代码随想录算法训练营20期|第五十八天|动态规划part16|● 583. 两个字符串的删除操作 ● 72. 编辑距离 ● 编辑距离总结篇

•  583. 两个字符串的删除操作 

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1 + 1][len2 + 1];

        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return len1 + len2 - dp[len1][len2] * 2;
    }
}

•  72. 编辑距离 

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp  = new int[m + 1][n + 1];

        for (int i = 1; i <= m; i++) {
            dp[i][0] = i;
        }

        for (int j = 1; j <= n; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; 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]), dp[i - 1][j]) +1;
                }
            }
        }
        return dp[m][n];
    }
}

•  编辑距离总结篇