【算法刷题day55】Leetcode：583. 两个字符串的删除操作、72. 编辑距离

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Leetcode 583. 两个字符串的删除操作

题目：583. 两个字符串的删除操作
解析：代码随想录解析

解题思路

dp数组的含义是，从word1从0到i-1，word2从0到j-1匹配上最少需要删除多少个字符

代码

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 = 0; i <= m; i++)
            dp[i][0] = i;
        for (int j = 0; 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(dp[i][j-1] + 1 , Math.min(dp[i-1][j] + 1, dp[i-1][j-1] + 2));
            }
        }
        return dp[m][n];
    }
}

总结

暂无

Leetcode 72. 编辑距离

题目：72. 编辑距离
解析：代码随想录解析

解题思路

和上面那题的区别是，改了一个dp[i-1][i-1] + 1，因为可以修改，而上一题只能删除两个

代码

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 = 0; i <= m; i++)
            dp[i][0] = i;
        for (int j = 0; 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(dp[i-1][j], Math.min(dp[i][j-1], dp[i-1][j-1])) + 1;
            }
        }
        return dp[m][n];
    }
}

总结

暂无