，1000 个线程对它进行自增操作之后，得到的值为 997 而不是 1000。
 
 为了方便讨论，将内存间的交互操作简化为 3 个：load、assign、store。
 
-下图演示了两个线程同时对 cnt 变量进行操作，load、assign、store 这一系列操作整体上看不具备原子性，那么在 T1 修改 cnt 并且还没有将修改后的值写入主内存，T2 依然可以读入该变量的值。可以看出，这两个线程虽然执行了两次自增运算，但是主内存中 cnt 的值最后为 1 而不是 2。因此对 int 类型读写操作满足原子性只是说明 load、assign、store 这些单个操作具备原子性。
+下图演示了两个线程同时对 cnt 进行操作，load、assign、store 这一系列操作整体上看不具备原子性，那么在 T1 修改 cnt 并且还没有将修改后的值写入主内存，T2 依然可以读入旧值。可以看出，这两个线程虽然执行了两次自增运算，但是主内存中 cnt 的值最后为 1 而不是 2。因此对 int 类型读写操作满足原子性只是说明 load、assign、store 这些单个操作具备原子性。
 
 <div align="center"> <img src="../pics//ef8eab00-1d5e-4d99-a7c2-d6d68ea7fe92.png" width=""/> </div><br>
 
@@ -1200,9 +1204,7 @@ public static void main(String[] args) throws InterruptedException {
 
 ### 3. 有序性
 
-有序性是指：在本线程内观察，所有操作都是有序的。在一个线程观察另一个线程，所有操作都是无序的，无序是因为发生了指令重排序。
-
-在 Java 内存模型中，允许编译器和处理器对指令进行重排序，重排序过程不会影响到单线程程序的执行，却会影响到多线程并发执行的正确性。
+有序性是指：在本线程内观察，所有操作都是有序的。在一个线程观察另一个线程，所有操作都是无序的，无序是因为发生了指令重排序。在 Java 内存模型中，允许编译器和处理器对指令进行重排序，重排序过程不会影响到单线程程序的执行，却会影响到多线程并发执行的正确性。
 
 volatile 关键字通过添加内存屏障的方式来禁止指令重排，即重排序时不能把后面的指令放到内存屏障之前。
 
@@ -1413,7 +1415,7 @@ synchronized 和 ReentrantLock。
 
 **（二）AtomicInteger** 
 
-J.U.C 包里面的整数原子类 AtomicInteger，其中的 compareAndSet() 和 getAndIncrement() 等方法都使用了 Unsafe 类的 CAS 操作。
+J.U.C 包里面的整数原子类 AtomicInteger 的方法调用了 Unsafe 类的 CAS 操作。
 
 以下代码使用了 AtomicInteger 执行了自增的操作。
 
@@ -1425,7 +1427,7 @@ public void add() {
 }
 ```
 
-以下代码是 incrementAndGet() 的源码，它调用了 unsafe 的 getAndAddInt() 。
+以下代码是 incrementAndGet() 的源码，它调用了 Unsafe 的 getAndAddInt() 。
 
 ```java
 public final int incrementAndGet() {
@@ -1463,9 +1465,6 @@ J.U.C 包提供了一个带有标记的原子引用类 AtomicStampedReference 
 多个线程访问同一个方法的局部变量时，不会出现线程安全问题，因为局部变量存储在虚拟机栈中，属于线程私有的。
 
 ```java
-import java.util.concurrent.ExecutorService;
-import java.util.concurrent.Executors;
-
 public class StackClosedExample {
     public void add100() {
         int cnt = 0;
@@ -1555,7 +1554,7 @@ public class ThreadLocalExample1 {
 
 <div align="center"> <img src="../pics//3646544a-cb57-451d-9e03-d3c4f5e4434a.png" width=""/> </div><br>
 
-每个 Thread 都有一个 ThreadLocal.ThreadLocalMap 对象，Thread 类中就定义了 ThreadLocal.ThreadLocalMap 成员。
+每个 Thread 都有一个 ThreadLocal.ThreadLocalMap 对象。
 
 ```java
 /* ThreadLocal values pertaining to this thread. This map is maintained
@@ -1686,15 +1685,15 @@ JDK 1.6 引入了偏向锁和轻量级锁，从而让锁拥有了四个状态：
 
 - 缩小同步范围，从而减少锁争用。例如对于 synchronized，应该尽量使用同步块而不是同步方法。
 
-- 多用同步工具少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现复杂控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善，使用这些更高等级的同步工具你的程序可以不费吹灰之力获得优化。
+- 多用同步工具少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现复杂控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善。
+
+- 使用 BlockingQueue 实现生产者消费者问题。
 
 - 多用并发集合少用同步集合，例如应该使用 ConcurrentHashMap 而不是 Hashtable。
 
 - 使用本地变量和不可变类来保证线程安全。
 
-- 使用线程池而不是直接创建 Thread 对象，这是因为创建线程代价很高，线程池可以有效地利用有限的线程来启动任务。
-
-- 使用 BlockingQueue 实现生产者消费者问题。
+- 使用线程池而不是直接创建线程，这是因为创建线程代价很高，线程池可以有效地利用有限的线程来启动任务。
 
 # 参考资料
 
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 40972bfd..1dc4d1d6 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -7,8 +7,6 @@
     * [方法区](#方法区)
     * [运行时常量池](#运行时常量池)
     * [直接内存](#直接内存)
-    * [类的创建过程](#类的创建过程)
-    * [对象的结构](#对象的结构)
 * [二、垃圾收集](#二垃圾收集)
     * [判断一个对象是否可被回收](#判断一个对象是否可被回收)
     * [引用类型](#引用类型)
@@ -104,28 +102,6 @@ Class 文件中的常量池（编译器生成的各种字面量和符号引用
 
 这样能在一些场景中显著提高性能，因为避免了在 Java 堆和 Native 堆中来回复制数据。
 
-## 对象的创建过程
-
-[图解JAVA对象的创建过程](https://www.cnblogs.com/chenyangyao/p/5296807.html)
-
-![111](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235423381-1926278401.png)
-
-## 对象的结构
-
-对象包含 3 个部分：
-1. Header(对象头)：如下图所示；
-2. InstanceData(实例数据)：将等宽的类型放在一起；
-3. Padding(对齐填充)：Hotspot 虚拟机的内存管理系统要求对象的起始地址要是 8 个字节的整数倍，而对象头就是 8 个字节的整数倍。padding 是用来填充实例数据不足 8 个字节整数倍的部分，可以理解为占位符。
-
-![112](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235633553-1910724119.png)
-
-## 对象的地址访问
-
-1. 直接指针：对象的引用直接指向堆中的内存地址 (Hotspot 采用的方式)；
-2. 使用句柄：Java 堆中划分出一块内存作为句柄池，对象的引用指向句柄池。
-
-![113](https://images2015.cnblogs.com/blog/592743/201603/592743-20160319235555303-769658219.jpg)
-
 # 二、垃圾收集
 
 垃圾收集主要是针对堆和方法区进行。
@@ -162,8 +138,8 @@ public class ReferenceCountingGC {
 
 Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC Roots 一般包含以下内容：
 
-- 虚拟机栈 (局部变量表) 中引用的对象
-- 本地方法栈中引用的对象
+- 虚拟机栈中局部变量表中引用的对象
+- 本地方法栈中 JNI 中引用的对象
 - 方法区中类静态属性引用的对象
 - 方法区中的常量引用的对象
 
@@ -292,7 +268,7 @@ HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了
 以上是 HotSpot 虚拟机中的 7 个垃圾收集器，连线表示垃圾收集器可以配合使用。
 
 - 单线程与多线程：单线程指的是垃圾收集器只使用一个线程进行收集，而多线程使用多个线程；
-- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并行指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
+- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并形指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
 
 ### 1. Serial 收集器
 
diff --git a/notes/Linux.md b/notes/Linux.md
index 73a2e9ec..4ba647a7 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -644,7 +644,7 @@ locate 使用 /var/lib/mlocate/ 这个数据库来进行搜索，它存储在内
 example: find . -name "shadow*"
 ```
 
-（一）与时间有关的选项
+**① 与时间有关的选项** 
 
 ```html
 -mtime  n ：列出在 n 天前的那一天修改过内容的文件
@@ -657,7 +657,7 @@ example: find . -name "shadow*"
 
 <div align="center"> <img src="../pics//658fc5e7-79c0-4247-9445-d69bf194c539.png" width=""/> </div><br>
 
-（二）与文件拥有者和所属群组有关的选项
+**② 与文件拥有者和所属群组有关的选项** 
 
 ```html
 -uid n
@@ -668,7 +668,7 @@ example: find . -name "shadow*"
 -nogroup：搜索所属群组不存在于 /etc/group 的文件
 ```
 
-（三）与文件权限和名称有关的选项
+**③ 与文件权限和名称有关的选项** 
 
 ```html
 -name filename
@@ -1038,9 +1038,7 @@ $ grep -n 'go\{2,5\}g' regular_express.txt
 
 ## printf
 
-用于格式化输出。
-
-它不属于管道命令，在给 printf 传数据时需要使用 $( ) 形式。
+用于格式化输出。它不属于管道命令，在给 printf 传数据时需要使用 $( ) 形式。
 
 ```html
 $ printf '%10s %5i %5i %5i %8.2f \n' $(cat printf.txt)
diff --git a/notes/分布式.md b/notes/分布式.md
index e98f2860..113837ce 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -29,18 +29,18 @@
 
 # 一、分布式锁
 
-在单机场景下，可以使用 Java 提供的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
+在单机场景下，可以使用语言的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
 
 阻塞锁通常使用互斥量来实现：
 
-- 互斥量为 1 表示有其它进程在使用锁，此时处于锁定状态；
-- 互斥量为 0 表示未锁定状态。
+- 互斥量为 0 表示有其它进程在使用锁，此时处于锁定状态；
+- 互斥量为 1 表示未锁定状态。
 
-1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示，存在表示互斥量为 1。
+1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示。
 
 ## 数据库的唯一索引
 
-当想要获得锁时，就向表中插入一条记录，释放锁时就删除这条记录。唯一索引可以保证该记录只被插入一次，那么就可以用这个记录是否存在来判断是否存于锁定状态。
+获得锁时向表中插入一条记录，释放锁时删除这条记录。唯一索引可以保证该记录只被插入一次，那么就可以用这个记录是否存在来判断是否存于锁定状态。
 
 存在以下几个问题：
 
@@ -91,11 +91,11 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ### 5. 会话超时
 
-如果一个已经获得锁的会话超时了，因为创建的是临时节点，所以该会话对应的临时节点会被删除，其它会话就可以获得锁了。可以看到，Zookeeper 分布式锁不会出现数据库的唯一索引实现分布式锁的释放锁失败问题。
+如果一个已经获得锁的会话超时了，因为创建的是临时节点，所以该会话对应的临时节点会被删除，其它会话就可以获得锁了。可以看到，Zookeeper 分布式锁不会出现数据库的唯一索引实现的分布式锁释放锁失败问题。
 
 ### 6. 羊群效应
 
-一个节点未获得锁，需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
+一个节点未获得锁，只需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
 
 # 二、分布式事务
 
@@ -159,9 +159,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ## 一致性
 
-一致性指的是多个数据副本是否能保持一致的特性。
-
-在一致性的条件下，系统在执行数据更新操作之后能够从一致性状态转移到另一个一致性状态。
+一致性指的是多个数据副本是否能保持一致的特性，在一致性的条件下，系统在执行数据更新操作之后能够从一致性状态转移到另一个一致性状态。
 
 对系统的一个数据更新成功之后，如果所有用户都能够读取到最新的值，该系统就被认为具有强一致性。
 
@@ -169,7 +167,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 可用性指分布式系统在面对各种异常时可以提供正常服务的能力，可以用系统可用时间占总时间的比值来衡量，4 个 9 的可用性表示系统 99.99% 的时间是可用的。
 
-在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
+在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操，请求总是能够在有限的时间内返回结果。
 
 ## 分区容忍性
 
@@ -179,9 +177,9 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ## 权衡
 
-在分布式系统中，分区容忍性必不可少，因为需要总是假设网络是不可靠的。因此，CAP 理论实际在是要在可用性和一致性之间做权衡。
+在分布式系统中，分区容忍性必不可少，因为需要总是假设网络是不可靠的。因此，CAP 理论实际上是要在可用性和一致性之间做权衡。
 
-可用性和一致性往往是冲突的，很难都使它们同时满足。在多个节点之间进行数据同步时，
+可用性和一致性往往是冲突的，很难使它们同时满足。在多个节点之间进行数据同步时，
 
 - 为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
 - 为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
@@ -204,7 +202,7 @@ BASE 理论是对 CAP 中一致性和可用性权衡的结果，它的核心思
 
 ## 软状态
 
-指允许系统中的数据存在中间状态，并认为该中间状态不会影响系统整体可用性，即允许系统不同节点的数据副本之间进行同步的过程存在延时。
+指允许系统中的数据存在中间状态，并认为该中间状态不会影响系统整体可用性，即允许系统不同节点的数据副本之间进行同步的过程存在时延。
 
 ## 最终一致性
 
diff --git a/notes/构建工具.md b/notes/构建工具.md
index ce157934..34fef54b 100644
--- a/notes/构建工具.md
+++ b/notes/构建工具.md
@@ -1,11 +1,12 @@
 <!-- GFM-TOC -->
-* [一、什么是构建工具](#一什么是构建工具)
+* [一、构建工具的作用](#一构建工具的作用)
 * [二、Java 主流构建工具](#二java-主流构建工具)
 * [三、Maven](#三maven)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、什么是构建工具
+# 一、构建工具的作用
 
 构建工具是用于构建项目的自动化工具，主要包含以下工作：
 
@@ -29,10 +30,6 @@
 
 不再需要通过 FTP 将 Jar 包上传到服务器上。
 
-参考资料：
-
-- [What is a build tool?](https://stackoverflow.com/questions/7249871/what-is-a-build-tool)
-
 # 二、Java 主流构建工具
 
 主要包括 Ant、Maven 和 Gradle。
@@ -72,12 +69,6 @@ dependencies {
 }
 ```
 
-参考资料：
-
-- [Java Build Tools Comparisons: Ant vs Maven vs Gradle](https://programmingmitra.blogspot.com/2016/05/java-build-tools-comparisons-ant-vs.html)
-- [maven 2 gradle](http://sagioto.github.io/maven2gradle/)
-- [新一代构建工具 gradle](https://www.imooc.com/learn/833)
-
 # 三、Maven
 
 ## 概述
@@ -114,7 +105,7 @@ POM 代表项目对象模型，它是一个 XML 文件，保存在项目根目
 
 ## 依赖原则
 
-### 依赖路径最短优先原则
+### 1. 依赖路径最短优先原则
 
 ```html
 A -> B -> C -> X(1.0)
@@ -122,7 +113,7 @@ A -> D -> X(2.0)
 ```
 由于 X(2.0) 路径最短，所以使用 X(2.0)。
 
-### 声明顺序优先原则
+### 2. 声明顺序优先原则
 
 ```html
 A -> B -> X(1.0)
@@ -131,7 +122,7 @@ A -> C -> X(2.0)
 
 在 POM 中最先声明的优先，上面的两个依赖如果先声明 B，那么最后使用 X(1.0)。
 
-### 覆写优先原则
+### 3. 覆写优先原则
 
 子 POM 内声明的依赖优先于父 POM 中声明的依赖。
 
@@ -139,9 +130,11 @@ A -> C -> X(2.0)
 
 找到 Maven 加载的 Jar 包版本，使用 `mvn dependency:tree` 查看依赖树，根据依赖原则来调整依赖在 POM 文件的声明顺序。
 
-参考资料：
+# 参考资料
 
 - [POM Reference](http://maven.apache.org/pom.html#Dependency_Version_Requirement_Specification)
-
-
+- [What is a build tool?](https://stackoverflow.com/questions/7249871/what-is-a-build-tool)
+- [Java Build Tools Comparisons: Ant vs Maven vs Gradle](https://programmingmitra.blogspot.com/2016/05/java-build-tools-comparisons-ant-vs.html)
+- [maven 2 gradle](http://sagioto.github.io/maven2gradle/)
+- [新一代构建工具 gradle](https://www.imooc.com/learn/833)
 
diff --git a/notes/消息队列.md b/notes/消息队列.md
index 209d962a..bc80894d 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -40,7 +40,7 @@
 
 发送者将消息发送给消息队列之后，不需要同步等待消息接收者处理完毕，而是立即返回进行其它操作。消息接收者从消息队列中订阅消息之后异步处理。
 
-例如在注册流程中通常需要发送验证邮件来确保注册用户的身份合法，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
+例如在注册流程中通常需要发送验证邮件来确保注册用户身份的合法性，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
 
 只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成注册的话，就不能再使用消息队列。
 
diff --git a/notes/算法.md b/notes/算法.md
index 53edc1c1..cb83830b 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -379,7 +379,7 @@ public class Insertion<T extends Comparable<T>> extends Sort<T> {
 
 对于大规模的数组，插入排序很慢，因为它只能交换相邻的元素，每次只能将逆序数量减少 1。
 
-希尔排序的出现就是为了改进插入排序的这种局限性，它通过交换不相邻的元素，每次可以将逆序数量减少大于 1。
+希尔排序的出现就是为了解决插入排序的这种局限性，它通过交换不相邻的元素，每次可以将逆序数量减少大于 1。
 
 希尔排序使用插入排序对间隔 h 的序列进行排序。通过不断减小 h，最后令 h=1，就可以使得整个数组是有序的。
 
@@ -571,15 +571,15 @@ private int partition(T[] nums, int l, int h) {
 
 ### 4. 算法改进
 
-（一）切换到插入排序
+#### 4.1 切换到插入排序
 
 因为快速排序在小数组中也会递归调用自己，对于小数组，插入排序比快速排序的性能更好，因此在小数组中可以切换到插入排序。
 
-（二）三数取中
+#### 4.2 三数取中
 
 最好的情况下是每次都能取数组的中位数作为切分元素，但是计算中位数的代价很高。人们发现取 3 个元素并将大小居中的元素作为切分元素的效果最好。
 
-（三）三向切分
+#### 4.3 三向切分
 
 对于有大量重复元素的数组，可以将数组切分为三部分，分别对应小于、等于和大于切分元素。
 
@@ -645,7 +645,7 @@ public T select(T[] nums, int k) {
 
 堆的某个节点的值总是大于等于子节点的值，并且堆是一颗完全二叉树。
 
-堆可以用数组来表示，因为堆是完全二叉树，而完全二叉树很容易就存储在数组中。位置 k 的节点的父节点位置为 k/2，而它的两个子节点的位置分别为 2k 和 2k+1。这里不使用数组索引为 0 的位置，是为了更清晰地描述节点的位置关系。
+堆可以用数组来表示，这是因为堆是完全二叉树，而完全二叉树很容易就存储在数组中。位置 k 的节点的父节点位置为 k/2，而它的两个子节点的位置分别为 2k 和 2k+1。这里不使用数组索引为 0 的位置，是为了更清晰地描述节点的位置关系。
 
