LeetCode动态规划系列 编辑距离 冲冲冲

温故而知新嘛！
其实讲道理，这个编辑距离其实蛮简单的啊！

class Solution {
    public int minDistance(String word1, String word2) {
        int N = word1.length();
        int M = word2.length();
        if(N==0)
            return M;
        if(M==0)
            return N;
        int dp[][] = new int[N+1][M+1];
        dp[0][0] = word1.charAt(0) == word2.charAt(0) ? 0 : 1;
        //标记是否已经有相同的值了
        boolean flag1 = false;
        boolean flag2 = false;
        if(word1.charAt(0)==word2.charAt(0)){
            dp[0][0] = 0;
        }
        else{
            dp[0][0] = 1;
        }
        for(int i = 1; i < M; i++){
            //遇到不等的时候或已经有相同的值了
            if(flag1 || word2.charAt(i) != word1.charAt(0)){
                dp[0][i] = dp[0][i-1] + 1;
            }
            else{
                dp[0][i] = dp[0][i-1];
                flag1 = true;
            }
        }
        for(int i = 1; i < N; i++){
            if(flag2 || word1.charAt(i) != word2.charAt(0)){
                dp[i][0] = dp[i-1][0] + 1;
            }
            else{
                dp[i][0] = dp[i-1][0];
                flag2 = true;
            }
        }
        
        for(int i = 1; i < N; i++){
            for(int j = 1; j < M; j++){
                if(word1.charAt(i) == word2.charAt(j)){
                    dp[i][j] = dp[i-1][j-1];
                }
                else{
                    dp[i][j] = Math.min(dp[i-1][j-1],Math.min(dp[i-1][j], dp[i][j-1])) + 1; 
                }
            }
        }
        return dp[N-1][M-1];  
    }
}

标签：charAt,冲冲,int,else,word1,word2,动态,LeetCode,dp
来源： https://blog.csdn.net/q1072118803/article/details/113729885