1261 lines
41 KiB
Java
1261 lines
41 KiB
Java
<!-- GFM-TOC -->
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* [斐波那契数列](#斐波那契数列)
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* [1. 爬楼梯](#1-爬楼梯)
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* [2. 强盗抢劫](#2-强盗抢劫)
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* [3. 强盗在环形街区抢劫](#3-强盗在环形街区抢劫)
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* [4. 信件错排](#4-信件错排)
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* [5. 母牛生产](#5-母牛生产)
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* [矩阵路径](#矩阵路径)
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* [1. 矩阵的最小路径和](#1-矩阵的最小路径和)
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* [2. 矩阵的总路径数](#2-矩阵的总路径数)
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* [数组区间](#数组区间)
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* [1. 数组区间和](#1-数组区间和)
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* [2. 数组中等差递增子区间的个数](#2-数组中等差递增子区间的个数)
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* [分割整数](#分割整数)
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* [1. 分割整数的最大乘积](#1-分割整数的最大乘积)
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* [2. 按平方数来分割整数](#2-按平方数来分割整数)
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* [3. 分割整数构成字母字符串](#3-分割整数构成字母字符串)
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* [最长递增子序列](#最长递增子序列)
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* [1. 最长递增子序列](#1-最长递增子序列)
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* [2. 一组整数对能够构成的最长链](#2-一组整数对能够构成的最长链)
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* [3. 最长摆动子序列](#3-最长摆动子序列)
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* [最长公共子序列](#最长公共子序列)
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* [0-1 背包](#0-1-背包)
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* [1. 划分数组为和相等的两部分](#1-划分数组为和相等的两部分)
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* [2. 改变一组数的正负号使得它们的和为一给定数](#2-改变一组数的正负号使得它们的和为一给定数)
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* [3. 01 字符构成最多的字符串](#3-01-字符构成最多的字符串)
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* [4. 找零钱的最少硬币数](#4-找零钱的最少硬币数)
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* [5. 找零钱的硬币数组合](#5-找零钱的硬币数组合)
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* [6. 字符串按单词列表分割](#6-字符串按单词列表分割)
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* [7. 组合总和](#7-组合总和)
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* [股票交易](#股票交易)
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* [1. 需要冷却期的股票交易](#1-需要冷却期的股票交易)
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* [2. 需要交易费用的股票交易](#2-需要交易费用的股票交易)
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* [3. 只能进行两次的股票交易](#3-只能进行两次的股票交易)
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* [4. 只能进行 k 次的股票交易](#4-只能进行-k-次的股票交易)
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* [字符串编辑](#字符串编辑)
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* [1. 删除两个字符串的字符使它们相等](#1-删除两个字符串的字符使它们相等)
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* [2. 编辑距离](#2-编辑距离)
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* [3. 复制粘贴字符](#3-复制粘贴字符)
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<!-- GFM-TOC -->
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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。
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# 斐波那契数列
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## 1. 爬楼梯
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[70. Climbing Stairs (Easy)](https://leetcode.com/problems/climbing-stairs/description/)
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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。
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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。
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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。
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<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-2]" class="mathjax-pic"/></div> <br>-->
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<div align="center"> <img src="pics/14fe1e71-8518-458f-a220-116003061a83.png" width="200px"> </div><br>
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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。
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```java
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public int climbStairs(int n) {
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if (n <= 2) {
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return n;
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}
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int pre2 = 1, pre1 = 2;
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for (int i = 2; i < n; i++) {
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int cur = pre1 + pre2;
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pre2 = pre1;
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pre1 = cur;
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}
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return pre1;
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}
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```
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## 2. 强盗抢劫
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[198. House Robber (Easy)](https://leetcode.com/problems/house-robber/description/)
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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。
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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。
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由于不能抢劫邻近住户，如果抢劫了第 i -1 个住户，那么就不能再抢劫第 i 个住户，所以
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<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=max(dp[i-2]+nums[i],dp[i-1])" class="mathjax-pic"/></div> <br>-->
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<div align="center"> <img src="pics/2de794ca-aa7b-48f3-a556-a0e2708cb976.jpg" width="350px"> </div><br>
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```java
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public int rob(int[] nums) {
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int pre2 = 0, pre1 = 0;
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for (int i = 0; i < nums.length; i++) {
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int cur = Math.max(pre2 + nums[i], pre1);
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pre2 = pre1;
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pre1 = cur;
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}
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return pre1;
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}
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```
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## 3. 强盗在环形街区抢劫
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[213. House Robber II (Medium)](https://leetcode.com/problems/house-robber-ii/description/)
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```java
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public int rob(int[] nums) {
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if (nums == null || nums.length == 0) {
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return 0;
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}
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int n = nums.length;
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if (n == 1) {
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return nums[0];
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}
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return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
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}
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private int rob(int[] nums, int first, int last) {
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int pre2 = 0, pre1 = 0;
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for (int i = first; i <= last; i++) {
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int cur = Math.max(pre1, pre2 + nums[i]);
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pre2 = pre1;
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pre1 = cur;
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}
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return pre1;
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}
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```
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## 4. 信件错排
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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。
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定义一个数组 dp 存储错误方式数量，dp[i] 表示前 i 个信和信封的错误方式数量。假设第 i 个信装到第 j 个信封里面，而第 j 个信装到第 k 个信封里面。根据 i 和 k 是否相等，有两种情况：
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- i==k，交换 i 和 j 的信后，它们的信和信封在正确的位置，但是其余 i-2 封信有 dp[i-2] 种错误装信的方式。由于 j 有 i-1 种取值，因此共有 (i-1)\*dp[i-2] 种错误装信方式。
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- i != k，交换 i 和 j 的信后，第 i 个信和信封在正确的位置，其余 i-1 封信有 dp[i-1] 种错误装信方式。由于 j 有 i-1 种取值，因此共有 (i-1)\*dp[i-1] 种错误装信方式。
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综上所述，错误装信数量方式数量为：
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<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=(i-1)*dp[i-2]+(i-1)*dp[i-1]" class="mathjax-pic"/></div> <br>-->
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<div align="center"> <img src="pics/da1f96b9-fd4d-44ca-8925-fb14c5733388.png" width="350px"> </div><br>
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## 5. 母牛生产
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[程序员代码面试指南-P181](#)
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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。
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第 i 年成熟的牛的数量为：
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<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i]=dp[i-1]+dp[i-3]" class="mathjax-pic"/></div> <br>-->
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<div align="center"> <img src="pics/879814ee-48b5-4bcb-86f5-dcc400cb81ad.png" width="250px"> </div><br>
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# 矩阵路径
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## 1. 矩阵的最小路径和
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[64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/)
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```html
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[[1,3,1],
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[1,5,1],
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[4,2,1]]
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Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.
