第一题、两个字符串的删除操作 力扣题目链接

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 = 1; i < word1.size()+1; i++){
            for(int j = 1; j < word2.size()+1; j++){
                if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
                else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
        return word1.size()+word2.size() - 2 * dp[word1.size()][word2.size()];
    }
};

第二题、编辑距离 力扣题目链接

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 j = 0; j <= word2.size(); j++) dp[0][j] = j;
        for(int i = 1; i < word1.size()+1; i++){
            for(int j = 1; j < word2.size()+1; 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], dp[i][j-1], dp[i-1][j-1]}) + 1;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

编辑距离总结：