题解 | #最长公共子序列-II#

import java.util.*;


public class Solution {
    /**
     * longest common subsequence
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    public String LCS (String s1, String s2) {
        // write code here
        int m = s1.length();
        int n = s2.length();
        // dp表示 s1前i个元素 与s2前j个元素的最长公共子序列串
        String[][] dp = new String[m+1][n+1];
        for (int i = 0; i <= s1.length(); i++) {
            dp[i][0] = "";
        }
        for (int i = 0; i <= s2.length(); i++) {
            dp[0][i] = "";
        }
        for(int i = 1; i <= m;i++){
            for(int j = 1; j <= n;j++){
                if(s1.charAt(i-1) == s2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + s1.charAt(i-1);
                }else{
                    dp[i][j] = dp[i-1][j].length() > dp[i][j-1].length() ? 
                        dp[i-1][j] : dp[i][j-1];
                }
            }
        }
        return dp[m][n].length() == 0 ? "-1" : dp[m][n];
    }
}