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```
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题目描述：求从矩阵的左上角到右下角的最小路径和，每次只能向右和向下移动。
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```java
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public int minPathSum(int[][] grid) {
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if (grid.length == 0 || grid[0].length == 0) {
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return 0;
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}
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int m = grid.length, n = grid[0].length;
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int[] dp = new int[n];
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (j == 0) {
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dp[j] = dp[j]; // 只能从上侧走到该位置
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} else if (i == 0) {
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dp[j] = dp[j - 1]; // 只能从左侧走到该位置
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} else {
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dp[j] = Math.min(dp[j - 1], dp[j]);
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}
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dp[j] += grid[i][j];
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}
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}
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return dp[n - 1];
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}
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```
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## 2. 矩阵的总路径数
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[62. Unique Paths (Medium)](https://leetcode.com/problems/unique-paths/description/)
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题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。
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<div align="center"> <img src="pics/dc82f0f3-c1d4-4ac8-90ac-d5b32a9bd75a.jpg" width=""> </div><br>
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```java
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public int uniquePaths(int m, int n) {
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int[] dp = new int[n];
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Arrays.fill(dp, 1);
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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dp[j] = dp[j] + dp[j - 1];
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}
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}
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return dp[n - 1];
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}
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```
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也可以直接用数学公式求解，这是一个组合问题。机器人总共移动的次数 S=m+n-2，向下移动的次数 D=m-1，那么问题可以看成从 S 中取出 D 个位置的组合数量，这个问题的解为 C(S, D)。
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```java
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public int uniquePaths(int m, int n) {
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int S = m + n - 2; // 总共的移动次数
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int D = m - 1; // 向下的移动次数
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long ret = 1;
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for (int i = 1; i <= D; i++) {
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ret = ret * (S - D + i) / i;
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}
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return (int) ret;
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}
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```
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# 数组区间
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## 1. 数组区间和
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[303. Range Sum Query - Immutable (Easy)](https://leetcode.com/problems/range-sum-query-immutable/description/)
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```html
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Given nums = [-2, 0, 3, -5, 2, -1]
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sumRange(0, 2) -> 1
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sumRange(2, 5) -> -1
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sumRange(0, 5) -> -3
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```
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求区间 i \~ j 的和，可以转换为 sum[j + 1] - sum[i]，其中 sum[i] 为 0 \~ i - 1 的和。
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```java
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class NumArray {
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private int[] sums;
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public NumArray(int[] nums) {
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sums = new int[nums.length + 1];
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for (int i = 1; i <= nums.length; i++) {
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sums[i] = sums[i - 1] + nums[i - 1];
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}
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}
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public int sumRange(int i, int j) {
