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471 lines
14 KiB
Markdown
Executable File
471 lines
14 KiB
Markdown
Executable File
* [做项目（多个C++、Java、Go、测开、前端项目）](https://www.programmercarl.com/other/kstar.html)
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* [刷算法（两个月高强度学算法）](https://www.programmercarl.com/xunlian/xunlianying.html)
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* [背八股（40天挑战高频面试题）](https://www.programmercarl.com/xunlian/bagu.html)
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# 583. 两个字符串的删除操作
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[力扣题目链接](https://leetcode.cn/problems/delete-operation-for-two-strings/)
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给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。
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示例：
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* 输入: "sea", "eat"
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* 输出: 2
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* 解释: 第一步将"sea"变为"ea"，第二步将"eat"变为"ea"
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[LeetCode：583.两个字符串的删除操](https://www.bilibili.com/video/BV1we4y157wB/)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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### 动态规划一
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本题和[动态规划：115.不同的子序列](https://programmercarl.com/0115.不同的子序列.html)相比，其实就是两个字符串都可以删除了，情况虽说复杂一些，但整体思路是不变的。
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这次是两个字符串可以相互删了，这种题目也知道用动态规划的思路来解，动规五部曲，分析如下：
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1. 确定dp数组（dp table）以及下标的含义
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dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，想要达到相等，所需要删除元素的最少次数。
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这里dp数组的定义有点点绕，大家要理清思路。
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2. 确定递推公式
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* 当word1[i - 1] 与 word2[j - 1]相同的时候
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* 当word1[i - 1] 与 word2[j - 1]不相同的时候
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当word1[i - 1] 与 word2[j - 1]相同的时候，dp[i][j] = dp[i - 1][j - 1];
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当word1[i - 1] 与 word2[j - 1]不相同的时候，有三种情况：
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情况一：删word1[i - 1]，最少操作次数为dp[i - 1][j] + 1
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情况二：删word2[j - 1]，最少操作次数为dp[i][j - 1] + 1
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情况三：同时删word1[i - 1]和word2[j - 1]，操作的最少次数为dp[i - 1][j - 1] + 2
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那最后当然是取最小值，所以当word1[i - 1] 与 word2[j - 1]不相同的时候，递推公式：dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
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因为 dp[i][j - 1] + 1 = dp[i - 1][j - 1] + 2，所以递推公式可简化为：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
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这里可能不少录友有点迷糊，从字面上理解 就是 当 同时删word1[i - 1]和word2[j - 1]，dp[i][j-1] 本来就不考虑 word2[j - 1]了，那么我在删 word1[i - 1]，是不是就达到两个元素都删除的效果，即 dp[i][j-1] + 1。
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3. dp数组如何初始化
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从递推公式中，可以看出来，dp[i][0] 和 dp[0][j]是一定要初始化的。
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dp[i][0]：word2为空字符串，以i-1为结尾的字符串word1要删除多少个元素，才能和word2相同呢，很明显dp[i][0] = i。
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dp[0][j]的话同理，所以代码如下：
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```CPP
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vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
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for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
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for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
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```
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4. 确定遍历顺序
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从递推公式 dp[i][j] = min(dp[i - 1][j - 1] + 2, min(dp[i - 1][j], dp[i][j - 1]) + 1); 和dp[i][j] = dp[i - 1][j - 1]可以看出dp[i][j]都是根据左上方、正上方、正左方推出来的。
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所以遍历的时候一定是从上到下，从左到右，这样保证dp[i][j]可以根据之前计算出来的数值进行计算。
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5. 举例推导dp数组
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以word1:"sea"，word2:"eat"为例，推导dp数组状态图如下：
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以上分析完毕，代码如下：
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```CPP
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class Solution {
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public:
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int minDistance(string word1, string word2) {
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vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
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for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
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for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
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for (int i = 1; i <= word1.size(); i++) {
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for (int j = 1; j <= word2.size(); j++) {
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if (word1[i - 1] == word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
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}
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}
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}
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return dp[word1.size()][word2.size()];
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}
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};
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```
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* 时间复杂度: O(n * m)
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* 空间复杂度: O(n * m)
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### 动态规划二
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本题和[动态规划：1143.最长公共子序列](https://programmercarl.com/1143.最长公共子序列.html)基本相同，只要求出两个字符串的最长公共子序列长度即可，那么除了最长公共子序列之外的字符都是必须删除的，最后用两个字符串的总长度减去两个最长公共子序列的长度就是删除的最少步数。
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代码如下：
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```CPP
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class Solution {
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public:
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int minDistance(string word1, string word2) {
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vector<vector<int>> dp(word1.size()+1, vector<int>(word2.size()+1, 0));
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for (int i=1; i<=word1.size(); i++){
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for (int j=1; j<=word2.size(); j++){
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if (word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
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else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
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}
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}
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return word1.size()+word2.size()-dp[word1.size()][word2.size()]*2;
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}
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};
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```
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* 时间复杂度: O(n * m)
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* 空间复杂度: O(n * m)
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## 其他语言版本
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### Java：
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```java
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// dp数组中存储word1和word2最长相同子序列的长度
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class Solution {
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public int minDistance(String word1, String word2) {
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int len1 = word1.length();
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int len2 = word2.length();
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int[][] dp = new int[len1 + 1][len2 + 1];
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for (int i = 1; i <= len1; i++) {
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for (int j = 1; j <= len2; j++) {
