【代码随想录Day56 动态规划 part16】583.两个字符串的删除操作、72.编辑距离

Day56

583.两个字符串的删除操作

class Solution {
    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int dp[][] = new int[n + 1][m + 1];
        for (int i = 1; i <= n; i++) {
            char char1 = word1.charAt(i - 1);
            for (int j = 1; j <= m; j++) {
                char char2 = word2.charAt(j - 1);
                if (char1 == char2) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return n + m - 2 * dp[n][m];
    }
}

72.编辑距离

class Solution {
    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int dp[][] = new int[n + 1][m + 1];
        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.charAt(i - 1) == word2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
                else {
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1), dp[i - 1][j - 1] + 1);
                }
             }
        }
        return dp[n][m];
    }
}