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return sums[j + 1] - sums[i];
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}
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}
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```
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## 2. 数组中等差递增子区间的个数
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[413. Arithmetic Slices (Medium)](https://leetcode.com/problems/arithmetic-slices/description/)
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```html
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A = [0, 1, 2, 3, 4]
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return: 6, for 3 arithmetic slices in A:
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[0, 1, 2],
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[1, 2, 3],
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[0, 1, 2, 3],
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[0, 1, 2, 3, 4],
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[ 1, 2, 3, 4],
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[2, 3, 4]
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```
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dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
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当 A[i] - A[i-1] == A[i-1] - A[i-2]，那么 [A[i-2], A[i-1], A[i]] 构成一个等差递增子区间。而且在以 A[i-1] 为结尾的递增子区间的后面再加上一个 A[i]，一样可以构成新的递增子区间。
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```html
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dp[2] = 1
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[0, 1, 2]
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dp[3] = dp[2] + 1 = 2
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[0, 1, 2, 3], // [0, 1, 2] 之后加一个 3
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[1, 2, 3] // 新的递增子区间
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dp[4] = dp[3] + 1 = 3
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[0, 1, 2, 3, 4], // [0, 1, 2, 3] 之后加一个 4
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[1, 2, 3, 4], // [1, 2, 3] 之后加一个 4
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[2, 3, 4] // 新的递增子区间
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```
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综上，在 A[i] - A[i-1] == A[i-1] - A[i-2] 时，dp[i] = dp[i-1] + 1。
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因为递增子区间不一定以最后一个元素为结尾，可以是任意一个元素结尾，因此需要返回 dp 数组累加的结果。
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```java
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public int numberOfArithmeticSlices(int[] A) {
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if (A == null || A.length == 0) {
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return 0;
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}
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int n = A.length;
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int[] dp = new int[n];
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for (int i = 2; i < n; i++) {
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if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
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dp[i] = dp[i - 1] + 1;
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}
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}
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int total = 0;
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for (int cnt : dp) {
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total += cnt;
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}
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return total;
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}
|
||
```
|
||
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# 分割整数
|
||
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## 1. 分割整数的最大乘积
|
||
|
||
[343. Integer Break (Medim)](https://leetcode.com/problems/integer-break/description/)
|
||
|
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题目描述：For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
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|
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```java
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public int integerBreak(int n) {
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int[] dp = new int[n + 1];
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dp[1] = 1;
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for (int i = 2; i <= n; i++) {
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for (int j = 1; j <= i - 1; j++) {
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dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
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}
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}
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return dp[n];
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}
|
||
```
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## 2. 按平方数来分割整数
|
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||
[279. Perfect Squares(Medium)](https://leetcode.com/problems/perfect-squares/description/)
|
||
|
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题目描述：For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
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```java