 <div align="center"> <img src="../pics//f3080f83-6239-459b-8e9c-03b6641f7815.png" width="200"/> </div><br>
 
@@ -739,15 +739,15 @@ public T delMax() {
 
 ### 5. 堆排序
 
-由于堆可以很容易得到最大的元素并删除它，不断地进行这种操作可以得到一个递减序列。如果把最大元素和当前堆中数组的最后一个元素交换位置，并且不删除它，那么就可以得到一个从尾到头的递减序列，从正向来看就是一个递增序列。因此很容易使用堆来进行排序。并且堆排序是原地排序，不占用额外空间。
+把最大元素和当前堆中数组的最后一个元素交换位置，并且不删除它，那么就可以得到一个从尾到头的递减序列，从正向来看就是一个递增序列，这就是堆排序。
 
-（一）构建堆
+#### 5.1 构建堆
 
-无序数组建立堆最直接的方法是从左到右遍历数组，然后进行上浮操作。一个更高效的方法是从右至左进行下沉操作，如果一个节点的两个节点都已经是堆有序，那么进行下沉操作可以使得这个节点为根节点的堆有序。叶子节点不需要进行下沉操作，可以忽略叶子节点的元素，因此只需要遍历一半的元素即可。
+无序数组建立堆最直接的方法是从左到右遍历数组进行上浮操作。一个更高效的方法是从右至左进行下沉操作，如果一个节点的两个节点都已经是堆有序，那么进行下沉操作可以使得这个节点为根节点的堆有序。叶子节点不需要进行下沉操作，可以忽略叶子节点的元素，因此只需要遍历一半的元素即可。
 
 <div align="center"> <img src="../pics//b84ba6fb-312b-4e69-8c77-fb6eb6fb38d4.png" width="300"/> </div><br>
 
-（二）交换堆顶元素与最后一个元素
+#### 5.2 交换堆顶元素与最后一个元素
 
 交换之后需要进行下沉操作维持堆的有序状态。
 
@@ -804,7 +804,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 ### 1. 排序算法的比较
 
-| 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
+| 算法 | 稳定性 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
 | 选择排序 | × | N<sup>2</sup> | 1 | |
 | 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
@@ -815,7 +815,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 | 归并排序 | √ |  NlogN | N | |
 | 堆排序 | ×  |  NlogN | 1 | | |
 
-快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其他线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
+快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其它线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
 
 ### 2. Java 的排序算法实现
 
@@ -882,7 +882,6 @@ public class QuickFindUF extends UF {
 
     @Override
     public void union(int p, int q) {
-
         int pID = find(p);
         int qID = find(q);
 
@@ -917,7 +916,6 @@ public class QuickUnionUF extends UF {
 
     @Override
     public int find(int p) {
-
         while (p != id[p]) {
             p = id[p];
         }
@@ -927,7 +925,6 @@ public class QuickUnionUF extends UF {
 
     @Override
     public void union(int p, int q) {
-
         int pRoot = find(p);
         int qRoot = find(q);
 
@@ -938,7 +935,7 @@ public class QuickUnionUF extends UF {
 }
 ```
 
-这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为触点的数目。
+这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为节点的数目。
 
 <div align="center"> <img src="../pics//bfbb11e2-d208-4efa-b97b-24cd40467cd8.png" width="150"/> </div><br>
 
@@ -1588,7 +1585,7 @@ public class BST<Key extends Comparable<Key>, Value> implements OrderedST<Key, V
 }
 ```
 
-（为了方便绘图，下文中二叉树的空链接不画出来。）
+为了方便绘图，下文中二叉树的空链接不画出来。
 
 ### 1. get()
 
@@ -1748,7 +1745,7 @@ public Node deleteMin(Node x) {
 
 ### 8. delete()
 
-- 如果待删除的节点只有一个子树，那么只需要让指向待删除节点的链接指向唯一的子树即可；
+- 如果待删除的节点只有一个子树，  那么只需要让指向待删除节点的链接指向唯一的子树即可；
 - 否则，让右子树的最小节点替换该节点。
 
 <div align="center"> <img src="../pics//fa568fac-ac58-48dd-a9bb-23b3065bf2dc.png" width="400"/> </div><br>
@@ -1806,9 +1803,9 @@ private List<Key> keys(Node x, Key l, Key h) {
 }
 ```
 
-### 10. 性能分析
+### 10. 分析
 
-复杂度：二叉查找树所有操作在最坏的情况下所需要的时间都和树的高度成正比。
+二叉查找树所有操作在最坏的情况下所需要的时间都和树的高度成正比。
 
 ## 2-3 查找树
 
@@ -1838,7 +1835,7 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ## 红黑树
 
-2-3 查找树需要用到 2- 节点和 3- 节点，红黑树使用红链接来实现 3- 节点。指向一个节点的链接颜色如果为红色，那么这个节点和上层节点表示的是一个 3- 节点，而黑色则是普通链接。
+红黑树是 2-3 查找树，但它不需要分别定义 2- 节点和 3- 节点，而是在普通的二叉查找树之上，为节点添加颜色。指向一个节点的链接颜色如果为红色，那么这个节点和上层节点表示的是一个 3- 节点，而黑色则是普通链接。
 
 <div align="center"> <img src="../pics//4f48e806-f90b-4c09-a55f-ac0cd641c047.png" width="250"/> </div><br>
 
@@ -1853,6 +1850,7 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ```java
 public class RedBlackBST<Key extends Comparable<Key>, Value> extends BST<Key, Value> {
+
     private static final boolean RED = true;
     private static final boolean BLACK = false;
 
@@ -2008,16 +2006,17 @@ int hash = (((day * R + month) % M) * R + year) % M;
 
 R 通常取 31。
 
-Java 中的 hashCode() 实现了 hash 函数，但是默认使用对象的内存地址值。在使用 hashCode() 函数时，应当结合除留余数法来使用。因为内存地址是 32 位整数，我们只需要 31 位的非负整数，因此应当屏蔽符号位之后再使用除留余数法。
+Java 中的 hashCode() 实现了哈希函数，但是默认使用对象的内存地址值。在使用 hashCode() 时，应当结合除留余数法来使用。因为内存地址是 32 位整数，我们只需要 31 位的非负整数，因此应当屏蔽符号位之后再使用除留余数法。
 
 ```java
 int hash = (x.hashCode() & 0x7fffffff) % M;
 ```
 
-使用 Java 自带的 HashMap 等自带的哈希表实现时，只需要去实现 Key 类型的 hashCode() 函数即可。Java 规定 hashCode() 能够将键均匀分布于所有的 32 位整数，Java 中的 String、Integer 等对象的 hashCode() 都能实现这一点。以下展示了自定义类型如何实现 hashCode()：
+使用 Java 的 HashMap 等自带的哈希表实现时，只需要去实现 Key 类型的 hashCode() 函数即可。Java 规定 hashCode() 能够将键均匀分布于所有的 32 位整数，Java 中的 String、Integer 等对象的 hashCode() 都能实现这一点。以下展示了自定义类型如何实现 hashCode()：
 
 ```java
 public class Transaction {
+
     private final String who;
     private final Date when;
     private final double amount;
@@ -2039,17 +2038,17 @@ public class Transaction {
 }
 ```
 
-### 2. 基于拉链法的散列表
+### 2. 拉链法
 
 拉链法使用链表来存储 hash 值相同的键，从而解决冲突。
 
 查找需要分两步，首先查找 Key 所在的链表，然后在链表中顺序查找。
 
-对于 N 个键，M 条链表 (N>M)，如果 hash 函数能够满足均匀性的条件，每条链表的大小趋向于 N/M，因此未命中的查找和插入操作所需要的比较次数为 \~N/M。
+对于 N 个键，M 条链表 (N>M)，如果哈希函数能够满足均匀性的条件，每条链表的大小趋向于 N/M，因此未命中的查找和插入操作所需要的比较次数为 \~N/M。
 
 <div align="center"> <img src="../pics//b4252c85-6fb0-4995-9a68-a1a5925fbdb1.png" width="300"/> </div><br>
 
-### 3. 基于线性探测法的散列表
+### 3. 线性探测法
 
 线性探测法使用空位来解决冲突，当冲突发生时，向前探测一个空位来存储冲突的键。
 
@@ -2059,6 +2058,7 @@ public class Transaction {
 
 ```java
 public class LinearProbingHashST<Key, Value> implements UnorderedST<Key, Value> {
+
     private int N = 0;
     private int M = 16;
     private Key[] keys;
@@ -2084,7 +2084,7 @@ public class LinearProbingHashST<Key, Value> implements UnorderedST<Key, Value>
 }
 ```
 
-**（一）查找** 
+#### 3.1 查找
 
 ```java
 public Value get(Key key) {
@@ -2096,7 +2096,7 @@ public Value get(Key key) {
 }
 ```
 
-**（二）插入** 
+#### 3.2 插入
 
 ```java
 public void put(Key key, Value value) {
@@ -2118,7 +2118,7 @@ private void putInternal(Key key, Value value) {
 }
 ```
 
-**（三）删除** 
+#### 3.3 删除
 
 删除操作应当将右侧所有相邻的键值对重新插入散列表中。
 
@@ -2151,7 +2151,7 @@ public void delete(Key key) {
 }
 ```
 
-**（四）调整数组大小** 
+#### 3.5 调整数组大小
 
 线性探测法的成本取决于连续条目的长度，连续条目也叫聚簇。当聚簇很长时，在查找和插入时也需要进行很多次探测。例如下图中 2\~5 位置就是一个聚簇。
 
@@ -2235,15 +2235,15 @@ public class SparseVector {
 
 这是一个经典的递归问题，分为三步求解：
 
-- 将 n-1 个圆盘从 from -> buffer
+① 将 n-1 个圆盘从 from -> buffer
 
 <div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
 
-- 将 1 个圆盘从 from -> to
+② 将 1 个圆盘从 from -> to
 
 <div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
 
-- 将 n-1 个圆盘从 buffer -> to
+③ 将 n-1 个圆盘从 buffer -> to
 
 <div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
 
@@ -2281,9 +2281,9 @@ from H1 to H3
 
 ## 哈夫曼编码
 
-哈夫曼编码根据数据出现的频率对数据进行编码，从而压缩原始数据。
+根据数据出现的频率对数据进行编码，从而压缩原始数据。
 
-例如对于文本文件，其中各种字符出现的次数如下：
+例如对于一个文本文件，其中各种字符出现的次数如下：
 
 - a : 10
 - b : 20
@@ -2294,7 +2294,7 @@ from H1 to H3
 
 首先生成一颗哈夫曼树，每次生成过程中选取频率最少的两个节点，生成一个新节点作为它们的父节点，并且新节点的频率为两个节点的和。选取频率最少的原因是，生成过程使得先选取的节点在树的最底层，那么需要的编码长度更长，频率更少可以使得总编码长度更少。
 
-生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到根节点，根节点代表的字符的编码就是这个路径编码。
+生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到叶子节点，叶子节点代表的字符的编码就是这个路径编码。
 
 <div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png" width="400"/> </div><br>
 
diff --git a/notes/缓存.md b/notes/缓存.md
index ed834253..26d96be1 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -26,16 +26,16 @@
 
 ## 淘汰策略
 
-- FIFO（First In First Out）：先进先出策略，在实时性的场景下，需要经常访问最新的数据，那么就可以使用 FIFO，使最先进入的数据（最晚的数据）被淘汰。
+- FIFO（First In First Out）：先进先出策略，在实时性的场景下，需要经常访问最新的数据，那么就可以使用 FIFO，使得最先进入的数据（最晚的数据）被淘汰。
 
-- LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最远的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
+- LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最久的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
 
 # 二、LRU
 
-以下是一个基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
+以下是基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
 
-- 最基本的思路是当访问某个节点时，将其从原来的位置删除，并重新插入到链表头部，这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就删除链表尾部的节点。
-- 为了使删除操作时间复杂度为 O(1)，那么就不能采用遍历的方式找到某个节点。HashMap 存储着 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
+- 访问某个节点时，将其从原来的位置删除，并重新插入到链表头部。这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就淘汰链表尾部的节点。
+- 为了使删除操作时间复杂度为 O(1)，就不能采用遍历的方式找到某个节点。HashMap 存储着 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
 
 ```java
 public class LRU<K, V> implements Iterable<K> {
@@ -152,7 +152,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 反向代理
 
-反向代理位于服务器之前，请求与响应都需要经过反向代理。通过将数据缓存在反向代理，在用户请求时就可以直接使用缓存进行响应。
+反向代理位于服务器之前，请求与响应都需要经过反向代理。通过将数据缓存在反向代理，在用户请求反向代理时就可以直接使用缓存进行响应。
 
 ## 本地缓存
 
@@ -166,7 +166,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 数据库缓存
 
-MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查询效率。
+MySQL 等数据库管理系统具有自己的查询缓存机制来提高查询效率。
 
 # 四、CDN
 
@@ -193,9 +193,9 @@ CDN 主要有以下优点：
 
 ## 缓存雪崩
 
-指的是由于数据没有被加载到缓存中，或者缓存数据在同一时间大面积失效（过期），又或者缓存服务器宕机，导致大量的请求都去到达数据库。
+指的是由于数据没有被加载到缓存中，或者缓存数据在同一时间大面积失效（过期），又或者缓存服务器宕机，导致大量的请求都到达数据库。
 
-在存在缓存的系统中，系统非常依赖于缓存，缓存分担了很大一部分的数据请求。当发生缓存雪崩时，数据库无法处理这么大的请求，导致数据库崩溃。
+在有缓存的系统中，系统非常依赖于缓存，缓存分担了很大一部分的数据请求。当发生缓存雪崩时，数据库无法处理这么大的请求，导致数据库崩溃。
 
 解决方案：
 
@@ -233,7 +233,7 @@ CDN 主要有以下优点：
 
 # 七、一致性哈希
 
-Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据失效的问题。
+Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据迁移的问题。
 
 ## 基本原理
 
@@ -241,7 +241,7 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 <div align="center"> <img src="../pics//68b110b9-76c6-4ee2-b541-4145e65adb3e.jpg"/> </div><br>
 
-一致性哈希在增加或者删除节点时只会影响到哈希环中相邻的节点，例如下图中新增节点 X，只需要将它前一个节点 C 上的数据重新进行分布即可，对于节点 A、B、D 都没有影响。
+一致性哈希在增加或者删除节点时只会影响到哈希环中相邻的节点，例如下图中新增节点 X，只需要将它后一个节点 C 上的数据重新进行分布即可，对于节点 A、B、D 都没有影响。
 
 <div align="center"> <img src="../pics//66402828-fb2b-418f-83f6-82153491bcfe.jpg"/> </div><br>
 
@@ -249,7 +249,9 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
 
-数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。
+
+解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得多，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
 
 # 参考资料
 
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index 99a6fb43..e5b8dfcf 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -59,7 +59,7 @@
 
 网络把主机连接起来，而互联网是把多种不同的网络连接起来，因此互联网是网络的网络。
 
-<div align="center"> <img src="../pics//network-of-networks.gif" width="400"/> </div><br>
+<div align="center"> <img src="../pics//network-of-networks.gif" width="450"/> </div><br>
 
 ## ISP
 
@@ -304,7 +304,7 @@ PPP 的帧格式：
 
 MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标识网络适配器（网卡）。
 
-一台主机拥有多少个适配器就有多少个 MAC 地址。例如笔记本电脑普遍存在无线网络适配器和有线网络适配器，因此就有两个 MAC 地址。
+一台主机拥有多少个网络适配器就有多少个 MAC 地址。例如笔记本电脑普遍存在无线网络适配器和有线网络适配器，因此就有两个 MAC 地址。
 
 ## 局域网
 
@@ -320,7 +320,7 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 以太网是一种星型拓扑结构局域网。
 
-早期使用集线器进行连接，集线器是一种物理层设备，作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离，之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
+早期使用集线器进行连接，集线器是一种物理层设备， 作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离，之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
 
 目前以太网使用交换机替代了集线器，交换机是一种链路层设备，它不会发生碰撞，能根据 MAC 地址进行存储转发。
 
diff --git a/notes/设计模式.md b/notes/设计模式.md
index 531595e3..d7ef735a 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -726,7 +726,12 @@ request2 is handle by ConcreteHandler2
 
 ### 意图
 
-将命令封装成对象中，以便使用命令来参数化其它对象，或者将命令对象放入队列中进行排队，或者将命令对象的操作记录到日志中，以及支持可撤销的操作。
+将命令封装成对象中，具有以下作用：
+
+- 使用命令来参数化其它对象
+- 将命令放入队列中进行排队
+- 将命令的操作记录到日志中
+- 支持可撤销的操作
 
 ### 类图
 
@@ -853,8 +858,8 @@ public class Client {
 
 ### 类图
 
-- TerminalExpression：终结符表达式，每个终结符都需要一个 TerminalExpression
-- Context：上下文，包含解释器之外的一些全局信息
+- TerminalExpression：终结符表达式，每个终结符都需要一个 TerminalExpression。
+- Context：上下文，包含解释器之外的一些全局信息。
 
 <div align="center"> <img src="../pics//794239e3-4baf-4aad-92df-f02f59b2a6fe.png"/> </div><br>
 
@@ -1851,7 +1856,7 @@ No gumball dispensed
 ### 类图
 
 - Strategy 接口定义了一个算法族，它们都具有 behavior() 方法。
-- Context 是使用到该算法族的类，其中的 doSomething() 方法会调用 behavior()，setStrategy(in Strategy) 方法可以动态地改变 strategy 对象，也就是说能动态地改变 Context 所使用的算法。
+- Context 是使用到该算法族的类，其中的 doSomething() 方法会调用 behavior()，setStrategy(Strategy) 方法可以动态地改变 strategy 对象，也就是说能动态地改变 Context 所使用的算法。
 
 <div align="center"> <img src="../pics//1fc969e4-0e7c-441b-b53c-01950d2f2be5.png"/> </div><br>
 
diff --git a/notes/重构.md b/notes/重构.md
index b5a834e0..e90fa84e 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -161,6 +161,7 @@ class Customer {
 
 ```java
 class Rental {
+
     private int daysRented;
 
     private Movie movie;
@@ -199,6 +200,7 @@ class Movie {
 
 ```java
 public class App {
+
     public static void main(String[] args) {
         Customer customer = new Customer();
         Rental rental1 = new Rental(1, new Movie(Movie.Type1));
@@ -236,6 +238,7 @@ public class App {
 
