题解 | 最长公共子序列(二)

#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 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 Str1="";
    string Str2="";
    string dfs(int i,int j,vector<vector<int>> &b){
        string ans="";
        if (i==0||j==0) return ans;
        if(b[i][j]==1){
            ans += dfs(i-1,j-1,b);
            ans += Str1[i-1];
        }else if(b[i][j]==2){
            ans += dfs(i-1,j,b);
        }else{
            ans += dfs(i,j-1,b);
        }
        return ans;
    }

    string LCS(string s1, string s2) {
        // write code here
        if(s1.empty() || s2.empty()) return "-1";
        int x=s1.size();
        int y=s2.size();
        Str1 = s1;
        Str2 = s2;
        vector<vector<int>> dp(x+1,vector<int>(y+1,0));//记录s1以i为结束，s2以j为结束的最长公共子序列
        vector<vector<int>> b(x+1,vector<int>(y+1,0));
        for(int i=1;i<=x;i++) for(int j=1;j<=y;j++){
            if(s1[i-1]==s2[j-1]){
                dp[i][j] = dp[i-1][j-1]+1;
                b[i][j] = 1; //来自左上
            }else {
                if(dp[i-1][j]>dp[i][j-1]){
                    dp[i][j] = dp[i-1][j];
                    b[i][j] = 2; //来自左侧
                }else{
                    dp[i][j] = dp[i][j-1];
                    b[i][j] = 3; //来自上侧
                }
            }
        }
        string res = dfs(x,y,b);
        return res ==""?"-1":res;
    }
};