536 lines
15 KiB
Markdown
536 lines
15 KiB
Markdown
# Leetcode 题解 - 数学
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<!-- GFM-TOC -->
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* [Leetcode 题解 - 数学](#leetcode-题解---数学)
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* [素数分解](#素数分解)
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* [整除](#整除)
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* [最大公约数最小公倍数](#最大公约数最小公倍数)
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* [1. 生成素数序列](#1-生成素数序列)
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* [2. 最大公约数](#2-最大公约数)
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* [3. 使用位操作和减法求解最大公约数](#3-使用位操作和减法求解最大公约数)
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* [进制转换](#进制转换)
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* [1. 7 进制](#1-7-进制)
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* [2. 16 进制](#2-16-进制)
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* [3. 26 进制](#3-26-进制)
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* [阶乘](#阶乘)
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* [1. 统计阶乘尾部有多少个 0](#1-统计阶乘尾部有多少个-0)
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* [字符串加法减法](#字符串加法减法)
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* [1. 二进制加法](#1-二进制加法)
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* [2. 字符串加法](#2-字符串加法)
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* [相遇问题](#相遇问题)
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* [1. 改变数组元素使所有的数组元素都相等](#1-改变数组元素使所有的数组元素都相等)
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* [多数投票问题](#多数投票问题)
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* [1. 数组中出现次数多于 n / 2 的元素](#1-数组中出现次数多于-n--2-的元素)
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* [其它](#其它)
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* [1. 平方数](#1-平方数)
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* [2. 3 的 n 次方](#2-3-的-n-次方)
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* [3. 乘积数组](#3-乘积数组)
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* [4. 找出数组中的乘积最大的三个数](#4-找出数组中的乘积最大的三个数)
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<!-- GFM-TOC -->
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## 素数分解
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每一个数都可以分解成素数的乘积，例如 84 = 2<sup>2</sup> \* 3<sup>1</sup> \* 5<sup>0</sup> \* 7<sup>1</sup> \* 11<sup>0</sup> \* 13<sup>0</sup> \* 17<sup>0</sup> \* …
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## 整除
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令 x = 2<sup>m0</sup> \* 3<sup>m1</sup> \* 5<sup>m2</sup> \* 7<sup>m3</sup> \* 11<sup>m4</sup> \* …
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令 y = 2<sup>n0</sup> \* 3<sup>n1</sup> \* 5<sup>n2</sup> \* 7<sup>n3</sup> \* 11<sup>n4</sup> \* …
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如果 x 整除 y（y mod x == 0），则对于所有 i，mi \<= ni。
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## 最大公约数最小公倍数
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x 和 y 的最大公约数为：gcd(x,y) = 2<sup>min(m0,n0)</sup> \* 3<sup>min(m1,n1)</sup> \* 5<sup>min(m2,n2)</sup> \* ...
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x 和 y 的最小公倍数为：lcm(x,y) = 2<sup>max(m0,n0)</sup> \* 3<sup>max(m1,n1)</sup> \* 5<sup>max(m2,n2)</sup> \* ...