 ```java
 class Customer {
+
     private List<Rental> rentals = new ArrayList<>();
 
     void addRental(Rental rental) {
@@ -254,6 +257,7 @@ class Customer {
 
 ```java
 class Rental {
+
     private int daysRented;
 
     private Movie movie;
@@ -294,8 +298,6 @@ class Price2 implements Price {
 ```
 
 ```java
-package imp2;
-
 class Price3 implements Price {
     @Override
     public double getCharge() {
diff --git a/notes/集群.md b/notes/集群.md
index 15657e43..338ddcb4 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -119,7 +119,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 首先了解一下正向代理与反向代理的区别：
 
 - 正向代理：发生在客户端，是由用户主动发起的。比如翻墙，客户端通过主动访问代理服务器，让代理服务器获得需要的外网数据，然后转发回客户端；
-- 反向代理：发生在服务器端，用户不知道代理的存在。
+- 反向代理：发生在服务器端，用户不知道反向代理的存在。
 
 反向代理服务器位于源服务器前面，用户的请求需要先经过反向代理服务器才能到达源服务器。反向代理可以用来进行缓存、日志记录等，同时也可以用来做为负载均衡服务器。
 
@@ -153,7 +153,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，可以直接转发给客户端，避免了负载均衡服务器的成为瓶颈。
 
-这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
+这是一种三角传输模式，被称为直接路由。对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
 
 这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
 
@@ -185,7 +185,6 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 - 占用过多内存；
 - 同步过程占用网络带宽以及服务器处理器时间。
 
-
 <div align="center"> <img src="../pics//MultiNode-SessionReplication.jpg"/> </div><br>
 
 ## Session Server
@@ -206,3 +205,4 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 - [Session Management using Spring Session with JDBC DataStore](https://sivalabs.in/2018/02/session-management-using-spring-session-jdbc-datastore/)
 
+
diff --git a/notes/面向对象思想.md b/notes/面向对象思想.md
index 203c7ebf..87b3b033 100644
--- a/notes/面向对象思想.md
+++ b/notes/面向对象思想.md
@@ -89,24 +89,28 @@ Animal animal = new Cat();
 
 ```java
 public class Instrument {
+
     public void play() {
         System.out.println("Instument is playing...");
     }
 }
 
 public class Wind extends Instrument {
+
     public void play() {
         System.out.println("Wind is playing...");
     }
 }
 
 public class Percussion extends Instrument {
+
     public void play() {
         System.out.println("Percussion is playing...");
     }
 }
 
 public class Music {
+
     public static void main(String[] args) {
         List<Instrument> instruments = new ArrayList<>();
         instruments.add(new Wind());
@@ -353,4 +357,3 @@ Vihicle .. N
 - [看懂 UML 类图和时序图](http://design-patterns.readthedocs.io/zh_CN/latest/read_uml.html#generalization)
 - [UML 系列——时序图（顺序图）sequence diagram](http://www.cnblogs.com/wolf-sun/p/UML-Sequence-diagram.html)
 - [面向对象编程三大特性 ------ 封装、继承、多态](http://blog.csdn.net/jianyuerensheng/article/details/51602015)
-

From b99d1d5237fd134e10b41eb18a269bc7fab596e8 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Thu, 30 Aug 2018 20:28:09 +0800
Subject: [PATCH 162/172] auto commit

---
 notes/Java 虚拟机.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 1dc4d1d6..333c1418 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -268,7 +268,7 @@ HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了
 以上是 HotSpot 虚拟机中的 7 个垃圾收集器，连线表示垃圾收集器可以配合使用。
 
 - 单线程与多线程：单线程指的是垃圾收集器只使用一个线程进行收集，而多线程使用多个线程；
-- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并形指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
+- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并行指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
 
 ### 1. Serial 收集器
 

From bb2592f820012d6220462f6a18a28d80983aa3f0 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Fri, 31 Aug 2018 21:46:11 +0800
Subject: [PATCH 163/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index b9f43ec8..27edaf24 100644
--- a/README.md
+++ b/README.md
@@ -4,7 +4,7 @@
 
 <br>
 <div align="center">
-    <img src="other/LogoMakr_0zpEzN.png" width="200px">
+    <img src="other/LogoMakr_0zpEzN.png" width="150px">
     <br>
     <a href="other/Group.md"> <img src="https://img.shields.io/badge/>-group-4ab8a1.svg"></a> <a href="https://legacy.gitbook.com/book/cyc2018/interview-notebook/details"> <img src="https://img.shields.io/badge/_-gitbook-4ab8a1.svg"></a> 
 </div>

From 0b2d8ac6fdab50c7e4ed961d43ea33e17d48769b Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 1 Sep 2018 01:22:24 +0800
Subject: [PATCH 164/172] auto commit

---
 notes/Docker.md       |  32 ++----
 notes/Git.md          |  32 +++---
 notes/HTTP.md         |  46 +++++----
 notes/Java 并发.md    |  33 +++---
 notes/Java 虚拟机.md  |   6 +-
 notes/Linux.md        |  10 +-
 notes/分布式.md       |  26 +++--
 notes/构建工具.md     |  29 ++----
 notes/消息队列.md     |   2 +-
 notes/算法.md         |  70 ++++++-------
 notes/缓存.md         |  26 ++---
 notes/计算机网络.md   |   8 +-
 notes/设计模式.md     | 228 +++++++++++++++++++-----------------------
 notes/重构.md         |   6 +-
 notes/集群.md         |   6 +-
 notes/面向对象思想.md |   5 +-
 16 files changed, 263 insertions(+), 302 deletions(-)

diff --git a/notes/Docker.md b/notes/Docker.md
index 80108896..1b61a7fe 100644
--- a/notes/Docker.md
+++ b/notes/Docker.md
@@ -4,6 +4,7 @@
 * [三、优势](#三优势)
 * [四、使用场景](#四使用场景)
 * [五、镜像与容器](#五镜像与容器)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
@@ -15,11 +16,6 @@
 
 Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程进行隔离，被隔离的进程独立于宿主操作系统和其它隔离的进程。使用 Docker 可以不修改应用程序代码，不需要开发人员学习特定环境下的技术，就能够将现有的应用程序部署在其他机器中。
 
-参考资料：
-
-- [DOCKER 101: INTRODUCTION TO DOCKER WEBINAR RECAP](https://blog.docker.com/2017/08/docker-101-introduction-docker-webinar-recap/)
-- [Docker 入门教程](http://www.ruanyifeng.com/blog/2018/02/docker-tutorial.html)
-
 # 二、与虚拟机的比较
 
 虚拟机也是一种虚拟化技术，它与 Docker 最大的区别在于它是通过模拟硬件，并在硬件上安装操作系统来实现。
@@ -40,30 +36,22 @@ Docker 主要解决环境配置问题，它是一种虚拟化技术，对进程
 
 而 Docker 只是一个进程，只需要将应用以及相关的组件打包，在运行时占用很少的资源，一台机器可以开启成千上万个 Docker。
 
-参考资料：
-
-- [Docker container vs Virtual machine](http://www.bogotobogo.com/DevOps/Docker/Docker_Container_vs_Virtual_Machine.php)
-
 # 三、优势
 
 除了启动速度快以及占用资源少之外，Docker 具有以下优势：
 
 ## 更容易迁移
 
-Docker 可以提供一致性的运行环境，可以在不同的机器上进行迁移，而不用担心环境变化导致无法运行。
+提供一致性的运行环境，可以在不同的机器上进行迁移，而不用担心环境变化导致无法运行。
 
 ## 更容易维护
 
-Docker 使用分层技术和镜像，使得应用可以更容易复用重复部分。复用程度越高，维护工作也越容易。
+使用分层技术和镜像，使得应用可以更容易复用重复部分。复用程度越高，维护工作也越容易。
 
 ## 更容易扩展
 
 可以使用基础镜像进一步扩展得到新的镜像，并且官方和开源社区提供了大量的镜像，通过扩展这些镜像可以非常容易得到我们想要的镜像。
 
-参考资料：
-
-- [为什么要使用 Docker？](https://yeasy.gitbooks.io/docker_practice/introduction/why.html)
-
 # 四、使用场景
 
 ## 持续集成
@@ -80,11 +68,6 @@ Docker 具有轻量级以及隔离性的特点，在将代码集成到一个 Doc
 
 Docker 轻量级的特点使得它很适合用于部署、维护、组合微服务。
 
-参考资料：
-
-- [What is Docker](https://www.docker.com/what-docker)
-- [持续集成是什么？](http://www.ruanyifeng.com/blog/2015/09/continuous-integration.html)
-
 # 五、镜像与容器
 
 镜像是一种静态的结构，可以看成面向对象里面的类，而容器是镜像的一个实例。
@@ -95,9 +78,14 @@ Docker 轻量级的特点使得它很适合用于部署、维护、组合微服
 
 <div align="center"> <img src="../pics//docker-filesystems-busyboxrw.png"/> </div><br>
 
-参考资料：
+# 参考资料
 
+- [DOCKER 101: INTRODUCTION TO DOCKER WEBINAR RECAP](https://blog.docker.com/2017/08/docker-101-introduction-docker-webinar-recap/)
+- [Docker 入门教程](http://www.ruanyifeng.com/blog/2018/02/docker-tutorial.html)
+- [Docker container vs Virtual machine](http://www.bogotobogo.com/DevOps/Docker/Docker_Container_vs_Virtual_Machine.php)
 - [How to Create Docker Container using Dockerfile](https://linoxide.com/linux-how-to/dockerfile-create-docker-container/)
 - [理解 Docker（2）：Docker 镜像](http://www.cnblogs.com/sammyliu/p/5877964.html)
-
+- [为什么要使用 Docker？](https://yeasy.gitbooks.io/docker_practice/introduction/why.html)
+- [What is Docker](https://www.docker.com/what-docker)
+- [持续集成是什么？](http://www.ruanyifeng.com/blog/2015/09/continuous-integration.html)
 
diff --git a/notes/Git.md b/notes/Git.md
index 4bb0a543..71ff502c 100644
--- a/notes/Git.md
+++ b/notes/Git.md
@@ -1,8 +1,7 @@
 <!-- GFM-TOC -->
-* [学习资料](#学习资料)
 * [集中式与分布式](#集中式与分布式)
-* [Git 的中心服务器](#git-的中心服务器)
-* [Git 工作流](#git-工作流)
+* [中心服务器](#中心服务器)
+* [工作流](#工作流)
 * [分支实现](#分支实现)
 * [冲突](#冲突)
 * [Fast forward](#fast-forward)
@@ -11,16 +10,10 @@
 * [SSH 传输设置](#ssh-传输设置)
 * [.gitignore 文件](#gitignore-文件)
 * [Git 命令一览](#git-命令一览)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 学习资料
-
-- [Git - 简明指南](http://rogerdudler.github.io/git-guide/index.zh.html)
-- [图解 Git](http://marklodato.github.io/visual-git-guide/index-zh-cn.html)
-- [廖雪峰 : Git 教程](https://www.liaoxuefeng.com/wiki/0013739516305929606dd18361248578c67b8067c8c017b000)
-- [Learn Git Branching](https://learngitbranching.js.org/)
-
 # 集中式与分布式
 
 Git 属于分布式版本控制系统，而 SVN 属于集中式。
@@ -33,11 +26,13 @@ Git 属于分布式版本控制系统，而 SVN 属于集中式。
 
 分布式版本控制新建分支、合并分支操作速度非常快，而集中式版本控制新建一个分支相当于复制一份完整代码。
 
-# Git 的中心服务器
+# 中心服务器
 
-Git 的中心服务器用来交换每个用户的修改。没有中心服务器也能工作，但是中心服务器能够 24 小时保持开机状态，这样就能更方便的交换修改。Github 就是一种 Git 中心服务器。
+中心服务器用来交换每个用户的修改，没有中心服务器也能工作，但是中心服务器能够 24 小时保持开机状态，这样就能更方便的交换修改。
 
-# Git 工作流
+Github 就是一个中心服务器。
+
+# 工作流
 
 <div align="center"> <img src="../pics//a1198642-9159-4d88-8aec-c3b04e7a2563.jpg"/> </div><br>
 
@@ -54,14 +49,14 @@ Git 版本库有一个称为 stage 的暂存区，还有自动创建的 master 
 
 <div align="center"> <img src="../pics//17976404-95f5-480e-9cb4-250e6aa1d55f.png"/> </div><br>
 
-可以跳过暂存区域直接从分支中取出修改或者直接提交修改到分支中
+可以跳过暂存区域直接从分支中取出修改，或者直接提交修改到分支中。
 
 - git commit -a 直接把所有文件的修改添加到暂缓区然后执行提交
 - git checkout HEAD -- files 取出最后一次修改，可以用来进行回滚操作
 
 # 分支实现
 
-Git 把每次提交都连成一条时间线。分支使用指针来实现，例如 master 分支指针指向时间线的最后一个节点，也就是最后一次提交。HEAD 指针指向的是当前分支。
+使用指针将每个提交连接成一条时间线，HEAD 指针指向当前分支指针。
 
 <div align="center"> <img src="../pics//fb546e12-e1fb-4b72-a1fb-8a7f5000dce6.jpg"/> </div><br>
 
@@ -69,7 +64,7 @@ Git 把每次提交都连成一条时间线。分支使用指针来实现，例
 
 <div align="center"> <img src="../pics//bc775758-89ab-4805-9f9c-78b8739cf780.jpg"/> </div><br>
 
-每次提交只会让当前分支向前移动，而其它分支不会移动。
+每次提交只会让当前分支指针向前移动，而其它分支指针不会移动。
 
 <div align="center"> <img src="../pics//5292faa6-0141-4638-bf0f-bb95b081dcba.jpg"/> </div><br>
 
@@ -155,4 +150,9 @@ $ ssh-keygen -t rsa -C "youremail@example.com"
 
 比较详细的地址：http://www.cheat-sheets.org/saved-copy/git-cheat-sheet.pdf
 
+# 参考资料
 
+- [Git - 简明指南](http://rogerdudler.github.io/git-guide/index.zh.html)
+- [图解 Git](http://marklodato.github.io/visual-git-guide/index-zh-cn.html)
+- [廖雪峰 : Git 教程](https://www.liaoxuefeng.com/wiki/0013739516305929606dd18361248578c67b8067c8c017b000)
+- [Learn Git Branching](https://learngitbranching.js.org/)
diff --git a/notes/HTTP.md b/notes/HTTP.md
index 537b24fe..aae807e6 100644
--- a/notes/HTTP.md
+++ b/notes/HTTP.md
@@ -25,6 +25,7 @@
     * [实体首部字段](#实体首部字段)
 * [五、具体应用](#五具体应用)
     * [Cookie](#cookie)
+    * [6. Secure](#6-secure)
     * [缓存](#缓存)
     * [连接管理](#连接管理)
     * [内容协商](#内容协商)
@@ -310,7 +311,7 @@ HTTP 协议是无状态的，主要是为了让 HTTP 协议尽可能简单，使
 
 Cookie 是服务器发送到用户浏览器并保存在本地的一小块数据，它会在浏览器之后向同一服务器再次发起请求时被携带上，用于告知服务端两个请求是否来自同一浏览器。由于之后每次请求都会需要携带 Cookie 数据，因此会带来额外的性能开销（尤其是在移动环境下）。
 
-Cookie 曾一度用于客户端数据的存储，因为当时并没有其它合适的存储办法而作为唯一的存储手段，但现在随着现代浏览器开始支持各种各样的存储方式，Cookie 渐渐被淘汰。新的浏览器 API 已经允许开发者直接将数据存储到本地，如使用 Web storage API （本地存储和会话存储）或 IndexedDB。
+Cookie 曾一度用于客户端数据的存储，因为当时并没有其它合适的存储办法而作为唯一的存储手段，但现在随着现代浏览器开始支持各种各样的存储方式，Cookie 渐渐被淘汰。新的浏览器 API 已经允许开发者直接将数据存储到本地，如使用 Web storage API（本地存储和会话存储）或 IndexedDB。
 
 ### 1. 用途
 
@@ -348,7 +349,17 @@ Cookie: yummy_cookie=choco; tasty_cookie=strawberry
 Set-Cookie: id=a3fWa; Expires=Wed, 21 Oct 2015 07:28:00 GMT;
 ```
 
-### 4. JavaScript 获取 Cookie
+### 4. 作用域
+
+Domain 标识指定了哪些主机可以接受 Cookie。如果不指定，默认为当前文档的主机（不包含子域名）。如果指定了 Domain，则一般包含子域名。例如，如果设置 Domain=mozilla.org，则 Cookie 也包含在子域名中（如 developer.mozilla.org）。
+
+Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径必须存在于请求 URL 中）。以字符 %x2F ("/") 作为路径分隔符，子路径也会被匹配。例如，设置 Path=/docs，则以下地址都会匹配：
+
+- /docs
+- /docs/Web/
+- /docs/Web/HTTP
+
+### 5. JavaScript
 
 通过 `Document.cookie` 属性可创建新的 Cookie，也可通过该属性访问非 HttpOnly 标记的 Cookie。
 
@@ -358,9 +369,7 @@ document.cookie = "tasty_cookie=strawberry";
 console.log(document.cookie);
 ```
 
-### 5. Secure 和 HttpOnly
-
-标记为 Secure 的 Cookie 只能通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
+### 6.   HttpOnly
 
 标记为 HttpOnly 的 Cookie 不能被 JavaScript 脚本调用。跨站脚本攻击 (XSS) 常常使用 JavaScript 的 `Document.cookie` API 窃取用户的 Cookie 信息，因此使用 HttpOnly 标记可以在一定程度上避免 XSS 攻击。
 
