题解 | #计算字符串的编辑距离#

import java.util.*;

import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        char[] a = br.readLine().toCharArray();
        char[] b = br.readLine().toCharArray();
        // dp[i][j] a字符串第i个字符b字符串第j个字符时的最小距离 dp[i][j]= Min(dp[i][j-1]+1,dp[i-1][j]+1,dp[i-1][j-1]+(a[i]==b[j]?0:1))
        int[][] dp = new int[a.length + 1][b.length + 1];
        //初始化，记录当一个字符串为空时，另一个字符串的距离即为其长度
        for (int i = 0; i < a.length + 1; i++) {
            dp[i][0] = i;
        }
        for (int i = 0; i < b.length + 1; i++) {
            dp[0][i] = i;
        }
        //动态规划依次取值，当前值取决于其 左 左上，上 三个的值
        for (int i = 1; i < a.length + 1; i++) {
            for (int j = 1; j < b.length + 1; j++) {
                int left = dp[i][j - 1] + 1;
                int up = dp[i - 1][j] + 1;
                int leftUp = dp[i - 1][j - 1] + (a[i - 1] == b[j - 1] ? 0 : 1);
                dp[i][j] = Math.min(Math.min(left, up), leftUp);
            }
        }
        System.out.print(dp[a.length][b.length]);
    }
}