题解 | #最长公共子序列-II# 无需倒序查找最终序列，一维数组保存即可

class Solution {
public:
    /**
     * longest common subsequence
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    string LCS(string s1, string s2) {
        // write code here
        int l1 = s1.size(),l2 = s2.size();
        vector<string> res(l2+1,"");
        for(int i=1;i<=l1;i++){
            string pre = "";
            for(int j=1;j<=l2;j++){
                string tem = res[j];
                if(s1[i-1] == s2[j-1]){
                    res[j] = pre+s1[i-1];
                }else{
                    res[j] = (res[j].size() > res[j-1].size() ? res[j] : res[j-1]);
                }
                pre = tem;
            }
        }
        
        if(res[l2] == "")
            return res[l2] = "-1";
        return res[l2];
    }
};