@@ -368,15 +377,9 @@ console.log(document.cookie);
 Set-Cookie: id=a3fWa; Expires=Wed, 21 Oct 2015 07:28:00 GMT; Secure; HttpOnly
 ```
 
-### 6. 作用域
+## 6. Secure
 
-Domain 标识指定了哪些主机可以接受 Cookie。如果不指定，默认为当前文档的主机（不包含子域名）。如果指定了 Domain，则一般包含子域名。例如，如果设置 Domain=mozilla.org，则 Cookie 也包含在子域名中（如 developer.mozilla.org）。
-
-Path 标识指定了主机下的哪些路径可以接受 Cookie（该 URL 路径必须存在于请求 URL 中）。以字符 %x2F ("/") 作为路径分隔符，子路径也会被匹配。例如，设置 Path=/docs，则以下地址都会匹配：
-
-- /docs
-- /docs/Web/
-- /docs/Web/HTTP
+标记为 Secure 的 Cookie 只能通过被 HTTPS 协议加密过的请求发送给服务端。但即便设置了 Secure 标记，敏感信息也不应该通过 Cookie 传输，因为 Cookie 有其固有的不安全性，Secure 标记也无法提供确实的安全保障。
 
 ### 7. Session
 
@@ -387,8 +390,7 @@ Session 可以存储在服务器上的文件、数据库或者内存中。也可
 使用 Session 维护用户登录状态的过程如下：
 
 - 用户进行登录时，用户提交包含用户名和密码的表单，放入 HTTP 请求报文中；
-- 服务器验证该用户名和密码；
-- 如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 Key 称为 Session ID；
+- 服务器验证该用户名和密码，如果正确则把用户信息存储到 Redis 中，它在 Redis 中的 Key 称为 Session ID；
 - 服务器返回的响应报文的 Set-Cookie 首部字段包含了这个 Session ID，客户端收到响应报文之后将该 Cookie 值存入浏览器中；
 - 客户端之后对同一个服务器进行请求时会包含该 Cookie 值，服务器收到之后提取出 Session ID，从 Redis 中取出用户信息，继续之前的业务操作。
 
@@ -462,13 +464,13 @@ Cache-Control: max-age=31536000
 
 Expires 首部字段也可以用于告知缓存服务器该资源什么时候会过期。
 
-- 在 HTTP/1.1 中，会优先处理 max-age 指令；
-- 在 HTTP/1.0 中，max-age 指令会被忽略掉。
-
 ```html
 Expires: Wed, 04 Jul 2012 08:26:05 GMT
 ```
 
+- 在 HTTP/1.1 中，会优先处理 max-age 指令；
+- 在 HTTP/1.0 中，max-age 指令会被忽略掉。
+
 ### 4. 缓存验证
 
 需要先了解 ETag 首部字段的含义，它是资源的唯一标识。URL 不能唯一表示资源，例如 `http://www.google.com/` 有中文和英文两个资源，只有 ETag 才能对这两个资源进行唯一标识。
@@ -727,7 +729,7 @@ HTTPs 的报文摘要功能之所以安全，是因为它结合了加密和认
 
 ## HTTP/1.x 缺陷
 
- HTTP/1.x 实现简单是以牺牲性能为代价的：
+HTTP/1.x 实现简单是以牺牲性能为代价的：
 
 - 客户端需要使用多个连接才能实现并发和缩短延迟；
 - 不会压缩请求和响应首部，从而导致不必要的网络流量；
@@ -741,9 +743,9 @@ HTTP/2.0 将报文分成 HEADERS 帧和 DATA 帧，它们都是二进制格式
 
 在通信过程中，只会有一个 TCP 连接存在，它承载了任意数量的双向数据流（Stream）。
 
-- 一个数据流都有一个唯一标识符和可选的优先级信息，用于承载双向信息。
-- 消息（Message）是与逻辑请求或响应消息对应的完整的一系列帧。
-- 帧（Fram）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
+- 一个数据流（Stream）都有一个唯一标识符和可选的优先级信息，用于承载双向信息。
+- 消息（Message）是与逻辑请求或响应对应的完整的一系列帧。
+- 帧（Frame）是最小的通信单位，来自不同数据流的帧可以交错发送，然后再根据每个帧头的数据流标识符重新组装。
 
 <div align="center"> <img src="../pics//af198da1-2480-4043-b07f-a3b91a88b815.png" width="600"/> </div><br>
 
diff --git a/notes/Java 并发.md b/notes/Java 并发.md
index 1403f764..03aaa9fd 100644
--- a/notes/Java 并发.md	
+++ b/notes/Java 并发.md	
@@ -637,6 +637,7 @@ B
 
 ```java
 public class WaitNotifyExample {
+
     public synchronized void before() {
         System.out.println("before");
         notifyAll();
@@ -674,12 +675,15 @@ after
 
 ## await() signal() signalAll()
 
-java.util.concurrent 类库中提供了 Condition 类来实现线程之间的协调，可以在 Condition 上调用 await() 方法使线程等待，其它线程调用 signal() 或 signalAll() 方法唤醒等待的线程。相比于 wait() 这种等待方式，await() 可以指定等待的条件，因此更加灵活。
+java.util.concurrent 类库中提供了 Condition 类来实现线程之间的协调，可以在 Condition 上调用 await() 方法使线程等待，其它线程调用 signal() 或 signalAll() 方法唤醒等待的线程。
+
+相比于 wait() 这种等待方式，await() 可以指定等待的条件，因此更加灵活。
 
 使用 Lock 来获取一个 Condition 对象。
 
 ```java
 public class AwaitSignalExample {
+
     private Lock lock = new ReentrantLock();
     private Condition condition = lock.newCondition();
 
@@ -809,7 +813,7 @@ before..before..before..before..before..before..before..before..before..before..
 
 ## Semaphore
 
-Semaphore 就是操作系统中的信号量，可以控制对互斥资源的访问线程数。
+Semaphore 类似于操作系统中的信号量，可以控制对互斥资源的访问线程数。
 
 <div align="center"> <img src="../pics//Semaphore.png" width=""/> </div><br>
 
@@ -1098,11 +1102,11 @@ Java 内存模型定义了 8 个操作来完成主内存和工作内存的交互
 
 Java 内存模型保证了 read、load、use、assign、store、write、lock 和 unlock 操作具有原子性，例如对一个 int 类型的变量执行 assign 赋值操作，这个操作就是原子性的。但是 Java 内存模型允许虚拟机将没有被 volatile 修饰的 64 位数据（long，double）的读写操作划分为两次 32 位的操作来进行，即 load、store、read 和 write 操作可以不具备原子性。
 
-有一个错误认识就是，int 等原子性的变量在多线程环境中不会出现线程安全问题。前面的线程不安全示例代码中，cnt 变量属于 int 类型变量，1000 个线程对它进行自增操作之后，得到的值为 997 而不是 1000。
+有一个错误认识就是，int 等原子性的类型在多线程环境中不会出现线程安全问题。前面的线程不安全示例代码中，cnt 属于 int 类型变量，1000 个线程对它进行自增操作之后，得到的值为 997 而不是 1000。
 
 为了方便讨论，将内存间的交互操作简化为 3 个：load、assign、store。
 
-下图演示了两个线程同时对 cnt 变量进行操作，load、assign、store 这一系列操作整体上看不具备原子性，那么在 T1 修改 cnt 并且还没有将修改后的值写入主内存，T2 依然可以读入该变量的值。可以看出，这两个线程虽然执行了两次自增运算，但是主内存中 cnt 的值最后为 1 而不是 2。因此对 int 类型读写操作满足原子性只是说明 load、assign、store 这些单个操作具备原子性。
+下图演示了两个线程同时对 cnt 进行操作，load、assign、store 这一系列操作整体上看不具备原子性，那么在 T1 修改 cnt 并且还没有将修改后的值写入主内存，T2 依然可以读入旧值。可以看出，这两个线程虽然执行了两次自增运算，但是主内存中 cnt 的值最后为 1 而不是 2。因此对 int 类型读写操作满足原子性只是说明 load、assign、store 这些单个操作具备原子性。
 
 <div align="center"> <img src="../pics//ef8eab00-1d5e-4d99-a7c2-d6d68ea7fe92.png" width=""/> </div><br>
 
@@ -1200,9 +1204,7 @@ public static void main(String[] args) throws InterruptedException {
 
 ### 3. 有序性
 
-有序性是指：在本线程内观察，所有操作都是有序的。在一个线程观察另一个线程，所有操作都是无序的，无序是因为发生了指令重排序。
-
-在 Java 内存模型中，允许编译器和处理器对指令进行重排序，重排序过程不会影响到单线程程序的执行，却会影响到多线程并发执行的正确性。
+有序性是指：在本线程内观察，所有操作都是有序的。在一个线程观察另一个线程，所有操作都是无序的，无序是因为发生了指令重排序。在 Java 内存模型中，允许编译器和处理器对指令进行重排序，重排序过程不会影响到单线程程序的执行，却会影响到多线程并发执行的正确性。
 
 volatile 关键字通过添加内存屏障的方式来禁止指令重排，即重排序时不能把后面的指令放到内存屏障之前。
 
@@ -1413,7 +1415,7 @@ synchronized 和 ReentrantLock。
 
 **（二）AtomicInteger** 
 
-J.U.C 包里面的整数原子类 AtomicInteger，其中的 compareAndSet() 和 getAndIncrement() 等方法都使用了 Unsafe 类的 CAS 操作。
+J.U.C 包里面的整数原子类 AtomicInteger 的方法调用了 Unsafe 类的 CAS 操作。
 
 以下代码使用了 AtomicInteger 执行了自增的操作。
 
@@ -1425,7 +1427,7 @@ public void add() {
 }
 ```
 
-以下代码是 incrementAndGet() 的源码，它调用了 unsafe 的 getAndAddInt() 。
+以下代码是 incrementAndGet() 的源码，它调用了 Unsafe 的 getAndAddInt() 。
 
 ```java
 public final int incrementAndGet() {
@@ -1463,9 +1465,6 @@ J.U.C 包提供了一个带有标记的原子引用类 AtomicStampedReference 
 多个线程访问同一个方法的局部变量时，不会出现线程安全问题，因为局部变量存储在虚拟机栈中，属于线程私有的。
 
 ```java
-import java.util.concurrent.ExecutorService;
-import java.util.concurrent.Executors;
-
 public class StackClosedExample {
     public void add100() {
         int cnt = 0;
@@ -1555,7 +1554,7 @@ public class ThreadLocalExample1 {
 
 <div align="center"> <img src="../pics//3646544a-cb57-451d-9e03-d3c4f5e4434a.png" width=""/> </div><br>
 
-每个 Thread 都有一个 ThreadLocal.ThreadLocalMap 对象，Thread 类中就定义了 ThreadLocal.ThreadLocalMap 成员。
+每个 Thread 都有一个 ThreadLocal.ThreadLocalMap 对象。
 
 ```java
 /* ThreadLocal values pertaining to this thread. This map is maintained
@@ -1686,15 +1685,15 @@ JDK 1.6 引入了偏向锁和轻量级锁，从而让锁拥有了四个状态：
 
 - 缩小同步范围，从而减少锁争用。例如对于 synchronized，应该尽量使用同步块而不是同步方法。
 
-- 多用同步工具少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现复杂控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善，使用这些更高等级的同步工具你的程序可以不费吹灰之力获得优化。
+- 多用同步工具少用 wait() 和 notify()。首先，CountDownLatch, CyclicBarrier, Semaphore 和 Exchanger 这些同步类简化了编码操作，而用 wait() 和 notify() 很难实现复杂控制流；其次，这些同步类是由最好的企业编写和维护，在后续的 JDK 中还会不断优化和完善。
+
+- 使用 BlockingQueue 实现生产者消费者问题。
 
 - 多用并发集合少用同步集合，例如应该使用 ConcurrentHashMap 而不是 Hashtable。
 
 - 使用本地变量和不可变类来保证线程安全。
 
-- 使用线程池而不是直接创建 Thread 对象，这是因为创建线程代价很高，线程池可以有效地利用有限的线程来启动任务。
-
-- 使用 BlockingQueue 实现生产者消费者问题。
+- 使用线程池而不是直接创建线程，这是因为创建线程代价很高，线程池可以有效地利用有限的线程来启动任务。
 
 # 参考资料
 
diff --git a/notes/Java 虚拟机.md b/notes/Java 虚拟机.md
index 2b565cb5..333c1418 100644
--- a/notes/Java 虚拟机.md	
+++ b/notes/Java 虚拟机.md	
@@ -138,8 +138,8 @@ public class ReferenceCountingGC {
 
 Java 虚拟机使用该算法来判断对象是否可被回收，在 Java 中 GC Roots 一般包含以下内容：
 
-- 虚拟机栈中引用的对象
-- 本地方法栈中引用的对象
+- 虚拟机栈中局部变量表中引用的对象
+- 本地方法栈中 JNI 中引用的对象
 - 方法区中类静态属性引用的对象
 - 方法区中的常量引用的对象
 
@@ -268,7 +268,7 @@ HotSpot 虚拟机的 Eden 和 Survivor 的大小比例默认为 8:1，保证了
 以上是 HotSpot 虚拟机中的 7 个垃圾收集器，连线表示垃圾收集器可以配合使用。
 
 - 单线程与多线程：单线程指的是垃圾收集器只使用一个线程进行收集，而多线程使用多个线程；
-- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并形指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
+- 串行与并行：串行指的是垃圾收集器与用户程序交替执行，这意味着在执行垃圾收集的时候需要停顿用户程序；并行指的是垃圾收集器和用户程序同时执行。除了 CMS 和 G1 之外，其它垃圾收集器都是以串行的方式执行。
 
 ### 1. Serial 收集器
 
diff --git a/notes/Linux.md b/notes/Linux.md
index 73a2e9ec..4ba647a7 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -644,7 +644,7 @@ locate 使用 /var/lib/mlocate/ 这个数据库来进行搜索，它存储在内
 example: find . -name "shadow*"
 ```
 
-（一）与时间有关的选项
+**① 与时间有关的选项** 
 
 ```html
 -mtime  n ：列出在 n 天前的那一天修改过内容的文件
@@ -657,7 +657,7 @@ example: find . -name "shadow*"
 
 <div align="center"> <img src="../pics//658fc5e7-79c0-4247-9445-d69bf194c539.png" width=""/> </div><br>
 
-（二）与文件拥有者和所属群组有关的选项
+**② 与文件拥有者和所属群组有关的选项** 
 
 ```html
 -uid n
@@ -668,7 +668,7 @@ example: find . -name "shadow*"
 -nogroup：搜索所属群组不存在于 /etc/group 的文件
 ```
 
-（三）与文件权限和名称有关的选项
+**③ 与文件权限和名称有关的选项** 
 
 ```html
 -name filename
@@ -1038,9 +1038,7 @@ $ grep -n 'go\{2,5\}g' regular_express.txt
 
 ## printf
 
-用于格式化输出。
-
-它不属于管道命令，在给 printf 传数据时需要使用 $( ) 形式。
+用于格式化输出。它不属于管道命令，在给 printf 传数据时需要使用 $( ) 形式。
 
 ```html
 $ printf '%10s %5i %5i %5i %8.2f \n' $(cat printf.txt)
diff --git a/notes/分布式.md b/notes/分布式.md
index e98f2860..113837ce 100644
--- a/notes/分布式.md
+++ b/notes/分布式.md
@@ -29,18 +29,18 @@
 
 # 一、分布式锁
 
-在单机场景下，可以使用 Java 提供的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
+在单机场景下，可以使用语言的内置锁来实现进程同步。但是在分布式场景下，需要同步的进程可能位于不同的节点上，那么就需要使用分布式锁。
 
 阻塞锁通常使用互斥量来实现：
 
-- 互斥量为 1 表示有其它进程在使用锁，此时处于锁定状态；
-- 互斥量为 0 表示未锁定状态。
+- 互斥量为 0 表示有其它进程在使用锁，此时处于锁定状态；
+- 互斥量为 1 表示未锁定状态。
 
-1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示，存在表示互斥量为 1。
+1 和 0 可以用一个整型值表示，也可以用某个数据是否存在表示。
 
 ## 数据库的唯一索引
 
-当想要获得锁时，就向表中插入一条记录，释放锁时就删除这条记录。唯一索引可以保证该记录只被插入一次，那么就可以用这个记录是否存在来判断是否存于锁定状态。
+获得锁时向表中插入一条记录，释放锁时删除这条记录。唯一索引可以保证该记录只被插入一次，那么就可以用这个记录是否存在来判断是否存于锁定状态。
 
 存在以下几个问题：
 
@@ -91,11 +91,11 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ### 5. 会话超时
 
-如果一个已经获得锁的会话超时了，因为创建的是临时节点，所以该会话对应的临时节点会被删除，其它会话就可以获得锁了。可以看到，Zookeeper 分布式锁不会出现数据库的唯一索引实现分布式锁的释放锁失败问题。
+如果一个已经获得锁的会话超时了，因为创建的是临时节点，所以该会话对应的临时节点会被删除，其它会话就可以获得锁了。可以看到，Zookeeper 分布式锁不会出现数据库的唯一索引实现的分布式锁释放锁失败问题。
 
 ### 6. 羊群效应
 
-一个节点未获得锁，需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
+一个节点未获得锁，只需要监听自己的前一个子节点，这是因为如果监听所有的子节点，那么任意一个子节点状态改变，其它所有子节点都会收到通知（羊群效应），而我们只希望它的后一个子节点收到通知。
 
 # 二、分布式事务
 
@@ -159,9 +159,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ## 一致性
 
-一致性指的是多个数据副本是否能保持一致的特性。
-
-在一致性的条件下，系统在执行数据更新操作之后能够从一致性状态转移到另一个一致性状态。
+一致性指的是多个数据副本是否能保持一致的特性，在一致性的条件下，系统在执行数据更新操作之后能够从一致性状态转移到另一个一致性状态。
 
 对系统的一个数据更新成功之后，如果所有用户都能够读取到最新的值，该系统就被认为具有强一致性。
 
@@ -169,7 +167,7 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 可用性指分布式系统在面对各种异常时可以提供正常服务的能力，可以用系统可用时间占总时间的比值来衡量，4 个 9 的可用性表示系统 99.99% 的时间是可用的。
 
-在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操作请求总是能够在有限的时间内返回结果。
+在可用性条件下，要求系统提供的服务一直处于可用的状态，对于用户的每一个操，请求总是能够在有限的时间内返回结果。
 
 ## 分区容忍性
 
@@ -179,9 +177,9 @@ Zookeeper 提供了一种树形结构级的命名空间，/app1/p_1 节点的父
 
 ## 权衡
 
-在分布式系统中，分区容忍性必不可少，因为需要总是假设网络是不可靠的。因此，CAP 理论实际在是要在可用性和一致性之间做权衡。
+在分布式系统中，分区容忍性必不可少，因为需要总是假设网络是不可靠的。因此，CAP 理论实际上是要在可用性和一致性之间做权衡。
 
-可用性和一致性往往是冲突的，很难都使它们同时满足。在多个节点之间进行数据同步时，
+可用性和一致性往往是冲突的，很难使它们同时满足。在多个节点之间进行数据同步时，
 
 - 为了保证一致性（CP），就需要让所有节点下线成为不可用的状态，等待同步完成；
 - 为了保证可用性（AP），在同步过程中允许读取所有节点的数据，但是数据可能不一致。
@@ -204,7 +202,7 @@ BASE 理论是对 CAP 中一致性和可用性权衡的结果，它的核心思
 
 ## 软状态
 
-指允许系统中的数据存在中间状态，并认为该中间状态不会影响系统整体可用性，即允许系统不同节点的数据副本之间进行同步的过程存在延时。
+指允许系统中的数据存在中间状态，并认为该中间状态不会影响系统整体可用性，即允许系统不同节点的数据副本之间进行同步的过程存在时延。
 
 ## 最终一致性
 
diff --git a/notes/构建工具.md b/notes/构建工具.md
index ce157934..34fef54b 100644
--- a/notes/构建工具.md
+++ b/notes/构建工具.md
@@ -1,11 +1,12 @@
 <!-- GFM-TOC -->
-* [一、什么是构建工具](#一什么是构建工具)
+* [一、构建工具的作用](#一构建工具的作用)
 * [二、Java 主流构建工具](#二java-主流构建工具)
 * [三、Maven](#三maven)
+* [参考资料](#参考资料)
 <!-- GFM-TOC -->
 
 
-# 一、什么是构建工具
+# 一、构建工具的作用
 
 构建工具是用于构建项目的自动化工具，主要包含以下工作：
 
@@ -29,10 +30,6 @@
 
 不再需要通过 FTP 将 Jar 包上传到服务器上。
 
-参考资料：
-
-- [What is a build tool?](https://stackoverflow.com/questions/7249871/what-is-a-build-tool)
-
 # 二、Java 主流构建工具
 
 主要包括 Ant、Maven 和 Gradle。
@@ -72,12 +69,6 @@ dependencies {
 }
 ```
 
-参考资料：
-
-- [Java Build Tools Comparisons: Ant vs Maven vs Gradle](https://programmingmitra.blogspot.com/2016/05/java-build-tools-comparisons-ant-vs.html)
-- [maven 2 gradle](http://sagioto.github.io/maven2gradle/)
-- [新一代构建工具 gradle](https://www.imooc.com/learn/833)
-
 # 三、Maven
 
