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Merge pull request #425 from youmark/master

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Fix issues in House Robber and Minimum Path Sum sample code
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郑永川 2018-09-19 09:38:25 +08:00 committed by GitHub
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1 changed files with 13 additions and 22 deletions
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notes/Leetcode 题解.md
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@ -2415,27 +2415,19 @@ public int climbStairs(int n) {
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。 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 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
dp[i] = max(dp[i-1], dp[i-2] + nums[i]) <br>
<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2],dp[i-3])+nums[i]"/></div> <br>
```java ```java
public int rob(int[] nums) { public int rob(int[] nums) {
int n = nums.length; int pre2 = 0, pre1 = 0;
if (n == 0) { for (int i = 0; i < nums.length; i++) {
return 0; int cur = Math.max(pre2 + nums[i], pre1);
}
if (n == 1) {
return nums[0];
}
int pre3 = 0, pre2 = 0, pre1 = 0;
for (int i = 0; i < n; i++) {
int cur = Math.max(pre2, pre3) + nums[i];
pre3 = pre2;
pre2 = pre1; pre2 = pre1;
pre1 = cur; pre1 = cur;
} }
return Math.max(pre1, pre2); return pre1;
} }
``` ```
**强盗在环形街区抢劫** **强盗在环形街区抢劫**
@ -2455,14 +2447,13 @@ public int rob(int[] nums) {
} }
private int rob(int[] nums, int first, int last) { private int rob(int[] nums, int first, int last) {
int pre3 = 0, pre2 = 0, pre1 = 0; int pre2 = 0, pre1 = 0;
for (int i = first; i <= last; i++) { for (int i = first; i <= last; i++) {
int cur = Math.max(pre3, pre2) + nums[i]; int cur = Math.max(pre1, pre2 + nums[i]);
pre3 = pre2;
pre2 = pre1; pre2 = pre1;
pre1 = cur; pre1 = cur;
} }
return Math.max(pre2, pre1); return pre1;
} }
``` ```
@ -2514,9 +2505,9 @@ public int minPathSum(int[][] grid) {
for (int i = 0; i < m; i++) { for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) { for (int j = 0; j < n; j++) {
if (i == 0) { if (i == 0) {
dp[j] = dp[j - 1]; if (j>0) dp[j] = dp[j - 1];
} else { } else {
dp[j] = Math.min(dp[j - 1], dp[j]); if (j>0) dp[j] = Math.min(dp[j - 1], dp[j]);
} }
dp[j] += grid[i][j]; dp[j] += grid[i][j];
} }