代码随想录二刷day56

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前言

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一、力扣583. 两个字符串的删除操作

class Solution {
    public int minDistance(String word1, String word2) {
        int row = word1.length()+1;
        int col = word2.length()+1;
        int[][] dp = new int[row][col];
        for(int i = 0; i < row; i ++){
            dp[i][0] = i;
        }
        for(int j = 0; j < col; j ++){
            dp[0][j] = j;
        }
        for(int i = 1;i < row; i ++){
            for(int j = 1; j < col; 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],dp[i][j-1])+1;
                }
            }
        }
        return dp[row-1][col-1];
    }
}

二、力扣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];
    // 初始化
    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++) {
            // 因为dp数组有效位从1开始
            // 所以当前遍历到的字符串的位置为i-1 | j-1
            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];
    }
}