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public int numSquares(int n) {
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List<Integer> squareList = generateSquareList(n);
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int[] dp = new int[n + 1];
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for (int i = 1; i <= n; i++) {
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int min = Integer.MAX_VALUE;
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for (int square : squareList) {
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if (square > i) {
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break;
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}
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min = Math.min(min, dp[i - square] + 1);
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}
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dp[i] = min;
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}
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return dp[n];
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}
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private List<Integer> generateSquareList(int n) {
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List<Integer> squareList = new ArrayList<>();
|
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int diff = 3;
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int square = 1;
|
||
while (square <= n) {
|
||
squareList.add(square);
|
||
square += diff;
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||
diff += 2;
|
||
}
|
||
return squareList;
|
||
}
|
||
```
|
||
|
||
## 3. 分割整数构成字母字符串
|
||
|
||
[91. Decode Ways (Medium)](https://leetcode.com/problems/decode-ways/description/)
|
||
|
||
题目描述：Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).
|
||
|
||
```java
|
||
public int numDecodings(String s) {
|
||
if (s == null || s.length() == 0) {
|
||
return 0;
|
||
}
|
||
int n = s.length();
|
||
int[] dp = new int[n + 1];
|
||
dp[0] = 1;
|
||
dp[1] = s.charAt(0) == '0' ? 0 : 1;
|
||
for (int i = 2; i <= n; i++) {
|
||
int one = Integer.valueOf(s.substring(i - 1, i));
|
||
if (one != 0) {
|
||
dp[i] += dp[i - 1];
|
||
}
|
||
if (s.charAt(i - 2) == '0') {
|
||
continue;
|
||
}
|
||
int two = Integer.valueOf(s.substring(i - 2, i));
|
||
if (two <= 26) {
|
||
dp[i] += dp[i - 2];
|
||
}
|
||
}
|
||
return dp[n];
|
||
}
|
||
```
|
||
|
||
# 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** 。
|
||
|
||
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} 。
|
||
|
||
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\}" class="mathjax-pic"/></div> <br>-->
|
||
|
||
<div align="center"> <img src="pics/ee994da4-0fc7-443d-ac56-c08caf00a204.jpg" width="350px"> </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，max{ dp[i] | 1 <= i <= N} 即为所求。
|
||
|
||
## 1. 最长递增子序列
|
||
|
||
[300. Longest Increasing Subsequence (Medium)](https://leetcode.com/problems/longest-increasing-subsequence/description/)
|
||
|
||
```java
|
||
public int lengthOfLIS(int[] nums) {
|
||
int n = nums.length;
|
||
int[] dp = new int[n];
|
||
for (int i = 0; i < n; i++) {
|
||
int max = 1;
|
||
for (int j = 0; j < i; j++) {
|
||
if (nums[i] > nums[j]) {
|
||
max = Math.max(max, dp[j] + 1);
|
||
}
|
||
}
|
||
dp[i] = max;
|
||
}
|
||
return Arrays.stream(dp).max().orElse(0);
|
||
}
|
||
```
|
||
|
||
使用 Stream 求最大值会导致运行时间过长，可以改成以下形式：
|
||
|
||
```java
|
||
int ret = 0;
|
||
for (int i = 0; i < n; i++) {
|
||
ret = Math.max(ret, dp[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)。
|
||
|
||
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，
|
||
|
||
- 如果它大于 tails 数组所有的值，那么把它添加到 tails 后面，表示最长递增子序列长度加 1；
|
||
- 如果 tails[i-1] < x <= tails[i]，那么更新 tails[i] = x。
|
||
|
||
例如对于数组 [4,3,6,5]，有：
|
||
|
||
```html
|
||
tails len num
|
||
[] 0 4
|
||
[4] 1 3
|
||
[3] 1 6
|
||
[3,6] 2 5
|
||
[3,5] 2 null
|
||
```
|
||
|
||
可以看出 tails 数组保持有序，因此在查找 S<sub>i</sub> 位于 tails 数组的位置时就可以使用二分查找。
|
||
|
||
```java
|
||
public int lengthOfLIS(int[] nums) {
|
||
int n = nums.length;
|
||
int[] tails = new int[n];
|
||
int len = 0;
|
||
for (int num : nums) {
|
||
int index = binarySearch(tails, len, num);
|
||
tails[index] = num;
|
||
if (index == len) {
|
||
len++;
|
||
}
|
||
}
|
||
return len;
|
||
}
|
||
|
||
private int binarySearch(int[] tails, int len, int key) {
|
||
int l = 0, h = len;
|
||
while (l < h) {
|
||
int mid = l + (h - l) / 2;
|
||
if (tails[mid] == key) {
|
||
return mid;
|
||
} else if (tails[mid] > key) {
|
||
h = mid;
|
||
} else {
|
||
l = mid + 1;
|
||
}
|
||
}
|
||
return l;
|
||
}
|
||
```
|
||
|
||
## 2. 一组整数对能够构成的最长链
|
||
|
||
[646. Maximum Length of Pair Chain (Medium)](https://leetcode.com/problems/maximum-length-of-pair-chain/description/)
|
||
|
||
```html
|
||
Input: [[1,2], [2,3], [3,4]]
|
||
Output: 2
|
||
Explanation: The longest chain is [1,2] -> [3,4]
|
||
```
|
||
|
||
题目描述：对于 (a, b) 和 (c, d) ，如果 b < c，则它们可以构成一条链。
|
||
|
||
```java
|
||
public int findLongestChain(int[][] pairs) {
|
||
if (pairs == null || pairs.length == 0) {
|
||
return 0;
|
||
}
|
||
Arrays.sort(pairs, (a, b) -> (a[0] - b[0]));
|
||
int n = pairs.length;
|
||
int[] dp = new int[n];
|
||
Arrays.fill(dp, 1);
|
||
for (int i = 1; i < n; i++) {
|
||
for (int j = 0; j < i; j++) {
|
||
if (pairs[j][1] < pairs[i][0]) {
|
