题解 | #查找两个字符串a,b中的最长公共子串#

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
    public static void main(String[]args) {
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        String a, b;
        try {
            a = r.readLine();
            b = r.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        char[] chs1 = a.toCharArray();
        char[] chs2 = b.toCharArray();
        char[] mid;
        int h1 = chs1.length, h2 = chs2.length, max = 0, start = 0, i = 0, j, k;
        if (h1 > h2) {
            mid = chs1;
            chs1 = chs2;
            chs2 = mid;
        }
        h1 = chs1.length;
        h2 = chs2.length;
        System.out.println(longString(chs1, chs2, h1 + 1, h2 + 1));
    }

    // 动态规划
    public static String longString(char[] chs1, char[] chs2, int m, int n) {
        // 表示在较短字符串str1以第i个字符结尾，str2中以第j个字符结尾时的公共子串长度。
        int[][] dp = new int[m][n];
        // 匹配字符，并记录最大值的str1的结尾下标
        int max = 0;
        int index = 0;
        // 从左向右递推，i为短字符串str1的结尾索引，j为str2的结尾索引
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (chs1[i - 1] == chs2[j - 1]) {
                    // 相等则计数
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    // 不断更新变量
                    if (dp[i][j] > max) {
                        max = dp[i][j];
                        index = i;
                    }
                }
            }
        }
        // 截取最大公共子串
        return new String(chs1, index - max, max);
    }
}