代码随想录算法训练营第五十五 | ● 583. 两个字符串的删除操作 ● 72. 编辑距离

本文主要是介绍代码随想录算法训练营第五十五 | ● 583. 两个字符串的删除操作 ● 72. 编辑距离，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

583. 两个字符串的删除操作

https://programmercarl.com/0583.%E4%B8%A4%E4%B8%AA%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9A%84%E5%88%A0%E9%99%A4%E6%93%8D%E4%BD%9C.html

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0));//初始化for(int i=0;i<=word1.size();i++)dp[i][0]=i;for(int i=0;i<=word2.size();i++)dp[0][i]=i;for(int i=1;i<=word1.size();i++) {for(int j=1;j<=word2.size();j++) {if(word1[i-1]==word2[j-1])dp[i][j] = dp[i-1][j-1];//这里的逻辑一开始考虑错了//dp[i][j] = min(dp[i-1][j]-1,min(dp[i][j-1]-1,dp[i-1][j-1]));elsedp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1);}}return dp[word1.size()][word2.size()];}
};

在这里插入图片描述

72. 编辑距离

https://programmercarl.com/0072.%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB.html

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0));for(int i=0;i<=word1.size();i++)dp[i][0]=i;for(int i=0;i<=word2.size();i++)dp[0][i]=i;for(int i=1;i<=word1.size();i++) {for(int j=1;j<=word2.size();j++) {if(word1[i-1]==word2[j-1])dp[i][j] = dp[i-1][j-1];else{dp[i][j]=min({dp[i-1][j-1],dp[i][j-1],dp[i-1][j]})+1;//dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1);}}}return dp[word1.size()][word2.size()];}
};