||
dp[i] = Math.max(dp[i], dp[j] + 1);
|
||
}
|
||
}
|
||
}
|
||
return Arrays.stream(dp).max().orElse(0);
|
||
}
|
||
```
|
||
|
||
## 3. 最长摆动子序列
|
||
|
||
[376. Wiggle Subsequence (Medium)](https://leetcode.com/problems/wiggle-subsequence/description/)
|
||
|
||
```html
|
||
Input: [1,7,4,9,2,5]
|
||
Output: 6
|
||
The entire sequence is a wiggle sequence.
|
||
|
||
Input: [1,17,5,10,13,15,10,5,16,8]
|
||
Output: 7
|
||
There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
|
||
|
||
Input: [1,2,3,4,5,6,7,8,9]
|
||
Output: 2
|
||
```
|
||
|
||
要求：使用 O(N) 时间复杂度求解。
|
||
|
||
```java
|
||
public int wiggleMaxLength(int[] nums) {
|
||
if (nums == null || nums.length == 0) {
|
||
return 0;
|
||
}
|
||
int up = 1, down = 1;
|
||
for (int i = 1; i < nums.length; i++) {
|
||
if (nums[i] > nums[i - 1]) {
|
||
up = down + 1;
|
||
} else if (nums[i] < nums[i - 1]) {
|
||
down = up + 1;
|
||
}
|
||
}
|
||
return Math.max(up, down);
|
||
}
|
||
```
|
||
|
||
# 最长公共子序列
|
||
|
||
对于两个子序列 S1 和 S2，找出它们最长的公共子序列。
|
||
|
||
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，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." class="mathjax-pic"/></div> <br>-->
|
||
|
||
<div align="center"> <img src="pics/ecd89a22-c075-4716-8423-e0ba89230e9a.jpg" width="450px"> </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。
|
||
|
||
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
|
||
public int lengthOfLCS(int[] nums1, int[] nums2) {
|
||
int n1 = nums1.length, n2 = nums2.length;
|
||
int[][] dp = new int[n1 + 1][n2 + 1];
|
||
for (int i = 1; i <= n1; i++) {
|
||
for (int j = 1; j <= n2; j++) {
|
||
if (nums1[i - 1] == nums2[j - 1]) {
|
||
dp[i][j] = dp[i - 1][j - 1] + 1;
|
||
} else {
|
||
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
|
||
}
|
||
}
|
||
}
|
||
return dp[n1][n2];
|
||
}
|
||
```
|
||
|
||
# 0-1 背包
|
||
|
||
有一个容量为 N 的背包，要用这个背包装下物品的价值最大，这些物品有两个属性：体积 w 和价值 v。
|
||
|
||
定义一个二维数组 dp 存储最大价值，其中 dp[i][j] 表示前 i 件物品体积不超过 j 的情况下能达到的最大价值。设第 i 件物品体积为 w，价值为 v，根据第 i 件物品是否添加到背包中，可以分两种情况讨论：
|
||
|
||
- 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。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)" class="mathjax-pic"/></div> <br>-->
|
||
|
||
<div align="center"> <img src="pics/8cb2be66-3d47-41ba-b55b-319fc68940d4.png" width="400px"> </div><br>
|
||
|
||
```java
|
||
// W 为背包总体积
|
||
// N 为物品数量
|
||
// weights 数组存储 N 个物品的重量
|
||
// values 数组存储 N 个物品的价值
|
||
public int knapsack(int W, int N, int[] weights, int[] values) {
|
||
int[][] dp = new int[N + 1][W + 1];
|
||
for (int i = 1; i <= N; i++) {
|
||
int w = weights[i - 1], v = values[i - 1];
|
||
for (int j = 1; j <= W; j++) {
|
||
if (j >= w) {
|
||
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - w] + v);
|
||
} else {
|
||
dp[i][j] = dp[i - 1][j];
|
||
}
|
||
}
|
||
}
|
||
return dp[N][W];
|
||
}
|
||
```
|
||
|
||
**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)" class="mathjax-pic"/></div> <br>-->
|
||
|
||
<div align="center"> <img src="pics/9ae89f16-7905-4a6f-88a2-874b4cac91f4.jpg" width="300px"> </div><br>
|
||
|
||
因为 dp[j-w] 表示 dp[i-1][j-w]，因此不能先求 dp[i][j-w]，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。
|
||
|
||
```java
|
||
public int knapsack(int W, int N, int[] weights, int[] values) {
|
||
int[] dp = new int[W + 1];
|
||
for (int i = 1; i <= N; i++) {
|
||
int w = weights[i - 1], v = values[i - 1];
|
||
for (int j = W; j >= 1; j--) {
|
||
if (j >= w) {
|
||
dp[j] = Math.max(dp[j], dp[j - w] + v);
|
||
}
|
||
}
|
||
}
|
||
return dp[W];
|
||
}
|
||
```
|
||
|
||
**无法使用贪心算法的解释**
|
||
|
||
0-1 背包问题无法使用贪心算法来求解，也就是说不能按照先添加性价比最高的物品来达到最优，这是因为这种方式可能造成背包空间的浪费，从而无法达到最优。考虑下面的物品和一个容量为 5 的背包，如果先添加物品 0 再添加物品 1，那么只能存放的价值为 16，浪费了大小为 2 的空间。最优的方式是存放物品 1 和物品 2，价值为 22.
|
||
|
||
| id | w | v | v/w |
|
||
| --- | --- | --- | --- |
|
||
| 0 | 1 | 6 | 6 |
|
||
| 1 | 2 | 10 | 5 |
|
||
| 2 | 3 | 12 | 4 |
|
||
|
||
**变种**
|
||
|
||
- 完全背包：物品数量为无限个
|
||
|
||
- 多重背包：物品数量有限制
|
||
|
||
- 多维费用背包：物品不仅有重量，还有体积，同时考虑这两种限制
|
||
|
||
- 其它：物品之间相互约束或者依赖
|
||
|
||
## 1. 划分数组为和相等的两部分
|
||
|
||
[416. Partition Equal Subset Sum (Medium)](https://leetcode.com/problems/partition-equal-subset-sum/description/)
|
||
|
||
```html
|
||
Input: [1, 5, 11, 5]
|
||
|
||
Output: true
|
||
|
||
Explanation: The array can be partitioned as [1, 5, 5] and [11].
|
||
```
|
||
|
||
可以看成一个背包大小为 sum/2 的 0-1 背包问题。
|
||
|
||
```java
|
||
public boolean canPartition(int[] nums) {
|
||
int sum = computeArraySum(nums);
|
||
if (sum % 2 != 0) {
|
||
return false;
|
||
}
|
||
int W = sum / 2;
|
||
boolean[] dp = new boolean[W + 1];
|
||
dp[0] = true;
|
||
for (int num : nums) { // 0-1 背包一个物品只能用一次
|
||
for (int i = W; i >= num; i--) { // 从后往前，先计算 dp[i] 再计算 dp[i-num]
|
||
dp[i] = dp[i] || dp[i - num];
|
||
}
|
||
}
|
||
return dp[W];
|
||
}
|
||
|
||
private int computeArraySum(int[] nums) {
|
||
int sum = 0;
|
||
for (int num : nums) {
|
||
sum += num;
|
||
}
|
||
return sum;
|
||
}
|
||
```
|
||
|
||
## 2. 改变一组数的正负号使得它们的和为一给定数
|
||
|
||
[494. Target Sum (Medium)](https://leetcode.com/problems/target-sum/description/)
|
||
|
||
```html
|
||
Input: nums is [1, 1, 1, 1, 1], S is 3.
|
||
Output: 5
|
||
Explanation:
|
||
|
||
-1+1+1+1+1 = 3
|
||
+1-1+1+1+1 = 3
|
||
+1+1-1+1+1 = 3
|
||
+1+1+1-1+1 = 3
|
||
+1+1+1+1-1 = 3
|
||
|
||
There are 5 ways to assign symbols to make the sum of nums be target 3.