 ## 概述
@@ -114,7 +105,7 @@ POM 代表项目对象模型，它是一个 XML 文件，保存在项目根目
 
 ## 依赖原则
 
-### 依赖路径最短优先原则
+### 1. 依赖路径最短优先原则
 
 ```html
 A -> B -> C -> X(1.0)
@@ -122,7 +113,7 @@ A -> D -> X(2.0)
 ```
 由于 X(2.0) 路径最短，所以使用 X(2.0)。
 
-### 声明顺序优先原则
+### 2. 声明顺序优先原则
 
 ```html
 A -> B -> X(1.0)
@@ -131,7 +122,7 @@ A -> C -> X(2.0)
 
 在 POM 中最先声明的优先，上面的两个依赖如果先声明 B，那么最后使用 X(1.0)。
 
-### 覆写优先原则
+### 3. 覆写优先原则
 
 子 POM 内声明的依赖优先于父 POM 中声明的依赖。
 
@@ -139,9 +130,11 @@ A -> C -> X(2.0)
 
 找到 Maven 加载的 Jar 包版本，使用 `mvn dependency:tree` 查看依赖树，根据依赖原则来调整依赖在 POM 文件的声明顺序。
 
-参考资料：
+# 参考资料
 
 - [POM Reference](http://maven.apache.org/pom.html#Dependency_Version_Requirement_Specification)
-
-
+- [What is a build tool?](https://stackoverflow.com/questions/7249871/what-is-a-build-tool)
+- [Java Build Tools Comparisons: Ant vs Maven vs Gradle](https://programmingmitra.blogspot.com/2016/05/java-build-tools-comparisons-ant-vs.html)
+- [maven 2 gradle](http://sagioto.github.io/maven2gradle/)
+- [新一代构建工具 gradle](https://www.imooc.com/learn/833)
 
diff --git a/notes/消息队列.md b/notes/消息队列.md
index 209d962a..bc80894d 100644
--- a/notes/消息队列.md
+++ b/notes/消息队列.md
@@ -40,7 +40,7 @@
 
 发送者将消息发送给消息队列之后，不需要同步等待消息接收者处理完毕，而是立即返回进行其它操作。消息接收者从消息队列中订阅消息之后异步处理。
 
-例如在注册流程中通常需要发送验证邮件来确保注册用户的身份合法，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
+例如在注册流程中通常需要发送验证邮件来确保注册用户身份的合法性，可以使用消息队列使发送验证邮件的操作异步处理，用户在填写完注册信息之后就可以完成注册，而将发送验证邮件这一消息发送到消息队列中。
 
 只有在业务流程允许异步处理的情况下才能这么做，例如上面的注册流程中，如果要求用户对验证邮件进行点击之后才能完成注册的话，就不能再使用消息队列。
 
diff --git a/notes/算法.md b/notes/算法.md
index 53edc1c1..cb83830b 100644
--- a/notes/算法.md
+++ b/notes/算法.md
@@ -379,7 +379,7 @@ public class Insertion<T extends Comparable<T>> extends Sort<T> {
 
 对于大规模的数组，插入排序很慢，因为它只能交换相邻的元素，每次只能将逆序数量减少 1。
 
-希尔排序的出现就是为了改进插入排序的这种局限性，它通过交换不相邻的元素，每次可以将逆序数量减少大于 1。
+希尔排序的出现就是为了解决插入排序的这种局限性，它通过交换不相邻的元素，每次可以将逆序数量减少大于 1。
 
 希尔排序使用插入排序对间隔 h 的序列进行排序。通过不断减小 h，最后令 h=1，就可以使得整个数组是有序的。
 
@@ -571,15 +571,15 @@ private int partition(T[] nums, int l, int h) {
 
 ### 4. 算法改进
 
-（一）切换到插入排序
+#### 4.1 切换到插入排序
 
 因为快速排序在小数组中也会递归调用自己，对于小数组，插入排序比快速排序的性能更好，因此在小数组中可以切换到插入排序。
 
-（二）三数取中
+#### 4.2 三数取中
 
 最好的情况下是每次都能取数组的中位数作为切分元素，但是计算中位数的代价很高。人们发现取 3 个元素并将大小居中的元素作为切分元素的效果最好。
 
-（三）三向切分
+#### 4.3 三向切分
 
 对于有大量重复元素的数组，可以将数组切分为三部分，分别对应小于、等于和大于切分元素。
 
@@ -645,7 +645,7 @@ public T select(T[] nums, int k) {
 
 堆的某个节点的值总是大于等于子节点的值，并且堆是一颗完全二叉树。
 
-堆可以用数组来表示，因为堆是完全二叉树，而完全二叉树很容易就存储在数组中。位置 k 的节点的父节点位置为 k/2，而它的两个子节点的位置分别为 2k 和 2k+1。这里不使用数组索引为 0 的位置，是为了更清晰地描述节点的位置关系。
+堆可以用数组来表示，这是因为堆是完全二叉树，而完全二叉树很容易就存储在数组中。位置 k 的节点的父节点位置为 k/2，而它的两个子节点的位置分别为 2k 和 2k+1。这里不使用数组索引为 0 的位置，是为了更清晰地描述节点的位置关系。
 
 <div align="center"> <img src="../pics//f3080f83-6239-459b-8e9c-03b6641f7815.png" width="200"/> </div><br>
 
@@ -739,15 +739,15 @@ public T delMax() {
 
 ### 5. 堆排序
 
-由于堆可以很容易得到最大的元素并删除它，不断地进行这种操作可以得到一个递减序列。如果把最大元素和当前堆中数组的最后一个元素交换位置，并且不删除它，那么就可以得到一个从尾到头的递减序列，从正向来看就是一个递增序列。因此很容易使用堆来进行排序。并且堆排序是原地排序，不占用额外空间。
+把最大元素和当前堆中数组的最后一个元素交换位置，并且不删除它，那么就可以得到一个从尾到头的递减序列，从正向来看就是一个递增序列，这就是堆排序。
 
-（一）构建堆
+#### 5.1 构建堆
 
-无序数组建立堆最直接的方法是从左到右遍历数组，然后进行上浮操作。一个更高效的方法是从右至左进行下沉操作，如果一个节点的两个节点都已经是堆有序，那么进行下沉操作可以使得这个节点为根节点的堆有序。叶子节点不需要进行下沉操作，可以忽略叶子节点的元素，因此只需要遍历一半的元素即可。
+无序数组建立堆最直接的方法是从左到右遍历数组进行上浮操作。一个更高效的方法是从右至左进行下沉操作，如果一个节点的两个节点都已经是堆有序，那么进行下沉操作可以使得这个节点为根节点的堆有序。叶子节点不需要进行下沉操作，可以忽略叶子节点的元素，因此只需要遍历一半的元素即可。
 
 <div align="center"> <img src="../pics//b84ba6fb-312b-4e69-8c77-fb6eb6fb38d4.png" width="300"/> </div><br>
 
-（二）交换堆顶元素与最后一个元素
+#### 5.2 交换堆顶元素与最后一个元素
 
 交换之后需要进行下沉操作维持堆的有序状态。
 
@@ -804,7 +804,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 
 ### 1. 排序算法的比较
 
-| 算法 | 稳定 | 时间复杂度 | 空间复杂度 | 备注 |
+| 算法 | 稳定性 | 时间复杂度 | 空间复杂度 | 备注 |
 | :---: | :---: |:---: | :---: | :---: |
 | 选择排序 | × | N<sup>2</sup> | 1 | |
 | 冒泡排序 | √ |  N<sup>2</sup> | 1 | |
@@ -815,7 +815,7 @@ public class HeapSort<T extends Comparable<T>> extends Sort<T> {
 | 归并排序 | √ |  NlogN | N | |
 | 堆排序 | ×  |  NlogN | 1 | | |
 
-快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其他线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
+快速排序是最快的通用排序算法，它的内循环的指令很少，而且它还能利用缓存，因为它总是顺序地访问数据。它的运行时间近似为 \~cNlogN，这里的 c 比其它线性对数级别的排序算法都要小。使用三向切分快速排序，实际应用中可能出现的某些分布的输入能够达到线性级别，而其它排序算法仍然需要线性对数时间。
 
 ### 2. Java 的排序算法实现
 
@@ -882,7 +882,6 @@ public class QuickFindUF extends UF {
 
     @Override
     public void union(int p, int q) {
-
         int pID = find(p);
         int qID = find(q);
 
@@ -917,7 +916,6 @@ public class QuickUnionUF extends UF {
 
     @Override
     public int find(int p) {
-
         while (p != id[p]) {
             p = id[p];
         }
@@ -927,7 +925,6 @@ public class QuickUnionUF extends UF {
 
     @Override
     public void union(int p, int q) {
-
         int pRoot = find(p);
         int qRoot = find(q);
 
@@ -938,7 +935,7 @@ public class QuickUnionUF extends UF {
 }
 ```
 
-这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为触点的数目。
+这种方法可以快速进行 union 操作，但是 find 操作和树高成正比，最坏的情况下树的高度为节点的数目。
 
 <div align="center"> <img src="../pics//bfbb11e2-d208-4efa-b97b-24cd40467cd8.png" width="150"/> </div><br>
 
@@ -1588,7 +1585,7 @@ public class BST<Key extends Comparable<Key>, Value> implements OrderedST<Key, V
 }
 ```
 
-（为了方便绘图，下文中二叉树的空链接不画出来。）
+为了方便绘图，下文中二叉树的空链接不画出来。
 
 ### 1. get()
 
@@ -1748,7 +1745,7 @@ public Node deleteMin(Node x) {
 
 ### 8. delete()
 
-- 如果待删除的节点只有一个子树，那么只需要让指向待删除节点的链接指向唯一的子树即可；
+- 如果待删除的节点只有一个子树，  那么只需要让指向待删除节点的链接指向唯一的子树即可；
 - 否则，让右子树的最小节点替换该节点。
 
 <div align="center"> <img src="../pics//fa568fac-ac58-48dd-a9bb-23b3065bf2dc.png" width="400"/> </div><br>
@@ -1806,9 +1803,9 @@ private List<Key> keys(Node x, Key l, Key h) {
 }
 ```
 
-### 10. 性能分析
+### 10. 分析
 
-复杂度：二叉查找树所有操作在最坏的情况下所需要的时间都和树的高度成正比。
+二叉查找树所有操作在最坏的情况下所需要的时间都和树的高度成正比。
 
 ## 2-3 查找树
 
@@ -1838,7 +1835,7 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ## 红黑树
 
-2-3 查找树需要用到 2- 节点和 3- 节点，红黑树使用红链接来实现 3- 节点。指向一个节点的链接颜色如果为红色，那么这个节点和上层节点表示的是一个 3- 节点，而黑色则是普通链接。
+红黑树是 2-3 查找树，但它不需要分别定义 2- 节点和 3- 节点，而是在普通的二叉查找树之上，为节点添加颜色。指向一个节点的链接颜色如果为红色，那么这个节点和上层节点表示的是一个 3- 节点，而黑色则是普通链接。
 
 <div align="center"> <img src="../pics//4f48e806-f90b-4c09-a55f-ac0cd641c047.png" width="250"/> </div><br>
 
@@ -1853,6 +1850,7 @@ private List<Key> keys(Node x, Key l, Key h) {
 
 ```java
 public class RedBlackBST<Key extends Comparable<Key>, Value> extends BST<Key, Value> {
+
     private static final boolean RED = true;
     private static final boolean BLACK = false;
 
@@ -2008,16 +2006,17 @@ int hash = (((day * R + month) % M) * R + year) % M;
 
 R 通常取 31。
 
-Java 中的 hashCode() 实现了 hash 函数，但是默认使用对象的内存地址值。在使用 hashCode() 函数时，应当结合除留余数法来使用。因为内存地址是 32 位整数，我们只需要 31 位的非负整数，因此应当屏蔽符号位之后再使用除留余数法。
+Java 中的 hashCode() 实现了哈希函数，但是默认使用对象的内存地址值。在使用 hashCode() 时，应当结合除留余数法来使用。因为内存地址是 32 位整数，我们只需要 31 位的非负整数，因此应当屏蔽符号位之后再使用除留余数法。
 
 ```java
 int hash = (x.hashCode() & 0x7fffffff) % M;
 ```
 
-使用 Java 自带的 HashMap 等自带的哈希表实现时，只需要去实现 Key 类型的 hashCode() 函数即可。Java 规定 hashCode() 能够将键均匀分布于所有的 32 位整数，Java 中的 String、Integer 等对象的 hashCode() 都能实现这一点。以下展示了自定义类型如何实现 hashCode()：
+使用 Java 的 HashMap 等自带的哈希表实现时，只需要去实现 Key 类型的 hashCode() 函数即可。Java 规定 hashCode() 能够将键均匀分布于所有的 32 位整数，Java 中的 String、Integer 等对象的 hashCode() 都能实现这一点。以下展示了自定义类型如何实现 hashCode()：
 
 ```java
 public class Transaction {
+
     private final String who;
     private final Date when;
     private final double amount;
@@ -2039,17 +2038,17 @@ public class Transaction {
 }
 ```
 
-### 2. 基于拉链法的散列表
+### 2. 拉链法
 
 拉链法使用链表来存储 hash 值相同的键，从而解决冲突。
 
 查找需要分两步，首先查找 Key 所在的链表，然后在链表中顺序查找。
 
-对于 N 个键，M 条链表 (N>M)，如果 hash 函数能够满足均匀性的条件，每条链表的大小趋向于 N/M，因此未命中的查找和插入操作所需要的比较次数为 \~N/M。
+对于 N 个键，M 条链表 (N>M)，如果哈希函数能够满足均匀性的条件，每条链表的大小趋向于 N/M，因此未命中的查找和插入操作所需要的比较次数为 \~N/M。
 
 <div align="center"> <img src="../pics//b4252c85-6fb0-4995-9a68-a1a5925fbdb1.png" width="300"/> </div><br>
 
-### 3. 基于线性探测法的散列表
+### 3. 线性探测法
 
 线性探测法使用空位来解决冲突，当冲突发生时，向前探测一个空位来存储冲突的键。
 
@@ -2059,6 +2058,7 @@ public class Transaction {
 
 ```java
 public class LinearProbingHashST<Key, Value> implements UnorderedST<Key, Value> {
+
     private int N = 0;
     private int M = 16;
     private Key[] keys;
@@ -2084,7 +2084,7 @@ public class LinearProbingHashST<Key, Value> implements UnorderedST<Key, Value>
 }
 ```
 
-**（一）查找** 
+#### 3.1 查找
 
 ```java
 public Value get(Key key) {
@@ -2096,7 +2096,7 @@ public Value get(Key key) {
 }
 ```
 
-**（二）插入** 
+#### 3.2 插入
 
 ```java
 public void put(Key key, Value value) {
@@ -2118,7 +2118,7 @@ private void putInternal(Key key, Value value) {
 }
 ```
 
-**（三）删除** 
+#### 3.3 删除
 
 删除操作应当将右侧所有相邻的键值对重新插入散列表中。
 
@@ -2151,7 +2151,7 @@ public void delete(Key key) {
 }
 ```
 
-**（四）调整数组大小** 
+#### 3.5 调整数组大小
 
 线性探测法的成本取决于连续条目的长度，连续条目也叫聚簇。当聚簇很长时，在查找和插入时也需要进行很多次探测。例如下图中 2\~5 位置就是一个聚簇。
 
@@ -2235,15 +2235,15 @@ public class SparseVector {
 
 这是一个经典的递归问题，分为三步求解：
 
-- 将 n-1 个圆盘从 from -> buffer
+① 将 n-1 个圆盘从 from -> buffer
 
 <div align="center"> <img src="../pics//8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
 
-- 将 1 个圆盘从 from -> to
+② 将 1 个圆盘从 from -> to
 
 <div align="center"> <img src="../pics//2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
 
-- 将 n-1 个圆盘从 buffer -> to
+③ 将 n-1 个圆盘从 buffer -> to
 
 <div align="center"> <img src="../pics//1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
 
@@ -2281,9 +2281,9 @@ from H1 to H3
 
 ## 哈夫曼编码
 
-哈夫曼编码根据数据出现的频率对数据进行编码，从而压缩原始数据。
+根据数据出现的频率对数据进行编码，从而压缩原始数据。
 
-例如对于文本文件，其中各种字符出现的次数如下：
+例如对于一个文本文件，其中各种字符出现的次数如下：
 
 - a : 10
 - b : 20
@@ -2294,7 +2294,7 @@ from H1 to H3
 
 首先生成一颗哈夫曼树，每次生成过程中选取频率最少的两个节点，生成一个新节点作为它们的父节点，并且新节点的频率为两个节点的和。选取频率最少的原因是，生成过程使得先选取的节点在树的最底层，那么需要的编码长度更长，频率更少可以使得总编码长度更少。
 
-生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到根节点，根节点代表的字符的编码就是这个路径编码。
+生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到叶子节点，叶子节点代表的字符的编码就是这个路径编码。
 
 <div align="center"> <img src="../pics//3ff4f00a-2321-48fd-95f4-ce6001332151.png" width="400"/> </div><br>
 
diff --git a/notes/缓存.md b/notes/缓存.md
index ed834253..26d96be1 100644
--- a/notes/缓存.md
+++ b/notes/缓存.md
@@ -26,16 +26,16 @@
 
 ## 淘汰策略
 
-- FIFO（First In First Out）：先进先出策略，在实时性的场景下，需要经常访问最新的数据，那么就可以使用 FIFO，使最先进入的数据（最晚的数据）被淘汰。
+- FIFO（First In First Out）：先进先出策略，在实时性的场景下，需要经常访问最新的数据，那么就可以使用 FIFO，使得最先进入的数据（最晚的数据）被淘汰。
 
-- LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最远的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
+- LRU（Least Recently Used）：最近最久未使用策略，优先淘汰最久未使用的数据，也就是上次被访问时间距离现在最久的数据。该策略可以保证内存中的数据都是热点数据，也就是经常被访问的数据，从而保证缓存命中率。
 
 # 二、LRU
 
-以下是一个基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
+以下是基于 双向链表 + HashMap 的 LRU 算法实现，对算法的解释如下：
 