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if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return len1 + len2 - dp[len1][len2] * 2;
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}
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}
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// dp数组中存储需要删除的字符个数
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class Solution {
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public int minDistance(String word1, String word2) {
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int[][] dp = new int[word1.length() + 1][word2.length() + 1];
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for (int i = 0; i < word1.length() + 1; i++) dp[i][0] = i;
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for (int j = 0; j < word2.length() + 1; j++) dp[0][j] = j;
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for (int i = 1; i < word1.length() + 1; i++) {
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for (int j = 1; j < word2.length() + 1; j++) {
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if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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}else{
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dp[i][j] = Math.min(dp[i - 1][j - 1] + 2,
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Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
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}
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}
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}
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return dp[word1.length()][word2.length()];
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}
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}
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```
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```java
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//DP - longest common subsequence (用最長公共子序列反推)
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class Solution {
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public int minDistance(String word1, String word2) {
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char[] char1 = word1.toCharArray();
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char[] char2 = word2.toCharArray();
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int len1 = char1.length;
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int len2 = char2.length;
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int dp[][] = new int [len1 + 1][len2 + 1];
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for(int i = 1; i <= len1; i++){
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for(int j = 1; j <= len2; j++){
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if(char1[i - 1] == char2[j - 1])
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dp[i][j] = dp[i - 1][j - 1] + 1;
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else
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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return len1 + len2 - (2 * dp[len1][len2]);//和leetcode 1143只差在這一行。
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}
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}
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```
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### Python：
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```python
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class Solution:
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def minDistance(self, word1: str, word2: str) -> int:
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dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
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for i in range(len(word1)+1):
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dp[i][0] = i
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for j in range(len(word2)+1):
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dp[0][j] = j
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for i in range(1, len(word1)+1):
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for j in range(1, len(word2)+1):
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if word1[i-1] == word2[j-1]:
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dp[i][j] = dp[i-1][j-1]
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else:
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dp[i][j] = min(dp[i-1][j-1] + 2, dp[i-1][j] + 1, dp[i][j-1] + 1)
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return dp[-1][-1]
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```
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> 版本 2
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```python
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class Solution(object):
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def minDistance(self, word1, word2):
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m, n = len(word1), len(word2)
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# dp 求解两字符串最长公共子序列
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dp = [[0] * (n+1) for _ in range(m+1)]
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for i in range(1, m+1):
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for j in range(1, n+1):
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if word1[i-1] == word2[j-1]:
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dp[i][j] = dp[i-1][j-1] + 1
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else:
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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# 删去最长公共子序列以外元素
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return m + n - 2 * dp[-1][-1]
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```
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### Go：
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动态规划一
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```go
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func minDistance(word1 string, word2 string) int {
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dp := make([][]int, len(word1)+1)
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for i := 0; i < len(dp); i++ {
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dp[i] = make([]int, len(word2)+1)
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}
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//初始化
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for i := 0; i < len(dp); i++ {
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dp[i][0] = i
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}
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for j := 0; j < len(dp[0]); j++ {
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dp[0][j] = j
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}
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for i := 1; i < len(dp); i++ {
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for j := 1; j < len(dp[i]); j++ {
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if word1[i-1] == word2[j-1] {
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dp[i][j] = dp[i-1][j-1]
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} else {
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dp[i][j] = min(min(dp[i-1][j]+1, dp[i][j-1]+1), dp[i-1][j-1]+2)
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}
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}
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}
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return dp[len(dp)-1][len(dp[0])-1]
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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动态规划二
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```go
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func minDistance(word1 string, word2 string) int {
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dp := make([][]int, len(word1) + 1)
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for i := range dp {
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dp[i] = make([]int, len(word2) + 1)
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}
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for i := 1; i <= len(word1); i++ {
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for j := 1; j <= len(word2); j++ {
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if word1[i-1] == word2[j-1] {
