294. Flip Game II-优快云博客

本文介绍了一个名为FlipGame的游戏算法，玩家需将字符串中的连续两个++翻转为--，并探讨了首位玩家是否能确保获胜的问题。文中提供了一个C++实现的示例，包括canWin函数和辅助函数generatePossibleNextMoves及canMove。

You are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.

Write a function to determine if the starting player can guarantee a win.

For example, given s = "++++", return true. The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:
Derive your algorithm's runtime complexity.

class Solution {
public:
    bool canWin(string s) {
        if (s == "") return false;
        return canMove(s);
    }    
private:
    vector<string> generatePossibleNextMoves(string s) {
        vector<string> res;
        if(s.size()==0) return res;
        for(int i = 0;i<s.size()-1;i++)
        {
            if(s[i]=='+' && s[i+1]=='+')
                res.push_back(s.substr(0,i)+"--"+s.substr(i+2)); 
        }
        return res;
    }
    
    bool canMove(string s)
    {
       vector<string> nextMoves = generatePossibleNextMoves(s);
       if(nextMoves.size()==0)  return false;
       for(auto nextMove:nextMoves)
       {
           if(!canMove(nextMove)) 
               return true;
       }
       return false;    
    }
};