题解 | #公共子串计算#

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);

        while(in.hasNext()){
            String A = in.nextLine();
            String B = in.nextLine();

            if(A.length() <= B.length()){
                // solution1(A, B);
                solution2(A, B);
            }else{
                // solution1(B, A);
                solution2(B, A);
            }
        }
    }


    /**
     * 滑动窗口
     * @param A
     * @param B
     */
    private static void solution1(String A, String B){
        int m = A.length();

        int result = 0;
        boolean found = false;
        for(int i = m; i>0; i--){
            for(int j=0; j+i<=m; j++){
                if(B.contains(A.substring(j, j+i))){
                    result = i;
                    found = true;
                    break;
                }
            }
            if(found){
                break;
            }
        }

        System.out.println(result);
    }


    /**
     * 动态规划 dp[i][j]表示在较短字符串A中以第i个字符结尾，较长字符串B中以第j个字符结尾时的当前局部公共子串长度
     * @param A
     * @param B
     */
    private static void solution2(String A, String B){
        int m = A.length();
        int n = B.length();

        int[][] dp = new int[m+1][n+1];

        int result = 0;
        for(int i=1; i<=m; i++){
            for(int j=1; j<=n; j++){
                if(A.charAt(i-1) == B.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + 1;
                    if(result < dp[i][j]){
                        result = dp[i][j];
                    }
                }
            }
        }
        
        System.out.println(result);
    }
}