-- 最基本的思路是当访问某个节点时，将其从原来的位置删除，并重新插入到链表头部，这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就删除链表尾部的节点。
-- 为了使删除操作时间复杂度为 O(1)，那么就不能采用遍历的方式找到某个节点。HashMap 存储着 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
+- 访问某个节点时，将其从原来的位置删除，并重新插入到链表头部。这样就能保证链表尾部存储的就是最近最久未使用的节点，当节点数量大于缓存最大空间时就淘汰链表尾部的节点。
+- 为了使删除操作时间复杂度为 O(1)，就不能采用遍历的方式找到某个节点。HashMap 存储着 Key 到节点的映射，通过 Key 就能以 O(1) 的时间得到节点，然后再以 O(1) 的时间将其从双向队列中删除。
 
 ```java
 public class LRU<K, V> implements Iterable<K> {
@@ -152,7 +152,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 反向代理
 
-反向代理位于服务器之前，请求与响应都需要经过反向代理。通过将数据缓存在反向代理，在用户请求时就可以直接使用缓存进行响应。
+反向代理位于服务器之前，请求与响应都需要经过反向代理。通过将数据缓存在反向代理，在用户请求反向代理时就可以直接使用缓存进行响应。
 
 ## 本地缓存
 
@@ -166,7 +166,7 @@ public class LRU<K, V> implements Iterable<K> {
 
 ## 数据库缓存
 
-MySQL 等数据库管理系统具有自己的查询缓存机制来提高 SQL 查询效率。
+MySQL 等数据库管理系统具有自己的查询缓存机制来提高查询效率。
 
 # 四、CDN
 
@@ -193,9 +193,9 @@ CDN 主要有以下优点：
 
 ## 缓存雪崩
 
-指的是由于数据没有被加载到缓存中，或者缓存数据在同一时间大面积失效（过期），又或者缓存服务器宕机，导致大量的请求都去到达数据库。
+指的是由于数据没有被加载到缓存中，或者缓存数据在同一时间大面积失效（过期），又或者缓存服务器宕机，导致大量的请求都到达数据库。
 
-在存在缓存的系统中，系统非常依赖于缓存，缓存分担了很大一部分的数据请求。当发生缓存雪崩时，数据库无法处理这么大的请求，导致数据库崩溃。
+在有缓存的系统中，系统非常依赖于缓存，缓存分担了很大一部分的数据请求。当发生缓存雪崩时，数据库无法处理这么大的请求，导致数据库崩溃。
 
 解决方案：
 
@@ -233,7 +233,7 @@ CDN 主要有以下优点：
 
 # 七、一致性哈希
 
-Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据失效的问题。
+Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了克服传统哈希分布在服务器节点数量变化时大量数据迁移的问题。
 
 ## 基本原理
 
@@ -241,7 +241,7 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 <div align="center"> <img src="../pics//68b110b9-76c6-4ee2-b541-4145e65adb3e.jpg"/> </div><br>
 
-一致性哈希在增加或者删除节点时只会影响到哈希环中相邻的节点，例如下图中新增节点 X，只需要将它前一个节点 C 上的数据重新进行分布即可，对于节点 A、B、D 都没有影响。
+一致性哈希在增加或者删除节点时只会影响到哈希环中相邻的节点，例如下图中新增节点 X，只需要将它后一个节点 C 上的数据重新进行分布即可，对于节点 A、B、D 都没有影响。
 
 <div align="center"> <img src="../pics//66402828-fb2b-418f-83f6-82153491bcfe.jpg"/> </div><br>
 
@@ -249,7 +249,9 @@ Distributed Hash Table（DHT） 是一种哈希分布方式，其目的是为了
 
 上面描述的一致性哈希存在数据分布不均匀的问题，节点存储的数据量有可能会存在很大的不同。
 
-数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得大，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
+数据不均匀主要是因为节点在哈希环上分布的不均匀，这种情况在节点数量很少的情况下尤其明显。
+
+解决方式是通过增加虚拟节点，然后将虚拟节点映射到真实节点上。虚拟节点的数量比真实节点来得多，那么虚拟节点在哈希环上分布的均匀性就会比原来的真实节点好，从而使得数据分布也更加均匀。
 
 # 参考资料
 
diff --git a/notes/计算机网络.md b/notes/计算机网络.md
index 99a6fb43..3a08e2a0 100644
--- a/notes/计算机网络.md
+++ b/notes/计算机网络.md
@@ -59,7 +59,7 @@
 
 网络把主机连接起来，而互联网是把多种不同的网络连接起来，因此互联网是网络的网络。
 
-<div align="center"> <img src="../pics//network-of-networks.gif" width="400"/> </div><br>
+<div align="center"> <img src="../pics//network-of-networks.gif" width="450"/> </div><br>
 
 ## ISP
 
@@ -304,7 +304,7 @@ PPP 的帧格式：
 
 MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标识网络适配器（网卡）。
 
-一台主机拥有多少个适配器就有多少个 MAC 地址。例如笔记本电脑普遍存在无线网络适配器和有线网络适配器，因此就有两个 MAC 地址。
+一台主机拥有多少个网络适配器就有多少个 MAC 地址。例如笔记本电脑普遍存在无线网络适配器和有线网络适配器，因此就有两个 MAC 地址。
 
 ## 局域网
 
@@ -320,7 +320,7 @@ MAC 地址是链路层地址，长度为 6 字节（48 位），用于唯一标
 
 以太网是一种星型拓扑结构局域网。
 
-早期使用集线器进行连接，集线器是一种物理层设备，作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离，之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
+早期使用集线器进行连接，集线器是一种物理层设备， 作用于比特而不是帧，当一个比特到达接口时，集线器重新生成这个比特，并将其能量强度放大，从而扩大网络的传输距离，之后再将这个比特发送到其它所有接口。如果集线器同时收到同时从两个不同接口的帧，那么就发生了碰撞。
 
 目前以太网使用交换机替代了集线器，交换机是一种链路层设备，它不会发生碰撞，能根据 MAC 地址进行存储转发。
 
@@ -488,7 +488,7 @@ Traceroute 发送的 IP 数据报封装的是无法交付的 UDP 用户数据报
 - 172.16.0.0 \~ 172.31.255.255
 - 192.168.0.0 \~ 192.168.255.255
 
-VPN 使用公用的互联网作为本机构各专用网之间的通信载体。专用指机构内的主机只与本机构内的其它主机通信；虚拟指“好像是”，而实际上并不是，它有经过公用的互联网。
+VPN 使用公用的互联网作为本机构各专用网之间的通信载体。专用指机构内的主机只与本机构内的其它主机通信；虚拟指好像是，而实际上并不是，它有经过公用的互联网。
 
 下图中，场所 A 和 B 的通信经过互联网，如果场所 A 的主机 X 要和另一个场所 B 的主机 Y 通信，IP 数据报的源地址是 10.1.0.1，目的地址是 10.2.0.3。数据报先发送到与互联网相连的路由器 R1，R1 对内部数据进行加密，然后重新加上数据报的首部，源地址是路由器 R1 的全球地址 125.1.2.3，目的地址是路由器 R2 的全球地址 194.4.5.6。路由器 R2 收到数据报后将数据部分进行解密，恢复原来的数据报，此时目的地址为 10.2.0.3，就交付给 Y。
 
diff --git a/notes/设计模式.md b/notes/设计模式.md
index 531595e3..82c09e6d 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -44,11 +44,11 @@
 
 ## 1. 单例（Singleton）
 
-### 意图
+### Intent
 
 确保一个类只有一个实例，并提供该实例的全局访问点。
 
-### 类图
+### Class Diagram
 
 使用一个私有构造函数、一个私有静态变量以及一个公有静态函数来实现。
 
@@ -56,13 +56,13 @@
 
 <div align="center"> <img src="../pics//562f2844-d77c-40e0-887a-28a7128abd42.png"/> </div><br>
 
-### 实现
+### Implementation
 
-（一）懒汉式-线程不安全
+#### Ⅰ 懒汉式-线程不安全
 
 以下实现中，私有静态变量 uniqueInstance 被延迟实例化，这样做的好处是，如果没有用到该类，那么就不会实例化 uniqueInstance，从而节约资源。
 
-这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么会有多个线程执行 `uniqueInstance = new Singleton();` 语句，这将导致多次实例化 uniqueInstance。
+这个实现在多线程环境下是不安全的，如果多个线程能够同时进入 `if (uniqueInstance == null)` ，并且此时 uniqueInstance 为 null，那么会有多个线程执行 `uniqueInstance = new Singleton();` 语句，这将导致实例化多次 uniqueInstance。
 
 ```java
 public class Singleton {
@@ -81,9 +81,9 @@ public class Singleton {
 }
 ```
 
-（二）饿汉式-线程安全
+#### Ⅱ 饿汉式-线程安全
 
-线程不安全问题主要是由于 uniqueInstance 被多次实例化，采取直接实例化 uniqueInstance 的方式就不会产生线程不安全问题。
+线程不安全问题主要是由于 uniqueInstance 被实例化多次，采取直接实例化 uniqueInstance 的方式就不会产生线程不安全问题。
 
 但是直接实例化的方式也丢失了延迟实例化带来的节约资源的好处。
 
@@ -91,11 +91,11 @@ public class Singleton {
 private static Singleton uniqueInstance = new Singleton();
 ```
 
-（三）懒汉式-线程安全
+#### Ⅲ 懒汉式-线程安全
 
-只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了多次实例化 uniqueInstance 的问题。
+只需要对 getUniqueInstance() 方法加锁，那么在一个时间点只能有一个线程能够进入该方法，从而避免了实例化多次 uniqueInstance。
 
-但是当一个线程进入该方法之后，其它试图进入该方法的线程都必须等待，因此性能上有一定的损耗。
+但是当一个线程进入该方法之后，其它试图进入该方法的线程都必须等待，即使 uniqueInstance 已经被实例化了。这会让线程阻塞时间过程，因此该方法有性能问题，不推荐使用。
 
 ```java
 public static synchronized Singleton getUniqueInstance() {
@@ -106,7 +106,7 @@ public static synchronized Singleton getUniqueInstance() {
 }
 ```
 
-（四）双重校验锁-线程安全
+#### Ⅳ 双重校验锁-线程安全
 
 uniqueInstance 只需要被实例化一次，之后就可以直接使用了。加锁操作只需要对实例化那部分的代码进行，只有当 uniqueInstance 没有被实例化时，才需要进行加锁。
 
@@ -133,7 +133,7 @@ public class Singleton {
 }
 ```
 
-考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程同时执行 if 语句，那么两个线程就会同时进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，那么就会进行两次实例化，从而产生了两个实例。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
+考虑下面的实现，也就是只使用了一个 if 语句。在 uniqueInstance == null 的情况下，如果两个线程都执行了 if 语句，那么两个线程都会进入 if 语句块内。虽然在 if 语句块内有加锁操作，但是两个线程都会执行 `uniqueInstance = new Singleton();` 这条语句，只是先后的问题，那么就会进行两次实例化。因此必须使用双重校验锁，也就是需要使用两个 if 语句。
 
 ```java
 if (uniqueInstance == null) {
@@ -145,19 +145,19 @@ if (uniqueInstance == null) {
 
 uniqueInstance 采用 volatile 关键字修饰也是很有必要的。`uniqueInstance = new Singleton();` 这段代码其实是分为三步执行。
 
-1. 分配内存空间
-2. 初始化对象
+1. 为 uniqueInstance 分配内存空间
+2. 初始化 uniqueInstance
 3. 将 uniqueInstance 指向分配的内存地址
 
-但是由于 JVM 具有指令重排的特性，有可能执行顺序变为了 1>3>2，这在单线程情况下自然是没有问题。但如果是多线程下，有可能获得是一个还没有被初始化的实例，以致于程序出错。
+但是由于 JVM 具有指令重排的特性，执行顺序有可能变成 1>3>2。指令重排在单线程环境下不会出先问题，但是在多线程环境下会导致一个线程获得还没有初始化的实例。例如，线程 T<sub>1</sub> 执行了 1 和 3，此时 T<sub>2</sub> 调用 getUniqueInstance() 后发现 uniqueInstance 不为空，因此返回 uniqueInstance，但此时 uniqueInstance 还未被初始化。
 
 使用 volatile 可以禁止 JVM 的指令重排，保证在多线程环境下也能正常运行。
 
-（五）静态内部类实现
+#### Ⅴ 静态内部类实现
 
-当 Singleton 类加载时，静态内部类 SingletonHolder 没有被加载进内存。只有当调用 `getUniqueInstance()` 方法从而触发 `SingletonHolder.INSTANCE` 时 SingletonHolder 才会被加载，此时初始化 INSTANCE 实例。
+当 Singleton 类加载时，静态内部类 SingletonHolder 没有被加载进内存。只有当调用 `getUniqueInstance()` 方法从而触发 `SingletonHolder.INSTANCE` 时 SingletonHolder 才会被加载，此时初始化 INSTANCE 实例，并且 JVM 能确保 INSTANCE 只被实例化一次。
 
-这种方式不仅具有延迟初始化的好处，而且由虚拟机提供了对线程安全的支持。
+这种方式不仅具有延迟初始化的好处，而且由 JVM 提供了对线程安全的支持。
 
 ```java
 public class Singleton {
@@ -175,40 +175,11 @@ public class Singleton {
 }
 ```
 
-（六）枚举实现
+该实现在多次序列化再进行反序列化之后，不会得到多个实例。而其它实现，为了保证不会出现反序列化之后出现多个实例，需要使用 transient 修饰所有字段，并且实现序列化和反序列化的方法。
 
-这是单例模式的最佳实践，它实现简单，并且在面对复杂的序列化或者反射攻击的时候，能够防止实例化多次。
+该实现可以防止反射攻击。在其它实现中，通过 setAccessible() 方法可以将私有构造函数的访问级别设置为 public，然后调用构造函数从而实例化对象，如果要防止这种攻击，需要在构造函数中添加防止实例化第二个对象的代码。但是该实现是由 JVM 保证只会实例化一次，因此不会出现上述的反射攻击。
 
-```java
-public enum Singleton {
-    uniqueInstance;
-}
-```
-
-考虑以下单例模式的实现，该 Singleton 在每次序列化的时候都会创建一个新的实例，为了保证只创建一个实例，必须声明所有字段都是 transient，并且提供一个 readResolve() 方法。
-
-```java
-public class Singleton implements Serializable {
-
-    private static Singleton uniqueInstance;
-
-    private Singleton() {
-    }
-
-    public static synchronized Singleton getUniqueInstance() {
-        if (uniqueInstance == null) {
-            uniqueInstance = new Singleton();
-        }
-        return uniqueInstance;
-    }
-}
-```
-
-如果不使用枚举来实现单例模式，会出现反射攻击，因为通过 setAccessible() 方法可以将私有构造函数的访问级别设置为 public，然后调用构造函数从而实例化对象。如果要防止这种攻击，需要在构造函数中添加防止实例化第二个对象的代码。
-
-从上面的讨论可以看出，解决序列化和反射攻击很麻烦，而枚举实现不会出现这两种问题，所以说枚举实现单例模式是最佳实践。
-
-### 使用场景
+### Examples
 
 - Logger Classes
 - Configuration Classes
@@ -223,11 +194,11 @@ public class Singleton implements Serializable {
 
 ## 2. 简单工厂（Simple Factory）
 
-### 意图
+### Intent
 
 在创建一个对象时不向客户暴露内部细节，并提供一个创建对象的通用接口。
 
-### 类图
+### Class Diagram
 
 简单工厂不是设计模式，更像是一种编程习惯。它把实例化的操作单独放到一个类中，这个类就成为简单工厂类，让简单工厂类来决定应该用哪个具体子类来实例化。
 
@@ -235,7 +206,7 @@ public class Singleton implements Serializable {
 
 <div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public interface Product {
@@ -303,11 +274,11 @@ public class Client {
 
 ## 3. 工厂方法（Factory Method）
 
-### 意图
+### Intent
 
 定义了一个创建对象的接口，但由子类决定要实例化哪个类。工厂方法把实例化操作推迟到子类。
 
-### 类图
+### Class Diagram
 
 在简单工厂中，创建对象的是另一个类，而在工厂方法中，是由子类来创建对象。
 
@@ -315,7 +286,7 @@ public class Client {
 
 <div align="center"> <img src="../pics//1818e141-8700-4026-99f7-900a545875f5.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public abstract class Factory {
@@ -363,11 +334,11 @@ public class ConcreteFactory2 extends Factory {
 
 ## 4. 抽象工厂（Abstract Factory）
 
-### 意图
+### Intent
 
 提供一个接口，用于创建  **相关的对象家族** 。
 
-### 类图
+### Class Diagram
 
 抽象工厂模式创建的是对象家族，也就是很多对象而不是一个对象，并且这些对象是相关的，也就是说必须一起创建出来。而工厂方法模式只是用于创建一个对象，这和抽象工厂模式有很大不同。
 
@@ -379,7 +350,7 @@ public class ConcreteFactory2 extends Factory {
 
 <div align="center"> <img src="../pics//8668a3e1-c9c7-4fcb-98b2-a96a5d841579.png"/> </div><br>
 