|
||
```
|
||
|
||
该问题可以转换为 Subset Sum 问题，从而使用 0-1 背包的方法来求解。
|
||
|
||
可以将这组数看成两部分，P 和 N，其中 P 使用正号，N 使用负号，有以下推导：
|
||
|
||
```html
|
||
sum(P) - sum(N) = target
|
||
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
|
||
2 * sum(P) = target + sum(nums)
|
||
```
|
||
|
||
因此只要找到一个子集，令它们都取正号，并且和等于 (target + sum(nums))/2，就证明存在解。
|
||
|
||
```java
|
||
public int findTargetSumWays(int[] nums, int S) {
|
||
int sum = computeArraySum(nums);
|
||
if (sum < S || (sum + S) % 2 == 1) {
|
||
return 0;
|
||
}
|
||
int W = (sum + S) / 2;
|
||
int[] dp = new int[W + 1];
|
||
dp[0] = 1;
|
||
for (int num : nums) {
|
||
for (int i = W; i >= num; i--) {
|
||
dp[i] = dp[i] + dp[i - num];
|
||
}
|
||
}
|
||
return dp[W];
|
||
}
|
||
|
||
private int computeArraySum(int[] nums) {
|
||
int sum = 0;
|
||
for (int num : nums) {
|
||
sum += num;
|
||
}
|
||
return sum;
|
||
}
|
||
```
|
||
|
||
DFS 解法：
|
||
|
||
```java
|
||
public int findTargetSumWays(int[] nums, int S) {
|
||
return findTargetSumWays(nums, 0, S);
|
||
}
|
||
|
||
private int findTargetSumWays(int[] nums, int start, int S) {
|
||
if (start == nums.length) {
|
||
return S == 0 ? 1 : 0;
|
||
}
|
||
return findTargetSumWays(nums, start + 1, S + nums[start])
|
||
+ findTargetSumWays(nums, start + 1, S - nums[start]);
|
||
}
|
||
```
|
||
|
||
## 3. 01 字符构成最多的字符串
|
||
|
||
[474. Ones and Zeroes (Medium)](https://leetcode.com/problems/ones-and-zeroes/description/)
|
||
|
||
```html
|
||
Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
|
||
Output: 4
|
||
|
||
Explanation: There are totally 4 strings can be formed by the using of 5 0s and 3 1s, which are "10","0001","1","0"
|
||
```
|
||
|
||
这是一个多维费用的 0-1 背包问题，有两个背包大小，0 的数量和 1 的数量。
|
||
|
||
```java
|
||
public int findMaxForm(String[] strs, int m, int n) {
|
||
if (strs == null || strs.length == 0) {
|
||
return 0;
|
||
}
|
||
int[][] dp = new int[m + 1][n + 1];
|
||
for (String s : strs) { // 每个字符串只能用一次
|
||
int ones = 0, zeros = 0;
|
||
for (char c : s.toCharArray()) {
|
||
if (c == '0') {
|
||
zeros++;
|
||
} else {
|
||
ones++;
|
||
}
|
||
}
|
||
for (int i = m; i >= zeros; i--) {
|
||
for (int j = n; j >= ones; j--) {
|
||
dp[i][j] = Math.max(dp[i][j], dp[i - zeros][j - ones] + 1);
|
||
}
|
||
}
|
||
}
|
||
return dp[m][n];
|
||
}
|
||
```
|
||
|
||
## 4. 找零钱的最少硬币数
|
||
|
||
[322. Coin Change (Medium)](https://leetcode.com/problems/coin-change/description/)
|
||
|
||
```html
|
||
Example 1:
|
||
coins = [1, 2, 5], amount = 11
|
||
return 3 (11 = 5 + 5 + 1)
|
||
|
||
Example 2:
|
||
coins = [2], amount = 3
|
||
return -1.
|
||
```
|
||
|
||
题目描述：给一些面额的硬币，要求用这些硬币来组成给定面额的钱数，并且使得硬币数量最少。硬币可以重复使用。
|
||
|
||
- 物品：硬币
|
||
- 物品大小：面额
|
||
- 物品价值：数量
|
||
|
||
因为硬币可以重复使用，因此这是一个完全背包问题。完全背包只需要将 0-1 背包的逆序遍历 dp 数组改为正序遍历即可。
|
||
|
||
```java
|
||
public int coinChange(int[] coins, int amount) {
|
||
int[] dp = new int[amount + 1];
|
||
for (int coin : coins) {
|
||
for (int i = coin; i <= amount; i++) { //将逆序遍历改为正序遍历
|
||
if (i == coin) {
|
||
dp[i] = 1;
|
||
} else if (dp[i] == 0 && dp[i - coin] != 0) {
|
||
dp[i] = dp[i - coin] + 1;
|
||
|
||
} else if (dp[i - coin] != 0) {
|
||
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
|
||
}
|
||
}
|
||
}
|
||
return dp[amount] == 0 ? -1 : dp[amount];
|
||
}
|
||
```
|
||
|
||
## 5. 找零钱的硬币数组合
|
||
|
||
[518\. Coin Change 2 (Medium)](https://leetcode.com/problems/coin-change-2/description/)
|
||
|
||
```text-html-basic
|
||
Input: amount = 5, coins = [1, 2, 5]
|
||
Output: 4
|
||
Explanation: there are four ways to make up the amount:
|
||
5=5
|
||
5=2+2+1
|
||
5=2+1+1+1
|
||
5=1+1+1+1+1
|
||
```
|
||
|
||
完全背包问题，使用 dp 记录可达成目标的组合数目。
|
||
|
||
```java
|
||
public int change(int amount, int[] coins) {
|
||
if (coins == null) {
|
||
return 0;
|
||
}
|
||
int[] dp = new int[amount + 1];
|
||
dp[0] = 1;
|
||
for (int coin : coins) {
|
||
for (int i = coin; i <= amount; i++) {
|
||
dp[i] += dp[i - coin];
|
||
}
|
||
}
|
||
return dp[amount];
|
||
}
|
||
```
|
||
|
||
## 6. 字符串按单词列表分割
|
||
|
||
[139. Word Break (Medium)](https://leetcode.com/problems/word-break/description/)
|
||
|
||
```html
|
||
s = "leetcode",
|
||
dict = ["leet", "code"].
