代码随想录二刷Day58

一天24小时在图书馆学习2023-11-04 12:23

583. 两个字符串的删除操作

知道了递推公式挺好写的，和之前那道题差不多

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

72. 编辑距离

删除操作和之前题目一样就不多说了；对一个字符串的增加操作就相当于给另一个字符串做删除操作，所以不需要考虑增的操作，只需要考虑删除就可以了。然后更改操作的目的是把最后一个数给改了然后就是dp $i-1$ $j-1$ +1这个是变数， dp $i - 1$ $j$ , dp $i$ $j - 1$ 是两种删除

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 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], dp[i - 1][j], dp[i][j - 1]}) + 1; //这句是难点，要再理解
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};