题目描述

blablabla

样例

blablabla

动态数组dp[word1.length+1][word2.length+1]

dp[i][j]表示从word1前i个字符转换到word2前j个字符最少的步骤数。

假设word1现在遍历到字符x，word2遍历到字符y（word1当前遍历到的长度为i，word2为j）。

以下两种可能性：

1. x==y，那么不用做任何编辑操作，所以dp[i][j] = dp[i-1][j-1]

2. x != y

(1) 在word1插入y， 那么dp[i][j] = dp[i][j-1] + 1

(2) 在word1删除x， 那么dp[i][j] = dp[i-1][j] + 1

(3) 把word1中的x用y来替换，那么dp[i][j] = dp[i-1][j-1] + 1

最少的步骤就是取这三个中的最小值。

最后返回 dp[word1.length+1][word2.length+1] 即可。

java 代码

import java.util.*;

public class Main {
    public static int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();

        // len1+1, len2+1, because finally return dp[len1][len2]
        int[][] dp = new int[len1 + 1][len2 + 1];

        for (int i = 0; i <= len1; i++)
            dp[i][0] = i;

        for (int j = 0; j <= len2; j++)
            dp[0][j] = j;


        //iterate though, and check last char
        for (int i = 1; i <= len1; i++) {
            char c1 = word1.charAt(i-1);
            for (int j = 1; j <= len2; j++) {
                char c2 = word2.charAt(j-1);

                //if last two chars equal
                if (c1 == c2) {
                    //update dp value for +1 length
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    int replace = dp[i-1][j-1] + 1;
                    int insert = dp[i-1][j] + 1;
                    int delete = dp[i][j-1] + 1;

                    int min = Math.min(replace, insert);
                    min = Math.min(min,delete);
                    dp[i][j] = min;
                }
            }
        }

        return dp[len1][len2];
    }

    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        int m=scanner.nextInt();
        String strs[]=new String[n];
        for (int i = 0; i < n; i++) {
            strs[i]=scanner.next();
        }
        while(m--!=0){
            String s=scanner.next();
            int a=scanner.nextInt();
            int ans=0;
            for (int i = 0; i <n ; i++) {
               if(a>=minDistance(strs[i],s))
                   ans++;
            }
            System.out.println(ans);
        }
    }
}