代码随想录三刷day47

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前言

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本周我们讲了动态规划之终极绝杀：编辑距离，为什么叫做终极绝杀呢？ 细心的录友应该知道，我们在前三篇动态规划的文章就一直为 编辑距离 这道题目做铺垫

一、力扣115. 不同的子序列

class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        int[][] dp = new int[m+1][n+1];
        for(int i = 0; i <= m; i ++){
            dp[i][0] = 1;
        }
        for(int i = 1; i <= m; i ++){
            for(int j = 1; j <= n; j ++){
                if(s.charAt(i-1) == t.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                }else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        return dp[m][n];
    }
}

二、力扣583. 两个字符串的删除操作

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m+1][n+1];
        for(int i = 0; i <= m; i ++){
            dp[i][0] = i;
        }
        for(int i = 0; i <= n; i ++){
            dp[0][i] = i;
        }
        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]+1),dp[i-1][j-1]+2);
                }
            }
        }
        return dp[m][n];
    }
}

三、力扣72. 编辑距离

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