How to solve the longest common subsequence problem in Java

1. Brief description:

Description

Given two strings str1 and str2, output the longest common subsequence of the two strings. If the longest common subsequence is empty, "-1" is returned. In the data currently given, there will only be one longest common subsequence

Data range:

Requirements: space complexity, time complexity

Example 1

Input:

"1A2C3D4B56","B1D23A456A"

Return value:

"123456"

Example 2

Input:

"abc","def"

Return value:

"-1"

Example 3

Input:

"abc","abc"

Return value:

"abc"

Example 4

Input:

"ab",""

Return value:

"-1"

2. Code implementation:

import java.util.*;
public class Solution {
    public String LCS (String s1, String s2) {
        //只要有一个空字符串便不会有子序列
        if(s1.length() == 0 || s2.length() == 0) 
            return "-1";
        int len1 = s1.length();
        int len2 = s2.length();
        //dp[i][j]表示第一个字符串到第i位，第二个字符串到第j位为止的最长公共子序列长度
        int[][] dp = new int[len1 + 1][len2 + 1]; 
        //遍历两个字符串每个位置求的最长长度
        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                //遇到两个字符相等
                if(s1.charAt(i - 1) == s2.charAt(j - 1))
                    //来自于左上方
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                //遇到的两个字符不同
                else
                    //来自左边或者上方的最大值
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        //从动态规划数组末尾开始
        int i = len1, j = len2;
        Stack s = new Stack();
        while(dp[i][j] != 0){
            //来自于左方向
            if(dp[i][j] == dp[i - 1][j])
                i--;
            //来自于上方向
            else if(dp[i][j] == dp[i][j - 1])
                j--;
            //来自于左上方向
            else if(dp[i][j] > dp[i - 1][j - 1]){
                i--;
                j--;
                //只有左上方向才是字符相等的情况，入栈，逆序使用
                s.push(s1.charAt(i)); 
           }
        }
        String res = "";
        //拼接子序列
        while(!s.isEmpty())
            res += s.pop();
        //如果两个完全不同，返回字符串为空，则要改成-1
        return !res.isEmpty() ? res : "-1";  
    }
}

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