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### 1. 生成素数序列
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204\. Count Primes (Easy)
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[Leetcode](https://leetcode.com/problems/count-primes/description/) / [力扣](https://leetcode-cn.com/problems/count-primes/description/)
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埃拉托斯特尼筛法在每次找到一个素数时，将能被素数整除的数排除掉。
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```java
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public int countPrimes(int n) {
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boolean[] notPrimes = new boolean[n + 1];
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int count = 0;
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for (int i = 2; i < n; i++) {
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if (notPrimes[i]) {
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continue;
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}
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count++;
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// 从 i * i 开始，因为如果 k < i，那么 k * i 在之前就已经被去除过了
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for (long j = (long) (i) * i; j < n; j += i) {
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notPrimes[(int) j] = true;
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}
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}
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return count;
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}
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```
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### 2. 最大公约数
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```java
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int gcd(int a, int b) {
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return b == 0 ? a : gcd(b, a % b);
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}
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```
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最小公倍数为两数的乘积除以最大公约数。
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```java
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int lcm(int a, int b) {
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return a * b / gcd(a, b);
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}
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```
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### 3. 使用位操作和减法求解最大公约数
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[编程之美：2.7](#)
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对于 a 和 b 的最大公约数 f(a, b)，有：
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- 如果 a 和 b 均为偶数，f(a, b) = 2\*f(a/2, b/2);
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- 如果 a 是偶数 b 是奇数，f(a, b) = f(a/2, b);
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- 如果 b 是偶数 a 是奇数，f(a, b) = f(a, b/2);
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- 如果 a 和 b 均为奇数，f(a, b) = f(b, a-b);
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乘 2 和除 2 都可以转换为移位操作。
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```java
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public int gcd(int a, int b) {
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if (a < b) {
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return gcd(b, a);
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}
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if (b == 0) {
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return a;
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}
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boolean isAEven = isEven(a), isBEven = isEven(b);
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if (isAEven && isBEven) {
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return 2 * gcd(a >> 1, b >> 1);
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} else if (isAEven && !isBEven) {
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return gcd(a >> 1, b);
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} else if (!isAEven && isBEven) {
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return gcd(a, b >> 1);
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} else {
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return gcd(b, a - b);
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}
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}
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```
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## 进制转换
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### 1. 7 进制
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504\. Base 7 (Easy)
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[Leetcode](https://leetcode.com/problems/base-7/description/) / [力扣](https://leetcode-cn.com/problems/base-7/description/)
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```java
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public String convertToBase7(int num) {
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if (num == 0) {
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return "0";
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}
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StringBuilder sb = new StringBuilder();
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boolean isNegative = num < 0;
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if (isNegative) {
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num = -num;
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}
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while (num > 0) {
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sb.append(num % 7);
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num /= 7;
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}
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String ret = sb.reverse().toString();
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return isNegative ? "-" + ret : ret;
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}
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```
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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。
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```java
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public String convertToBase7(int num) {
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return Integer.toString(num, 7);
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}
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```
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### 2. 16 进制
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405\. Convert a Number to Hexadecimal (Easy)
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[Leetcode](https://leetcode.com/problems/convert-a-number-to-hexadecimal/description/) / [力扣](https://leetcode-cn.com/problems/convert-a-number-to-hexadecimal/description/)
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```html
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Input:
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26
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Output:
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"1a"
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Input:
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-1
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Output:
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"ffffffff"
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```
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负数要用它的补码形式。
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```java
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public String toHex(int num) {
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char[] map = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
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if (num == 0) return "0";
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StringBuilder sb = new StringBuilder();
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while (num != 0) {
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sb.append(map[num & 0b1111]);
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num >>>= 4; // 因为考虑的是补码形式，因此符号位就不能有特殊的意义，需要使用无符号右移，左边填 0
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}
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return sb.reverse().toString();
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}
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```
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### 3. 26 进制
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168\. Excel Sheet Column Title (Easy)
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[Leetcode](https://leetcode.com/problems/excel-sheet-column-title/description/) / [力扣](https://leetcode-cn.com/problems/excel-sheet-column-title/description/)
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```html
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1 -> A
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2 -> B
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3 -> C
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...
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26 -> Z
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27 -> AA
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28 -> AB
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```
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因为是从 1 开始计算的，而不是从 0 开始，因此需要对 n 执行 -1 操作。
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```java
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public String convertToTitle(int n) {
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if (n == 0) {
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return "";
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}
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n--;
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return convertToTitle(n / 26) + (char) (n % 26 + 'A');
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}
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```
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## 阶乘
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### 1. 统计阶乘尾部有多少个 0
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172\. Factorial Trailing Zeroes (Easy)
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[Leetcode](https://leetcode.com/problems/factorial-trailing-zeroes/description/) / [力扣](https://leetcode-cn.com/problems/factorial-trailing-zeroes/description/)
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尾部的 0 由 2 * 5 得来，2 的数量明显多于 5 的数量，因此只要统计有多少个 5 即可。
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对于一个数 N，它所包含 5 的个数为：N/5 + N/5<sup>2</sup> + N/5<sup>3</sup> + ...，其中 N/5 表示不大于 N 的数中 5 的倍数贡献一个 5，N/5<sup>2</sup> 表示不大于 N 的数中 5<sup>2</sup> 的倍数再贡献一个 5 ...。
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```java
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public int trailingZeroes(int n) {
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return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
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}
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```
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如果统计的是 N! 的二进制表示中最低位 1 的位置，只要统计有多少个 2 即可，该题目出自 [编程之美：2.2](#) 。和求解有多少个 5 一样，2 的个数为 N/2 + N/2<sup>2</sup> + N/2<sup>3</sup> + ...
