代码随想录算法训练营第56天 | 583. 两个字符串的删除操作、72. 编辑距离

583. 两个字符串的删除操作

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector> dp(n + 1, vector(m + 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 <= n; i++) {
            for (int j = 1; j <= m; 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[n][m];
    }
};

72. 编辑距离

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