力扣第45天----第583题、第72题

文章目录

一、第583题--两个字符串的删除操作
二、第72题--编辑距离

一、第583题--两个字符串的删除操作

c++ 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 j = 0; j<=word2.size(); j++) dp[0][j] = j;
        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, min(dp[i][j-1]+1, dp[i-1][j-1]+2));
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

二、第72题--编辑距离

很不好想，但是代码跟上一题，差别很小。分为增、删、改3种，增、删的逻辑一致。

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