算法分析

时间复杂度 $O(n^2)$

Java 代码

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
    static int N = 1010;
    static int[][] f = new int[N][N];

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        String a = " " + scan.next();
        int m = scan.nextInt();
        String b = " " + scan.next();
        for(int i = 0;i <= n;i++) f[i][0] = i;//若a长度为i，b长度为0，则需要进行i次删除操作
        for(int i = 0;i <= m;i++) f[0][i] = i;//若a长度为0，b长度为i，则需要进行i次添加操作

        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= m;j++)
            {   //删除和添加
                f[i][j] = Math.min(f[i - 1][j] + 1, f[i][j - 1] + 1);
                //修改
                if(a.charAt(i) == b.charAt(j)) f[i][j] = Math.min(f[i][j], f[i - 1][j - 1]);
                else f[i][j] = Math.min(f[i][j], f[i - 1][j - 1] + 1);
            }
        System.out.println(f[n][m]);
    }
}