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## 字符串加法减法
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### 1. 二进制加法
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67\. Add Binary (Easy)
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[Leetcode](https://leetcode.com/problems/add-binary/description/) / [力扣](https://leetcode-cn.com/problems/add-binary/description/)
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```html
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a = "11"
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b = "1"
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Return "100".
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```
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```java
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public String addBinary(String a, String b) {
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int i = a.length() - 1, j = b.length() - 1, carry = 0;
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StringBuilder str = new StringBuilder();
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while (carry == 1 || i >= 0 || j >= 0) {
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if (i >= 0 && a.charAt(i--) == '1') {
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carry++;
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}
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if (j >= 0 && b.charAt(j--) == '1') {
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carry++;
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}
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str.append(carry % 2);
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carry /= 2;
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}
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return str.reverse().toString();
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}
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```
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### 2. 字符串加法
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415\. Add Strings (Easy)
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[Leetcode](https://leetcode.com/problems/add-strings/description/) / [力扣](https://leetcode-cn.com/problems/add-strings/description/)
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字符串的值为非负整数。
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```java
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public String addStrings(String num1, String num2) {
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StringBuilder str = new StringBuilder();
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int carry = 0, i = num1.length() - 1, j = num2.length() - 1;
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while (carry == 1 || i >= 0 || j >= 0) {
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int x = i < 0 ? 0 : num1.charAt(i--) - '0';
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int y = j < 0 ? 0 : num2.charAt(j--) - '0';
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str.append((x + y + carry) % 10);
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carry = (x + y + carry) / 10;
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}
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return str.reverse().toString();
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}
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```
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## 相遇问题
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### 1. 改变数组元素使所有的数组元素都相等
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462\. Minimum Moves to Equal Array Elements II (Medium)
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[Leetcode](https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/description/) / [力扣](https://leetcode-cn.com/problems/minimum-moves-to-equal-array-elements-ii/description/)
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```html
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Input:
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[1,2,3]
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Output:
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2
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Explanation:
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Only two moves are needed (remember each move increments or decrements one element):
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[1,2,3] => [2,2,3] => [2,2,2]
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```
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每次可以对一个数组元素加一或者减一，求最小的改变次数。
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这是个典型的相遇问题，移动距离最小的方式是所有元素都移动到中位数。理由如下：
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设 m 为中位数。a 和 b 是 m 两边的两个元素，且 b \> a。要使 a 和 b 相等，它们总共移动的次数为 b - a，这个值等于 (b - m) + (m - a)，也就是把这两个数移动到中位数的移动次数。
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设数组长度为 N，则可以找到 N/2 对 a 和 b 的组合，使它们都移动到 m 的位置。
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**解法 1**
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先排序，时间复杂度：O(NlogN)
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```java
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public int minMoves2(int[] nums) {
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Arrays.sort(nums);
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int move = 0;
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int l = 0, h = nums.length - 1;
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while (l <= h) {
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move += nums[h] - nums[l];
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l++;
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h--;
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}
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return move;
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}
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```
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**解法 2**
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使用快速选择找到中位数，时间复杂度 O(N)
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```java
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public int minMoves2(int[] nums) {
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int move = 0;
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int median = findKthSmallest(nums, nums.length / 2);
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for (int num : nums) {
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move += Math.abs(num - median);
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}
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return move;
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}
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private int findKthSmallest(int[] nums, int k) {
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int l = 0, h = nums.length - 1;
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while (l < h) {
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int j = partition(nums, l, h);
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if (j == k) {
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break;
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}
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if (j < k) {
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l = j + 1;
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} else {
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h = j - 1;
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}
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}
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return nums[k];
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}
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private int partition(int[] nums, int l, int h) {
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int i = l, j = h + 1;
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while (true) {
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while (nums[++i] < nums[l] && i < h) ;
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while (nums[--j] > nums[l] && j > l) ;
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if (i >= j) {
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break;
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}
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swap(nums, i, j);
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}
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swap(nums, l, j);
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return j;
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}
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private void swap(int[] nums, int i, int j) {
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int tmp = nums[i];
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nums[i] = nums[j];
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nums[j] = tmp;
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}
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```
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## 多数投票问题
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### 1. 数组中出现次数多于 n / 2 的元素
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169\. Majority Element (Easy)
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[Leetcode](https://leetcode.com/problems/majority-element/description/) / [力扣](https://leetcode-cn.com/problems/majority-element/description/)
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先对数组排序，最中间那个数出现次数一定多于 n / 2。
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```java
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public int majorityElement(int[] nums) {
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Arrays.sort(nums);
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return nums[nums.length / 2];
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}
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```
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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)。可以这么理解该算法：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。
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```java
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public int majorityElement(int[] nums) {
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int cnt = 0, majority = nums[0];
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for (int num : nums) {
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majority = (cnt == 0) ? num : majority;
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cnt = (majority == num) ? cnt + 1 : cnt - 1;
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}
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return majority;
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}
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```
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## 其它
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### 1. 平方数
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367\. Valid Perfect Square (Easy)
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[Leetcode](https://leetcode.com/problems/valid-perfect-square/description/) / [力扣](https://leetcode-cn.com/problems/valid-perfect-square/description/)
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```html
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Input: 16
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Returns: True
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```
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平方序列：1,4,9,16,..