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dp[i][j] = dp[i-1][j-1] + 1
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} else {
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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}
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}
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}
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return len(word1) + len(word2) - dp[len(word1)][len(word2)] * 2
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}
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```
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### JavaScript：
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```javascript
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// 方法一
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var minDistance = (word1, word2) => {
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let dp = Array.from(new Array(word1.length + 1), () =>
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Array(word2.length + 1).fill(0)
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);
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for (let i = 1; i <= word1.length; i++) {
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dp[i][0] = i;
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}
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for (let j = 1; j <= word2.length; j++) {
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dp[0][j] = j;
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}
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for (let i = 1; i <= word1.length; i++) {
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for (let j = 1; j <= word2.length; j++) {
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if (word1[i - 1] === word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(
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dp[i - 1][j] + 1,
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dp[i][j - 1] + 1,
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dp[i - 1][j - 1] + 2
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);
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}
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}
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}
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return dp[word1.length][word2.length];
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};
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// 方法二
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var minDistance = function (word1, word2) {
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let dp = new Array(word1.length + 1)
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.fill(0)
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.map((_) => new Array(word2.length + 1).fill(0));
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for (let i = 1; i <= word1.length; i++)
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for (let j = 1; j <= word2.length; j++)
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if (word1[i - 1] === word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
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else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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return word1.length + word2.length - dp[word1.length][word2.length] * 2;
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};
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```
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### TypeScript：
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> dp版本一：
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```typescript
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function minDistance(word1: string, word2: string): number {
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/**
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dp[i][j]: word1前i个字符，word2前j个字符，所需最小步数
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dp[0][0]=0: word1前0个字符为'', word2前0个字符为''
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*/
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const length1: number = word1.length,
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length2: number = word2.length;
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const dp: number[][] = new Array(length1 + 1).fill(0)
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.map(_ => new Array(length2 + 1).fill(0));
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for (let i = 0; i <= length1; i++) {
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dp[i][0] = i;
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}
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for (let i = 0; i <= length2; i++) {
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dp[0][i] = i;
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}
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for (let i = 1; i <= length1; i++) {
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for (let j = 1; j <= length2; j++) {
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if (word1[i - 1] === word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
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}
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}
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}
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return dp[length1][length2];
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};
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```
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> dp版本二：
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```typescript
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function minDistance(word1: string, word2: string): number {
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/**
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dp[i][j]: word1前i个字符，word2前j个字符，最长公共子序列的长度
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dp[0][0]=0: word1前0个字符为'', word2前0个字符为''
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*/
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const length1: number = word1.length,
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length2: number = word2.length;
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const dp: number[][] = new Array(length1 + 1).fill(0)
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.map(_ => new Array(length2 + 1).fill(0));
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for (let i = 1; i <= length1; i++) {
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for (let j = 1; j <= length2; j++) {
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if (word1[i - 1] === word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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} else {
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dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
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}
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}
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}
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const maxLen: number = dp[length1][length2];
|
||
return length1 + length2 - maxLen * 2;
|
||
};
|
||
```
|
||
|
||
Rust:
|
||
|
||
```rust
|
||
impl Solution {
|
||
pub fn min_distance(word1: String, word2: String) -> i32 {
|
||
let mut dp = vec![vec![0; word2.len() + 1]; word1.len() + 1];
|
||
for i in 0..word1.len() {
|
||
dp[i + 1][0] = i + 1;
|
||
}
|
||
for j in 0..word2.len() {
|
||
dp[0][j + 1] = j + 1;
|
||
}
|
||
for (i, char1) in word1.chars().enumerate() {
|
||
for (j, char2) in word2.chars().enumerate() {
|
||
if char1 == char2 {
|
||
dp[i + 1][j + 1] = dp[i][j];
|
||
continue;
|
||
}
|
||
dp[i + 1][j + 1] = dp[i][j + 1].min(dp[i + 1][j]) + 1;
|
||
}
|
||
}
|
||
dp[word1.len()][word2.len()] as i32
|
||
}
|
||
}
|
||
```
|
||
|
||
> 版本 2
|
||
|
||
```rust
|
||
impl Solution {
|
||
pub fn min_distance(word1: String, word2: String) -> i32 {
|
||
let mut dp = vec![vec![0; word2.len() + 1]; word1.len() + 1];
|
||
for (i, char1) in word1.chars().enumerate() {
|
||
for (j, char2) in word2.chars().enumerate() {
|
||
if char1 == char2 {
|
||
dp[i + 1][j + 1] = dp[i][j] + 1;
|
||
continue;
|
||
}
|
||
dp[i + 1][j + 1] = dp[i][j + 1].max(dp[i + 1][j]);
|
||
}
|
||
}
|
||
(word1.len() + word2.len() - 2 * dp[word1.len()][word2.len()]) as i32
|
||
}
|
||
}
|
||
```
|
||
|
||
|