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

每天努力进步！2024-06-18 19:48

https://programmercarl.com/0583.两个字符串的删除操作.html

cpp 复制代码

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]));
                else
                    dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1);
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

72. 编辑距离

https://programmercarl.com/0072.编辑距离.html

cpp 复制代码

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()];
    }
};

这类题目的代码大致思路：