-### 代码实现
+### Implementation
 
 ```java
 public class AbstractProductA {
@@ -461,15 +432,15 @@ public class Client {
 
 ## 5. 生成器（Builder）
 
-### 意图
+### Intent
 
 封装一个对象的构造过程，并允许按步骤构造。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//13b0940e-d1d7-4b17-af4f-b70cb0a75e08.png"/> </div><br>
 
-### 实现
+### Implementation
 
 以下是一个简易的 StringBuilder 实现，参考了 JDK 1.8 源码。
 
@@ -551,15 +522,15 @@ abcdefghijklmnopqrstuvwxyz
 
 ## 6. 原型模式（Prototype）
 
-### 意图
+### Intent
 
 使用原型实例指定要创建对象的类型，通过复制这个原型来创建新对象。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//a40661e4-1a71-46d2-a158-ff36f7fc3331.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public abstract class Prototype {
@@ -610,17 +581,17 @@ abc
 
 ## 1. 责任链（Chain Of Responsibility）
 
-### 意图
+### Intent
 
 使多个对象都有机会处理请求，从而避免请求的发送者和接收者之间的耦合关系。将这些对象连成一条链，并沿着这条链发送该请求，直到有一个对象处理它为止。
 
-### 类图
+### Class Diagram
 
 - Handler：定义处理请求的接口，并且实现后继链（successor）
 
 <div align="center"> <img src="../pics//691f11eb-31a7-46be-9de1-61f433c4b3c7.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public abstract class Handler {
@@ -724,11 +695,16 @@ request2 is handle by ConcreteHandler2
 
 ## 2. 命令（Command）
 
-### 意图
+### Intent
 
-将命令封装成对象中，以便使用命令来参数化其它对象，或者将命令对象放入队列中进行排队，或者将命令对象的操作记录到日志中，以及支持可撤销的操作。
+将命令封装成对象中，具有以下作用：
 
-### 类图
+- 使用命令来参数化其它对象
+- 将命令放入队列中进行排队
+- 将命令的操作记录到日志中
+- 支持可撤销的操作
+
+### Class Diagram
 
 - Command：命令
 - Receiver：命令接收者，也就是命令真正的执行者
@@ -737,7 +713,7 @@ request2 is handle by ConcreteHandler2
 
 <div align="center"> <img src="../pics//ae1b27b8-bc13-42e7-ac12-a2242e125499.png"/> </div><br>
 
-### 实现
+### Implementation
 
 设计一个遥控器，可以控制电灯开关。
 
@@ -847,18 +823,18 @@ public class Client {
 
 ## 3. 解释器（Interpreter）
 
-### 意图
+### Intent
 
 为语言创建解释器，通常由语言的语法和语法分析来定义。
 
-### 类图
+### Class Diagram
 
-- TerminalExpression：终结符表达式，每个终结符都需要一个 TerminalExpression
-- Context：上下文，包含解释器之外的一些全局信息
+- TerminalExpression：终结符表达式，每个终结符都需要一个 TerminalExpression。
+- Context：上下文，包含解释器之外的一些全局信息。
 
 <div align="center"> <img src="../pics//794239e3-4baf-4aad-92df-f02f59b2a6fe.png"/> </div><br>
 
-### 实现
+### Implementation
 
 以下是一个规则检验器实现，具有 and 和 or 规则，通过规则可以构建一颗解析树，用来检验一个文本是否满足解析树定义的规则。
 
@@ -971,11 +947,11 @@ false
 
 ## 4. 迭代器（Iterator）
 
-### 意图
+### Intent
 
 提供一种顺序访问聚合对象元素的方法，并且不暴露聚合对象的内部表示。
 
-### 类图
+### Class Diagram
 
 - Aggregate 是聚合类，其中 createIterator() 方法可以产生一个 Iterator；
 - Iterator 主要定义了 hasNext() 和 next() 方法。
@@ -983,7 +959,7 @@ false
 
 <div align="center"> <img src="../pics//b0f61ac2-a4b6-4042-9cf0-ccf4238c1ff7.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public interface Aggregate {
@@ -1059,18 +1035,18 @@ public class Client {
 
 ## 5. 中介者（Mediator）
 
-### 意图
+### Intent
 
 集中相关对象之间复杂的沟通和控制方式。
 
-### 类图
+### Class Diagram
 
 - Mediator：中介者，定义一个接口用于与各同事（Colleague）对象通信。
 - Colleague：同事，相关对象
 
 <div align="center"> <img src="../pics//d0afdd23-c9a5-4d1c-9b3d-404bff3bd0d1.png"/> </div><br>
 
-### 实现
+### Implementation
 
 Alarm（闹钟）、CoffeePot（咖啡壶）、Calendar（日历）、Sprinkler（喷头）是一组相关的对象，在某个对象的事件产生时需要去操作其它对象，形成了下面这种依赖结构：
 
@@ -1228,11 +1204,11 @@ doSprinkler()
 
 ## 6. 备忘录（Memento）
 
-### 意图
+### Intent
 
 在不违反封装的情况下获得对象的内部状态，从而在需要时可以将对象恢复到最初状态。
 
-### 类图
+### Class Diagram
 
 - Originator：原始对象
 - Caretaker：负责保存好备忘录
@@ -1240,7 +1216,7 @@ doSprinkler()
 
 <div align="center"> <img src="../pics//867e93eb-3161-4f39-b2d2-c0cd3788e194.png"/> </div><br>
 
-### 实现
+### Implementation
 
 以下实现了一个简单计算器程序，可以输入两个值，然后计算这两个值的和。备忘录模式允许将这两个值存储起来，然后在某个时刻用存储的状态进行恢复。
 
@@ -1405,7 +1381,7 @@ public class Client {
 
 ## 7. 观察者（Observer）
 
-### 意图
+### Intent
 
 定义对象之间的一对多依赖，当一个对象状态改变时，它的所有依赖都会收到通知并且自动更新状态。
 
@@ -1413,7 +1389,7 @@ public class Client {
 
 <div align="center"> <img src="../pics//7a3c6a30-c735-4edb-8115-337288a4f0f2.jpg" width="600"/> </div><br>
 
-### 类图
+### Class Diagram
 
 主题（Subject）具有注册和移除观察者、并通知所有观察者的功能，主题是通过维护一张观察者列表来实现这些操作的。
 
@@ -1421,7 +1397,7 @@ public class Client {
 
 <div align="center"> <img src="../pics//0df5d84c-e7ca-4e3a-a688-bb8e68894467.png"/> </div><br>
 
-### 实现
+### Implementation
 
 天气数据布告板会在天气信息发生改变时更新其内容，布告板有多个，并且在将来会继续增加。
 
@@ -1540,15 +1516,15 @@ StatisticsDisplay.update: 1.0 1.0 1.0
 
 ## 8. 状态（State）
 
-### 意图
+### Intent
 
 允许对象在内部状态改变时改变它的行为，对象看起来好像修改了它所属的类。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//c5085437-54df-4304-b62d-44b961711ba7.png"/> </div><br>
 
-### 实现
+### Implementation
 
 糖果销售机有多种状态，每种状态下销售机有不同的行为，状态可以发生转移，使得销售机的行为也发生改变。
 
@@ -1842,16 +1818,16 @@ No gumball dispensed
 
 ## 9. 策略（Strategy）
 
-### 意图
+### Intent
 
 定义一系列算法，封装每个算法，并使它们可以互换。
 
 策略模式可以让算法独立于使用它的客户端。
 
-### 类图
+### Class Diagram
 
 - Strategy 接口定义了一个算法族，它们都具有 behavior() 方法。
-- Context 是使用到该算法族的类，其中的 doSomething() 方法会调用 behavior()，setStrategy(in Strategy) 方法可以动态地改变 strategy 对象，也就是说能动态地改变 Context 所使用的算法。
+- Context 是使用到该算法族的类，其中的 doSomething() 方法会调用 behavior()，setStrategy(Strategy) 方法可以动态地改变 strategy 对象，也就是说能动态地改变 Context 所使用的算法。
 
 <div align="center"> <img src="../pics//1fc969e4-0e7c-441b-b53c-01950d2f2be5.png"/> </div><br>
 
@@ -1861,7 +1837,7 @@ No gumball dispensed
 
 状态模式主要是用来解决状态转移的问题，当状态发生转移了，那么 Context 对象就会改变它的行为；而策略模式主要是用来封装一组可以互相替代的算法族，并且可以根据需要动态地去替换 Context 使用的算法。
 
-### 实现
+### Implementation
 
 设计一个鸭子，它可以动态地改变叫声。这里的算法族是鸭子的叫声行为。
 
@@ -1930,17 +1906,17 @@ quack!
 
 ## 10. 模板方法（Template Method）
 
-### 意图
+### Intent
 
 定义算法框架，并将一些步骤的实现延迟到子类。
 
 通过模板方法，子类可以重新定义算法的某些步骤，而不用改变算法的结构。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//c3c1c0e8-3a78-4426-961f-b46dd0879dd8.png"/> </div><br>
 
-### 实现
+### Implementation
 
 冲咖啡和冲茶都有类似的流程，但是某些步骤会有点不一样，要求复用那些相同步骤的代码。
 
@@ -2031,11 +2007,11 @@ Tea.addCondiments
 
 ## 11. 访问者（Visitor）
 
-### 意图
+### Intent
 
 为一个对象结构（比如组合结构）增加新能力。
 
-### 类图
+### Class Diagram
 
 - Visitor：访问者，为每一个 ConcreteElement 声明一个 visit 操作
 - ConcreteVisitor：具体访问者，存储遍历过程中的累计结果
@@ -2043,7 +2019,7 @@ Tea.addCondiments
 
 <div align="center"> <img src="../pics//ec923dc7-864c-47b0-a411-1f2c48d084de.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public interface Element {
@@ -2238,17 +2214,17 @@ Number of items:     6
 
 ## 12. 空对象（Null）
 
-### 意图
+### Intent
 
 使用什么都不做的空对象来代替 NULL。
 
 一个方法返回 NULL，意味着方法的调用端需要去检查返回值是否是 NULL，这么做会导致非常多的冗余的检查代码。并且如果某一个调用端忘记了做这个检查返回值，而直接使用返回的对象，那么就有可能抛出空指针异常。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//dd3b289c-d90e-44a6-a44c-4880517eb1de.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public abstract class AbstractOperation {
@@ -2294,17 +2270,17 @@ public class Client {
 
 ## 1. 适配器（Adapter）
 
-### 意图
+### Intent
 
 把一个类接口转换成另一个用户需要的接口。
 
 <div align="center"> <img src="../pics//3d5b828e-5c4d-48d8-a440-281e4a8e1c92.png"/> </div><br>
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//0f754c1d-b5cb-48cd-90e0-4a86034290a1.png"/> </div><br>
 
-### 实现
+### Implementation
 
 鸭子（Duck）和火鸡（Turkey）拥有不同的叫声，Duck 的叫声调用 quack() 方法，而 Turkey 调用 gobble() 方法。
 
@@ -2365,18 +2341,18 @@ public class Client {
 
 ## 2. 桥接（Bridge）
 
-### 意图
+### Intent
 
 将抽象与实现分离开来，使它们可以独立变化。
 
-### 类图
+### Class Diagram
 
 - Abstraction：定义抽象类的接口
 - Implementor：定义实现类接口
 
 <div align="center"> <img src="../pics//c2cbf5d2-82af-4c78-bd43-495da5adf55f.png"/> </div><br>
 
-### 实现
+### Implementation
 
 RemoteControl 表示遥控器，指代 Abstraction。
 
@@ -2518,11 +2494,11 @@ public class Client {
 
 ## 3. 组合（Composite）
 
-### 意图
+### Intent
 
 将对象组合成树形结构来表示“整体/部分”层次关系，允许用户以相同的方式处理单独对象和组合对象。
 
-### 类图
+### Class Diagram
 
 组件（Component）类是组合类（Composite）和叶子类（Leaf）的父类，可以把组合类看成是树的中间节点。
 
@@ -2530,7 +2506,7 @@ public class Client {
 
 <div align="center"> <img src="../pics//3fb5b255-b791-45b6-8754-325c8741855a.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public abstract class Component {
@@ -2652,17 +2628,17 @@ Composite:root
 
 ## 4. 装饰（Decorator）
 
-### 意图
+### Intent
 
 为对象动态添加功能。
 
-### 类图
+### Class Diagram
 
 装饰者（Decorator）和具体组件（ConcreteComponent）都继承自组件（Component），具体组件的方法实现不需要依赖于其它对象，而装饰者组合了一个组件，这样它可以装饰其它装饰者或者具体组件。所谓装饰，就是把这个装饰者套在被装饰者之上，从而动态扩展被装饰者的功能。装饰者的方法有一部分是自己的，这属于它的功能，然后调用被装饰者的方法实现，从而也保留了被装饰者的功能。可以看到，具体组件应当是装饰层次的最低层，因为只有具体组件的方法实现不需要依赖于其它对象。
 
 <div align="center"> <img src="../pics//137c593d-0a9e-47b8-a9e6-b71f540b82dd.png"/> </div><br>
 
-### 实现
+### Implementation
 
 设计不同种类的饮料，饮料可以添加配料，比如可以添加牛奶，并且支持动态添加新配料。每增加一种配料，该饮料的价格就会增加，要求计算一种饮料的价格。
 
@@ -2759,15 +2735,15 @@ public class Client {
 
 ## 5. 外观（Facade）
 
-### 意图
+### Intent
 
 提供了一个统一的接口，用来访问子系统中的一群接口，从而让子系统更容易使用。
 
-### 类图
+### Class Diagram
 
 <div align="center"> <img src="../pics//f9978fa6-9f49-4a0f-8540-02d269ac448f.png"/> </div><br>
 
-### 实现
+### Implementation
 
 观看电影需要操作很多电器，使用外观模式实现一键看电影功能。
 
@@ -2814,11 +2790,11 @@ public class Client {
 
 ## 6. 享元（Flyweight）
 
-### 意图
+### Intent
 
 利用共享的方式来支持大量细粒度的对象，这些对象一部分内部状态是相同的。
 
-### 类图
+### Class Diagram
 
 - Flyweight：享元对象
 - IntrinsicState：内部状态，享元对象共享内部状态
@@ -2826,7 +2802,7 @@ public class Client {
 
 <div align="center"> <img src="../pics//d52270b4-9097-4667-9f18-f405fc661c99.png"/> </div><br>
 
-### 实现
+### Implementation
 
 ```java
 public interface Flyweight {
@@ -2899,11 +2875,11 @@ Java 利用缓存来加速大量小对象的访问时间。
 
 ## 7. 代理（Proxy）
 
-### 意图
+### Intent
 
 控制对其它对象的访问。
 
-### 类图
+### Class Diagram
 
 代理有以下四类：
 
@@ -2914,7 +2890,7 @@ Java 利用缓存来加速大量小对象的访问时间。
 
 <div align="center"> <img src="../pics//a6c20f60-5eba-427d-9413-352ada4b40fe.png"/> </div><br>
 
-### 实现
+### Implementation
 
 以下是一个虚拟代理的实现，模拟了图片延迟加载的情况下使用与图片大小相等的临时内容去替换原始图片，直到图片加载完成才将图片显示出来。
 
diff --git a/notes/重构.md b/notes/重构.md
index b5a834e0..e90fa84e 100644
--- a/notes/重构.md
+++ b/notes/重构.md
@@ -161,6 +161,7 @@ class Customer {
 
 ```java
 class Rental {
+
     private int daysRented;
 
     private Movie movie;
@@ -199,6 +200,7 @@ class Movie {
 
 ```java
 public class App {
+
     public static void main(String[] args) {
         Customer customer = new Customer();
         Rental rental1 = new Rental(1, new Movie(Movie.Type1));
@@ -236,6 +238,7 @@ public class App {
 
 ```java
 class Customer {
+
     private List<Rental> rentals = new ArrayList<>();
 
     void addRental(Rental rental) {
@@ -254,6 +257,7 @@ class Customer {
 
 ```java
 class Rental {
+
     private int daysRented;
 
     private Movie movie;
@@ -294,8 +298,6 @@ class Price2 implements Price {
 ```
 
 ```java
-package imp2;
-
 class Price3 implements Price {
     @Override
     public double getCharge() {
diff --git a/notes/集群.md b/notes/集群.md
index 15657e43..338ddcb4 100644
--- a/notes/集群.md
+++ b/notes/集群.md
@@ -119,7 +119,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 首先了解一下正向代理与反向代理的区别：
 
 - 正向代理：发生在客户端，是由用户主动发起的。比如翻墙，客户端通过主动访问代理服务器，让代理服务器获得需要的外网数据，然后转发回客户端；
-- 反向代理：发生在服务器端，用户不知道代理的存在。
+- 反向代理：发生在服务器端，用户不知道反向代理的存在。
 
 反向代理服务器位于源服务器前面，用户的请求需要先经过反向代理服务器才能到达源服务器。反向代理可以用来进行缓存、日志记录等，同时也可以用来做为负载均衡服务器。
 
@@ -153,7 +153,7 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 通过配置源服务器的虚拟 IP 地址和负载均衡服务器的 IP 地址一致，从而不需要修改 IP 地址就可以进行转发。也正因为 IP 地址一样，所以源服务器的响应不需要转发回负载均衡服务器，可以直接转发给客户端，避免了负载均衡服务器的成为瓶颈。
 
-这是一种三角传输模式，被称为直接路由，对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
+这是一种三角传输模式，被称为直接路由。对于提供下载和视频服务的网站来说，直接路由避免了大量的网络传输数据经过负载均衡服务器。
 
 这是目前大型网站使用最广负载均衡转发方式，在 Linux 平台可以使用的负载均衡服务器为 LVS（Linux Virtual Server）。
 
@@ -185,7 +185,6 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 - 占用过多内存；
 - 同步过程占用网络带宽以及服务器处理器时间。
 
-
 <div align="center"> <img src="../pics//MultiNode-SessionReplication.jpg"/> </div><br>
 
 ## Session Server
@@ -206,3 +205,4 @@ HTTP 重定向负载均衡服务器使用某种负载均衡算法计算得到服
 
 - [Session Management using Spring Session with JDBC DataStore](https://sivalabs.in/2018/02/session-management-using-spring-session-jdbc-datastore/)
 
+
diff --git a/notes/面向对象思想.md b/notes/面向对象思想.md
index 203c7ebf..87b3b033 100644
--- a/notes/面向对象思想.md
+++ b/notes/面向对象思想.md
@@ -89,24 +89,28 @@ Animal animal = new Cat();
 
 ```java
 public class Instrument {
+
     public void play() {
         System.out.println("Instument is playing...");
     }
 }
 
 public class Wind extends Instrument {
+
     public void play() {
         System.out.println("Wind is playing...");
     }
 }
 
 public class Percussion extends Instrument {
+
     public void play() {
         System.out.println("Percussion is playing...");
     }
 }
 
 public class Music {
+
     public static void main(String[] args) {
         List<Instrument> instruments = new ArrayList<>();
         instruments.add(new Wind());
@@ -353,4 +357,3 @@ Vihicle .. N
 - [看懂 UML 类图和时序图](http://design-patterns.readthedocs.io/zh_CN/latest/read_uml.html#generalization)
 - [UML 系列——时序图（顺序图）sequence diagram](http://www.cnblogs.com/wolf-sun/p/UML-Sequence-diagram.html)
 - [面向对象编程三大特性 ------ 封装、继承、多态](http://blog.csdn.net/jianyuerensheng/article/details/51602015)
-

From f3d69c363fad4363863fb8436e59a41619ffbf5d Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 1 Sep 2018 01:23:33 +0800
Subject: [PATCH 165/172] auto commit

---
 notes/设计模式.md | 5 -----
 1 file changed, 5 deletions(-)

diff --git a/notes/设计模式.md b/notes/设计模式.md
index eb838da7..82c09e6d 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -699,11 +699,6 @@ request2 is handle by ConcreteHandler2
 
 将命令封装成对象中，具有以下作用：
 
-<<<<<<< HEAD
-=======
-将命令封装成对象中，具有以下作用：
-
->>>>>>> bb2592f820012d6220462f6a18a28d80983aa3f0
 - 使用命令来参数化其它对象
 - 将命令放入队列中进行排队
 - 将命令的操作记录到日志中

From 66bc75d5d17cf09f8a0d747459b8c3336be28adb Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 1 Sep 2018 15:07:50 +0800
Subject: [PATCH 166/172] auto commit

---
 notes/设计模式.md | 12 +++++++++---
 1 file changed, 9 insertions(+), 3 deletions(-)

diff --git a/notes/设计模式.md b/notes/设计模式.md
index 82c09e6d..9d9ab38f 100644
--- a/notes/设计模式.md
+++ b/notes/设计模式.md
@@ -202,7 +202,7 @@ public class Singleton {
 
 简单工厂不是设计模式，更像是一种编程习惯。它把实例化的操作单独放到一个类中，这个类就成为简单工厂类，让简单工厂类来决定应该用哪个具体子类来实例化。
 
-这样做能把客户类和具体子类的实现解耦，客户类不再需要知道有哪些子类以及应当实例化哪个子类。因为客户类往往有多个，如果不使用简单工厂，所有的客户类都要知道所有子类的细节。而且一旦子类发生改变，例如增加子类，那么所有的客户类都要进行修改。
+这样做能把客户类和具体子类的实现解耦，客户类不再需要知道有哪些子类以及应当实例化哪个子类。客户类往往有多个，如果不使用简单工厂，那么所有的客户类都要知道所有子类的细节。而且一旦子类发生改变，例如增加子类，那么所有的客户类都要进行修改。
 
 <div align="center"> <img src="../pics//c79da808-0f28-4a36-bc04-33ccc5b83c13.png"/> </div><br>
 
@@ -228,7 +228,7 @@ public class ConcreteProduct2 implements Product {
 }
 ```
 