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间隔：3,5,7,...
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间隔为等差数列，使用这个特性可以得到从 1 开始的平方序列。
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```java
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public boolean isPerfectSquare(int num) {
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int subNum = 1;
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while (num > 0) {
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num -= subNum;
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subNum += 2;
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}
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return num == 0;
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}
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```
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### 2. 3 的 n 次方
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326\. Power of Three (Easy)
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[Leetcode](https://leetcode.com/problems/power-of-three/description/) / [力扣](https://leetcode-cn.com/problems/power-of-three/description/)
|
||
|
||
```java
|
||
public boolean isPowerOfThree(int n) {
|
||
return n > 0 && (1162261467 % n == 0);
|
||
}
|
||
```
|
||
|
||
### 3. 乘积数组
|
||
|
||
238\. Product of Array Except Self (Medium)
|
||
|
||
[Leetcode](https://leetcode.com/problems/product-of-array-except-self/description/) / [力扣](https://leetcode-cn.com/problems/product-of-array-except-self/description/)
|
||
|
||
```html
|
||
For example, given [1,2,3,4], return [24,12,8,6].
|
||
```
|
||
|
||
给定一个数组，创建一个新数组，新数组的每个元素为原始数组中除了该位置上的元素之外所有元素的乘积。
|
||
|
||
要求时间复杂度为 O(N)，并且不能使用除法。
|
||
|
||
```java
|
||
public int[] productExceptSelf(int[] nums) {
|
||
int n = nums.length;
|
||
int[] products = new int[n];
|
||
Arrays.fill(products, 1);
|
||
int left = 1;
|
||
for (int i = 1; i < n; i++) {
|
||
left *= nums[i - 1];
|
||
products[i] *= left;
|
||
}
|
||
int right = 1;
|
||
for (int i = n - 2; i >= 0; i--) {
|
||
right *= nums[i + 1];
|
||
products[i] *= right;
|
||
}
|
||
return products;
|
||
}
|
||
```
|
||
|
||
### 4. 找出数组中的乘积最大的三个数
|
||
|
||
628\. Maximum Product of Three Numbers (Easy)
|
||
|
||
[Leetcode](https://leetcode.com/problems/maximum-product-of-three-numbers/description/) / [力扣](https://leetcode-cn.com/problems/maximum-product-of-three-numbers/description/)
|
||
|
||
```html
|
||
Input: [1,2,3,4]
|
||
Output: 24
|
||
```
|
||
|
||
```java
|
||
public int maximumProduct(int[] nums) {
|
||
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE, min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
|
||
for (int n : nums) {
|
||
if (n > max1) {
|
||
max3 = max2;
|
||
max2 = max1;
|
||
max1 = n;
|
||
} else if (n > max2) {
|
||
max3 = max2;
|
||
max2 = n;
|
||
} else if (n > max3) {
|
||
max3 = n;
|
||
}
|
||
|
||
if (n < min1) {
|
||
min2 = min1;
|
||
min1 = n;
|
||
} else if (n < min2) {
|
||
min2 = n;
|
||
}
|
||
}
|
||
return Math.max(max1*max2*max3, max1*min1*min2);
|
||
}
|
||
```
|