|
||
Return true because "leetcode" can be segmented as "leet code".
|
||
```
|
||
|
||
dict 中的单词没有使用次数的限制，因此这是一个完全背包问题。
|
||
|
||
该问题涉及到字典中单词的使用顺序，也就是说物品必须按一定顺序放入背包中，例如下面的 dict 就不够组成字符串 "leetcode"：
|
||
|
||
```html
|
||
["lee", "tc", "cod"]
|
||
```
|
||
|
||
求解顺序的完全背包问题时，对物品的迭代应该放在最里层，对背包的迭代放在外层，只有这样才能让物品按一定顺序放入背包中。
|
||
|
||
```java
|
||
public boolean wordBreak(String s, List<String> wordDict) {
|
||
int n = s.length();
|
||
boolean[] dp = new boolean[n + 1];
|
||
dp[0] = true;
|
||
for (int i = 1; i <= n; i++) {
|
||
for (String word : wordDict) { // 对物品的迭代应该放在最里层
|
||
int len = word.length();
|
||
if (len <= i && word.equals(s.substring(i - len, i))) {
|
||
dp[i] = dp[i] || dp[i - len];
|
||
}
|
||
}
|
||
}
|
||
return dp[n];
|
||
}
|
||
```
|
||
|
||
## 7. 组合总和
|
||
|
||
[377. Combination Sum IV (Medium)](https://leetcode.com/problems/combination-sum-iv/description/)
|
||
|
||
```html
|
||
nums = [1, 2, 3]
|
||
target = 4
|
||
|
||
The possible combination ways are:
|
||
(1, 1, 1, 1)
|
||
(1, 1, 2)
|
||
(1, 2, 1)
|
||
(1, 3)
|
||
(2, 1, 1)
|
||
(2, 2)
|
||
(3, 1)
|
||
|
||
Note that different sequences are counted as different combinations.
|
||
|
||
Therefore the output is 7.
|
||
```
|
||
|
||
涉及顺序的完全背包。
|
||
|
||
```java
|
||
public int combinationSum4(int[] nums, int target) {
|
||
if (nums == null || nums.length == 0) {
|
||
return 0;
|
||
}
|
||
int[] maximum = new int[target + 1];
|
||
maximum[0] = 1;
|
||
Arrays.sort(nums);
|
||
for (int i = 1; i <= target; i++) {
|
||
for (int j = 0; j < nums.length && nums[j] <= i; j++) {
|
||
maximum[i] += maximum[i - nums[j]];
|
||
}
|
||
}
|
||
return maximum[target];
|
||
}
|
||
```
|
||
|
||
# 股票交易
|
||
|
||
## 1. 需要冷却期的股票交易
|
||
|
||
[309. Best Time to Buy and Sell Stock with Cooldown(Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/description/)
|
||
|
||
题目描述：交易之后需要有一天的冷却时间。
|
||
|
||
<div align="center"> <img src="pics/ffd96b99-8009-487c-8e98-11c9d44ef14f.png" width="300px"> </div><br>
|
||
|
||
```java
|
||
public int maxProfit(int[] prices) {
|
||
if (prices == null || prices.length == 0) {
|
||
return 0;
|
||
}
|
||
int N = prices.length;
|
||
int[] buy = new int[N];
|
||
int[] s1 = new int[N];
|
||
int[] sell = new int[N];
|
||
int[] s2 = new int[N];
|
||
s1[0] = buy[0] = -prices[0];
|
||
sell[0] = s2[0] = 0;
|
||
for (int i = 1; i < N; i++) {
|
||
buy[i] = s2[i - 1] - prices[i];
|
||
s1[i] = Math.max(buy[i - 1], s1[i - 1]);
|
||
sell[i] = Math.max(buy[i - 1], s1[i - 1]) + prices[i];
|
||
s2[i] = Math.max(s2[i - 1], sell[i - 1]);
|
||
}
|
||
return Math.max(sell[N - 1], s2[N - 1]);
|
||
}
|
||
```
|
||
|
||
## 2. 需要交易费用的股票交易
|
||
|
||
[714. Best Time to Buy and Sell Stock with Transaction Fee (Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)
|
||
|
||
```html
|
||
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
|
||
Output: 8
|
||
Explanation: The maximum profit can be achieved by:
|
||
Buying at prices[0] = 1
|
||
Selling at prices[3] = 8
|
||
Buying at prices[4] = 4
|
||
Selling at prices[5] = 9
|
||
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
|
||
```
|
||
|
||
题目描述：每交易一次，都要支付一定的费用。
|
||
|
||
<div align="center"> <img src="pics/1e2c588c-72b7-445e-aacb-d55dc8a88c29.png" width="300px"> </div><br>
|
||
|
||
```java
|
||
public int maxProfit(int[] prices, int fee) {
|
||
int N = prices.length;
|
||
int[] buy = new int[N];
|
||
int[] s1 = new int[N];
|
||
int[] sell = new int[N];
|
||
int[] s2 = new int[N];
|
||
s1[0] = buy[0] = -prices[0];
|
||
sell[0] = s2[0] = 0;
|
||
for (int i = 1; i < N; i++) {
|
||