-以下的 Client 类中包含了实例化的代码，这是一种错误的实现，如果在客户类中存在实例化代码，就需要将代码放到简单工厂中。
+以下的 Client 类包含了实例化的代码，这是一种错误的实现。如果在客户类中存在这种实例化代码，就需要考虑将代码放到简单工厂中。
 
 ```java
 public class Client {
@@ -988,6 +988,7 @@ public class ConcreteAggregate implements Aggregate {
 
 ```java
 public interface Iterator<Item> {
+
     Item next();
 
     boolean hasNext();
@@ -1018,6 +1019,7 @@ public class ConcreteIterator<Item> implements Iterator {
 
 ```java
 public class Client {
+
     public static void main(String[] args) {
         Aggregate aggregate = new ConcreteAggregate();
         Iterator<Integer> iterator = aggregate.createIterator();
@@ -1826,7 +1828,7 @@ No gumball dispensed
 
 ### Class Diagram
 
-- Strategy 接口定义了一个算法族，它们都具有 behavior() 方法。
+- Strategy 接口定义了一个算法族，它们都实现了  behavior() 方法。
 - Context 是使用到该算法族的类，其中的 doSomething() 方法会调用 behavior()，setStrategy(Strategy) 方法可以动态地改变 strategy 对象，也就是说能动态地改变 Context 所使用的算法。
 
 <div align="center"> <img src="../pics//1fc969e4-0e7c-441b-b53c-01950d2f2be5.png"/> </div><br>
@@ -1867,6 +1869,7 @@ public class Squeak implements QuackBehavior{
 
 ```java
 public class Duck {
+
     private QuackBehavior quackBehavior;
 
     public void performQuack() {
@@ -1883,6 +1886,7 @@ public class Duck {
 
 ```java
 public class Client {
+
     public static void main(String[] args) {
         Duck duck = new Duck();
         duck.setQuackBehavior(new Squeak());
@@ -2706,6 +2710,7 @@ public class Mocha extends CondimentDecorator {
 
 ```java
 public class Client {
+
     public static void main(String[] args) {
         Beverage beverage = new HouseBlend();
         beverage = new Mocha(beverage);
@@ -2845,6 +2850,7 @@ public class FlyweightFactory {
 
 ```java
 public class Client {
+
     public static void main(String[] args) {
         FlyweightFactory factory = new FlyweightFactory();
         Flyweight flyweight1 = factory.getFlyweight("aa");

From 7ec044276a91bdd6edecdac26a4ac356ab626549 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 1 Sep 2018 17:09:59 +0800
Subject: [PATCH 167/172] auto commit

---
 notes/计算机操作系统.md | 43 +++++++++++++++++++++++++++--------------
 notes/面向对象思想.md   |  4 ++--
 2 files changed, 30 insertions(+), 17 deletions(-)

diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 73f67c62..44f8eb35 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -15,6 +15,10 @@
 * [三、死锁](#三死锁)
     * [死锁的必要条件](#死锁的必要条件)
     * [死锁的处理方法](#死锁的处理方法)
+    * [鸵鸟策略](#鸵鸟策略)
+    * [死锁检测与死锁恢复](#死锁检测与死锁恢复)
+    * [死锁预防](#死锁预防)
+    * [死锁避免](#死锁避免)
 * [四、内存管理](#四内存管理)
     * [虚拟内存](#虚拟内存)
     * [分页系统地址映射](#分页系统地址映射)
@@ -603,19 +607,28 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 ## 死锁的处理方法
 
-### 1. 鸵鸟策略
+主要有以下四种方法：
+
+- 鸵鸟策略；
+- 死锁检测与死锁恢复
+- 死锁预防
+- 死锁避免
+
+## 鸵鸟策略
 
 把头埋在沙子里，假装根本没发生问题。
 
-因为解决死锁问题的代价很高，因此鸵鸟策略这种不采取任务措施的方案会获得更高的性能。当发生死锁时不会对用户造成多大影响，或发生死锁的概率很低，可以采用鸵鸟策略。
+因为解决死锁问题的代价很高，因此鸵鸟策略这种不采取任务措施的方案会获得更高的性能。
+
+当发生死锁时不会对用户造成多大影响，或发生死锁的概率很低，可以采用鸵鸟策略。
 
 大多数操作系统，包括 Unix，Linux 和 Windows，处理死锁问题的办法仅仅是忽略它。
 
-### 2. 死锁检测与死锁恢复
+## 死锁检测与死锁恢复
 
 不试图阻止死锁，而是当检测到死锁发生时，采取措施进行恢复。
 
-（一）每种类型一个资源的死锁检测
+### 1. 每种类型一个资源的死锁检测
 
 <div align="center"> <img src="../pics//b1fa0453-a4b0-4eae-a352-48acca8fff74.png"/> </div><br>
 
@@ -625,7 +638,7 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 每种类型一个资源的死锁检测算法是通过检测有向图是否存在环来实现，从一个节点出发进行深度优先搜索，对访问过的节点进行标记，如果访问了已经标记的节点，就表示有向图存在环，也就是检测到死锁的发生。
 
-（二）每种类型多个资源的死锁检测
+### 2. 每种类型多个资源的死锁检测
 
 <div align="center"> <img src="../pics//e1eda3d5-5ec8-4708-8e25-1a04c5e11f48.png"/> </div><br>
 
@@ -646,35 +659,35 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 2. 如果找到了这样一个进程，那么将 C 矩阵的第 i 行向量加到 A 中，标记该进程，并转回 1。
 3. 如果没有这样一个进程，算法终止。
 
-（三）死锁恢复
+### 3. 死锁恢复
 
 - 利用抢占恢复
 - 利用回滚恢复
 - 通过杀死进程恢复
 
-### 3. 死锁预防
+## 死锁预防
 
 在程序运行之前预防发生死锁。
 
-（一）破坏互斥条件
+### 1. 破坏互斥条件
 
 例如假脱机打印机技术允许若干个进程同时输出，唯一真正请求物理打印机的进程是打印机守护进程。
 
-（二）破坏占有和等待条件
+### 2. 破坏占有和等待条件
 
 一种实现方式是规定所有进程在开始执行前请求所需要的全部资源。
 
-（三）破坏不可抢占条件
+### 3. 破坏不可抢占条件
 
-（四）破坏环路等待
+### 4. 破坏环路等待
 
 给资源统一编号，进程只能按编号顺序来请求资源。
 
-### 4. 死锁避免
+## 死锁避免
 
 在程序运行时避免发生死锁。
 
-（一）安全状态
+### 1. 安全状态
 
 <div align="center"> <img src="../pics//ed523051-608f-4c3f-b343-383e2d194470.png"/> </div><br>
 
@@ -684,7 +697,7 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 安全状态的检测与死锁的检测类似，因为安全状态必须要求不能发生死锁。下面的银行家算法与死锁检测算法非常类似，可以结合着做参考对比。
 
-（二）单个资源的银行家算法
+### 2. 单个资源的银行家算法
 
 一个小城镇的银行家，他向一群客户分别承诺了一定的贷款额度，算法要做的是判断对请求的满足是否会进入不安全状态，如果是，就拒绝请求；否则予以分配。
 
@@ -692,7 +705,7 @@ FIFO 常用于客户-服务器应用程序中，FIFO 用作汇聚点，在客户
 
 上图 c 为不安全状态，因此算法会拒绝之前的请求，从而避免进入图 c 中的状态。
 
-（三）多个资源的银行家算法
+### 3. 多个资源的银行家算法
 
 <div align="center"> <img src="../pics//62e0dd4f-44c3-43ee-bb6e-fedb9e068519.png"/> </div><br>
 
diff --git a/notes/面向对象思想.md b/notes/面向对象思想.md
index 87b3b033..667ba3e1 100644
--- a/notes/面向对象思想.md
+++ b/notes/面向对象思想.md
@@ -235,9 +235,9 @@ School "1" - "n" Student
 
 和关联关系不同的是，依赖关系是在运行过程中起作用的。A 类和 B 类是依赖关系主要有三种形式：
 
-- A 类是 B 类中的（某中方法的）局部变量；
+- A 类是 B 类方法的局部变量；
 - A 类是 B 类方法当中的一个参数；
-- A 类向 B 类发送消息，从而影响 B 类发生变化；
+- A 类向 B 类发送消息，从而影响 B 类发生变化。
 
 <div align="center"> <img src="../pics//LOun2W9134NxVugmbJPp15d4LalxC4O.png"/> </div><br>
 

From b0b9fbe8c6bdad302ecf83055187d4e40c128686 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Sat, 1 Sep 2018 17:23:28 +0800
Subject: [PATCH 168/172] Update README.md

---
 README.md | 8 +++++---
 1 file changed, 5 insertions(+), 3 deletions(-)

diff --git a/README.md b/README.md
index 27edaf24..6a301d34 100644
--- a/README.md
+++ b/README.md
@@ -165,11 +165,13 @@
 
 #### About
 
-这个仓库是笔者的一个学习笔记，主要总结一些比较重要的知识点，希望对大家有所帮助。
+本仓库主要是根据计算机经典书籍以及官方技术文档进行总结的学习笔记，希望对大家有所帮助。
 
-笔记不是从网上到处复制粘贴拼凑而来，虽然有少部分内容会直接引入书上原文或者官方技术文档的原文，但是没有直接摘抄其他人的博客文章，只做了参考，参考的文章会在最后给出链接。
+学习笔记不是从网上到处拼凑而来，除了少部分引用书上和技术文档的原文，其余都是笔者的原创。在您引用本仓库内容或者对内容进行修改演绎时，请遵循文末的开源协议，谢谢。
 
-[BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)，这个书单是笔者至今看的一些比较好的技术书籍，虽然没有全都看完，但每本书多多少少都看了一部分。
+#### BookList
+
+本仓库参考的书目：[BOOKLIST](https://github.com/CyC2018/Interview-Notebook/blob/master/BOOKLIST.md)。
 
 #### How To Contribute
 

From e7af66ace189bf1defbfc85ce7641a44d3099ab2 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=83=91=E6=B0=B8=E5=B7=9D?= <1029579233@qq.com>
Date: Sat, 1 Sep 2018 17:24:36 +0800
Subject: [PATCH 169/172] Update README.md

---
 README.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/README.md b/README.md
index 6a301d34..7ee0bf57 100644
--- a/README.md
+++ b/README.md
@@ -175,7 +175,7 @@
 
 #### How To Contribute
 
-笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接在相应文档进行编辑修改。
+笔记内容是笔者一个字一个字打上去的，难免会有一些笔误，如果发现笔误可直接对相应文档进行编辑修改。
 
 如果想要提交一个仓库现在还没有的全新内容，可以先将相应的文档放到 other 目录下。
 

From 9308a2a496e78f6be405c8105659439cc4aac99a Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sat, 1 Sep 2018 17:34:48 +0800
Subject: [PATCH 170/172] auto commit

---
 notes/计算机操作系统.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/notes/计算机操作系统.md b/notes/计算机操作系统.md
index 44f8eb35..4125d86d 100644
--- a/notes/计算机操作系统.md
+++ b/notes/计算机操作系统.md
@@ -167,19 +167,19 @@ QQ 和浏览器是两个进程，浏览器进程里面有很多线程，例如 H
 
 ### 3. 区别
 
-（一）拥有资源
+Ⅰ 拥有资源
 
 进程是资源分配的基本单位，但是线程不拥有资源，线程可以访问隶属进程的资源。
 
-（二）调度
+Ⅱ 调度
 
 线程是独立调度的基本单位，在同一进程中，线程的切换不会引起进程切换，从一个进程内的线程切换到另一个进程中的线程时，会引起进程切换。
 
-（三）系统开销
+Ⅲ 系统开销
 
 由于创建或撤销进程时，系统都要为之分配或回收资源，如内存空间、I/O 设备等，所付出的开销远大于创建或撤销线程时的开销。类似地，在进行进程切换时，涉及当前执行进程 CPU 环境的保存及新调度进程 CPU 环境的设置，而线程切换时只需保存和设置少量寄存器内容，开销很小。
 
-（四）通信方面
+Ⅳ 通信方面
 
 进程间通信 (IPC) 需要进程同步和互斥手段的辅助，以保证数据的一致性。而线程间可以通过直接读/写同一进程中的数据段（如全局变量）来进行通信。
 

From 39965a71dbf1c7d6d0773001e3ca0b4103f0d9bd Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 2 Sep 2018 10:59:51 +0800
Subject: [PATCH 171/172] auto commit

---
 notes/Linux.md           | 34 +++++++++++++++++-----------------
 notes/剑指 offer 题解.md |  2 +-
 notes/系统设计基础.md    |  2 +-
 3 files changed, 19 insertions(+), 19 deletions(-)

diff --git a/notes/Linux.md b/notes/Linux.md
index 4ba647a7..5e9528e6 100644
--- a/notes/Linux.md
+++ b/notes/Linux.md
@@ -45,7 +45,7 @@
     * [变量操作](#变量操作)
     * [指令搜索顺序](#指令搜索顺序)
     * [数据流重定向](#数据流重定向)
-* [八、管线指令](#八管线指令)
+* [八、管道指令](#八管道指令)
     * [提取指令](#提取指令)
     * [排序指令](#排序指令)
     * [双向输出重定向](#双向输出重定向)
@@ -823,7 +823,7 @@ $ echo ${array[1]}
 
 - 以绝对或相对路径来执行指令，例如 /bin/ls 或者 ./ls ；
 - 由别名找到该指令来执行；
-- 由 Bash 内建的指令来执行；
+- 由 Bash 内置的指令来执行；
 - 按 \$PATH 变量指定的搜索路径的顺序找到第一个指令来执行。
 
 ## 数据流重定向
@@ -846,11 +846,11 @@ $ echo ${array[1]}
 $ find /home -name .bashrc > list 2>&1
 ```
 
-# 八、管线指令
+# 八、管道指令
 
-管线是将一个命令的标准输出作为另一个命令的标准输入，在数据需要经过多个步骤的处理之后才能得到我们想要的内容时就可以使用管线。
+管道是将一个命令的标准输出作为另一个命令的标准输入，在数据需要经过多个步骤的处理之后才能得到我们想要的内容时就可以使用管道。
 
-在命令之间使用 | 分隔各个管线命令。
+在命令之间使用 | 分隔各个管道命令。
 
 ```bash
 $ ls -al /etc | less
@@ -1053,7 +1053,7 @@ $ printf '%10s %5i %5i %5i %8.2f \n' $(cat printf.txt)
 
 awk 每次处理一行，处理的最小单位是字段，每个字段的命名方式为：\$n，n 为字段号，从 1 开始，\$0 表示一整行。
 
-示例：取出登录用户的用户名和 IP
+示例：取出最近五个登录用户的用户名和 IP
 
 ```html
 $ last -n 5
@@ -1128,17 +1128,7 @@ dmtsai lines: 5 columns: 9
 # ps aux | grep threadx
 ```
 
-### 2. top
-
-实时显示进程信息
-
-示例：两秒钟刷新一次
-
-```sh
-# top -d 2
-```
-
-### 3. pstree
+### 2. pstree
 
 查看进程树
 
@@ -1148,6 +1138,16 @@ dmtsai lines: 5 columns: 9
 # pstree -A
 ```
 
+### 3. top
+
+实时显示进程信息
+
+示例：两秒钟刷新一次
+
+```sh
+# top -d 2
+```
+
 ### 4. netstat
 
 查看占用端口的进程
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index 63cf170e..af4c1842 100644
--- a/notes/剑指 offer 题解.md	
+++ b/notes/剑指 offer 题解.md	
@@ -2244,7 +2244,7 @@ public int GetUglyNumber_Solution(int N) {
 
 ## 题目描述
 
-在一个字符串 中找到第一个只出现一次的字符，并返回它的位置。
+在一个字符串中找到第一个只出现一次的字符，并返回它的位置。
 
 ## 解题思路
 
diff --git a/notes/系统设计基础.md b/notes/系统设计基础.md
index b6a3ddf5..4a961140 100644
--- a/notes/系统设计基础.md
+++ b/notes/系统设计基础.md
@@ -13,7 +13,7 @@
 
 ### 1. 响应时间
 
-指从某个请求从发出到接收到响应消耗的时间。
+指某个请求从发出到接收到响应消耗的时间。
 
 在对响应时间进行测试时，通常采用重复请求方式，然后计算平均响应时间。
 

From 6ea965a9ef36fb877ec4b392b0ad16f2b20907e8 Mon Sep 17 00:00:00 2001
From: CyC2018 <1029579233@qq.com>
Date: Sun, 2 Sep 2018 22:31:20 +0800
Subject: [PATCH 172/172] auto commit

---
 notes/CyC 学习交流群 问题汇总.md | 16 ++++++++++++++++
 notes/MySQL.md                   | 10 +++++-----
 2 files changed, 21 insertions(+), 5 deletions(-)
 create mode 100644 notes/CyC 学习交流群 问题汇总.md

diff --git a/notes/CyC 学习交流群 问题汇总.md b/notes/CyC 学习交流群 问题汇总.md
new file mode 100644
index 00000000..806d48d0
--- /dev/null
+++ b/notes/CyC 学习交流群 问题汇总.md	
@@ -0,0 +1,16 @@
+<!-- GFM-TOC -->
+* [0. 进程内存空间中，堆和栈的区别](#0-进程内存空间中，堆和栈的区别)
+<!-- GFM-TOC -->
+
+
+# 0. 进程内存空间中，堆和栈的区别
+
+> C++
+
+堆：动态、malloc()、new、链式分配、向上生长；栈：函数调用、编译器分配回收、向下生长。
+
+https://www.cnblogs.com/sunziying/p/6510030.html
+
+By @CyC
+
+---
diff --git a/notes/MySQL.md b/notes/MySQL.md
index f115ce92..3922b8a6 100644
--- a/notes/MySQL.md
+++ b/notes/MySQL.md
@@ -22,7 +22,7 @@
     * [水平切分](#水平切分)
     * [垂直切分](#垂直切分)
     * [Sharding 策略](#sharding-策略)
-    * [Sharding 存在的问题及解决方案](#sharding-存在的问题及解决方案)
+    * [Sharding 存在的问题](#sharding-存在的问题)
 * [六、复制](#六复制)
     * [主从复制](#主从复制)
     * [读写分离](#读写分离)
@@ -366,19 +366,19 @@ MySQL 提供了 FROM_UNIXTIME() 函数把 UNIX 时间戳转换为日期，并提
 - 范围：可以是 ID 范围也可以是时间范围；
 - 映射表：使用单独的一个数据库来存储映射关系。
 
-## Sharding 存在的问题及解决方案
+## Sharding 存在的问题
 
 ### 1. 事务问题
 
 使用分布式事务来解决，比如 XA 接口。
 
-### 2. JOIN
+### 2. 连接
 
-可以将原来的 JOIN 分解成多个单表 JOIN 查询，然后在用户程序中进行 JOIN。
+可以将原来的连接分解成多个单表连接查询，然后在用户程序中进行连接。
 
 ### 3. ID 唯一性
 
-- 使用全局唯一 ID：GUID
+- 使用全局唯一 ID（GUID）
 - 为每个分片指定一个 ID 范围
 - 分布式 ID 生成器 (如 Twitter 的 Snowflake 算法)