buy[i] = Math.max(sell[i - 1], s2[i - 1]) - prices[i];
|
||
s1[i] = Math.max(buy[i - 1], s1[i - 1]);
|
||
sell[i] = Math.max(buy[i - 1], s1[i - 1]) - fee + prices[i];
|
||
s2[i] = Math.max(s2[i - 1], sell[i - 1]);
|
||
}
|
||
return Math.max(sell[N - 1], s2[N - 1]);
|
||
}
|
||
```
|
||
|
||
|
||
## 3. 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;
|
||
}
|
||
```
|
||
|
||
## 4. 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
|
||
|
||
## 1. 删除两个字符串的字符使它们相等
|
||
|
||
[583. Delete Operation for Two Strings (Medium)](https://leetcode.com/problems/delete-operation-for-two-strings/description/)
|
||
|
||
```html
|
||
Input: "sea", "eat"
|
||
Output: 2
|
||
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
|
||
```
|
||
|
||
可以转换为求两个字符串的最长公共子序列问题。
|
||
|
||
```java
|
||
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 = 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 {
|
||
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
|
||
}
|
||
}
|
||
}
|
||
return m + n - 2 * dp[m][n];
|
||
}
|
||
```
|
||
|
||
## 2. 编辑距离
|
||
|
||
[72. Edit Distance (Hard)](https://leetcode.com/problems/edit-distance/description/)
|
||
|
||
```html
|
||
Example 1:
|
||
|
||
Input: word1 = "horse", word2 = "ros"
|
||
Output: 3
|
||
Explanation:
|
||
horse -> rorse (replace 'h' with 'r')
|
||
rorse -> rose (remove 'r')
|
||
rose -> ros (remove 'e')
|
||
Example 2:
|
||
|
||
Input: word1 = "intention", word2 = "execution"
|
||
Output: 5
|
||
Explanation:
|
||
intention -> inention (remove 't')
|
||
inention -> enention (replace 'i' with 'e')
|
||
enention -> exention (replace 'n' with 'x')
|
||
exention -> exection (replace 'n' with 'c')
|
||
exection -> execution (insert 'u')
|
||
```
|
||
|
||
题目描述：修改一个字符串成为另一个字符串，使得修改次数最少。一次修改操作包括：插入一个字符、删除一个字符、替换一个字符。
|
||
|
||
```java
|
||
public int minDistance(String word1, String word2) {
|
||
if (word1 == null || word2 == null) {
|
||
return 0;
|
||
}
|
||
int m = word1.length(), n = word2.length();
|
||
int[][] dp = new int[m + 1][n + 1];
|
||
for (int i = 1; i <= m; i++) {
|
||
dp[i][0] = i;
|
||
}
|
||
for (int i = 1; i <= n; i++) {
|
||
dp[0][i] = i;
|
||
}
|
||
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];
|
||
} else {
|
||
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1;
|
||
}
|
||
}
|
||
}
|
||
return dp[m][n];
|
||
}
|
||
```
|
||
|
||
## 3. 复制粘贴字符
|
||
|
||
[650. 2 Keys Keyboard (Medium)](https://leetcode.com/problems/2-keys-keyboard/description/)
|
||
|
||
题目描述：最开始只有一个字符 A，问需要多少次操作能够得到 n 个字符 A，每次操作可以复制当前所有的字符，或者粘贴。
|
||
|
||
```
|
||
Input: 3
|
||
Output: 3
|
||
Explanation:
|
||
Intitally, we have one character 'A'.
|
||
In step 1, we use Copy All operation.
|
||
In step 2, we use Paste operation to get 'AA'.
|
||
In step 3, we use Paste operation to get 'AAA'.
|
||
```
|
||
|
||
```java
|
||
public int minSteps(int n) {
|
||
if (n == 1) return 0;
|
||
for (int i = 2; i <= Math.sqrt(n); i++) {
|
||
if (n % i == 0) return i + minSteps(n / i);
|
||
}
|
||
return n;
|
||
}
|
||
```
|
||
|
||
```java
|
||
public int minSteps(int n) {
|
||
int[] dp = new int[n + 1];
|
||
int h = (int) Math.sqrt(n);
|
||
for (int i = 2; i <= n; i++) {
|
||
dp[i] = i;
|
||
for (int j = 2; j <= h; j++) {
|
||
if (i % j == 0) {
|
||
dp[i] = dp[j] + dp[i / j];
|
||
break;
|
||
}
|
||
}
|
||
}
|
||
return dp[n];
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
# 微信公众号
|
||
|
||
|
||
更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
|
||
|
||
|
||
<br><div align="center"><img width="510px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号海报7.png"></img></div>
|