[Locked] Flip Game I & II

Flip Game I

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 compute all possible states of the string after one valid move.

For example, given s = "++++", after one move, it may become one of the following states:

[
  "--++",
  "+--+",
  "++--"
]

If there is no valid move, return an empty list [].

分析：

　　一层处理即可

代码：

vector<string> flipGame(string str) {
    vector<string> vs;
    for(int i = ; i < str.length() - ; i++) {
        if(str[i] == '+' && str[i + ] == '+') {
            str[i] = str[i + ] = '-';
            vs.push_back(str);
            str[i] = str[i + ] = '+';
        }
    }
    return vs;
}

Flip Game II

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.

分析：

　　第一想法是找到合适的规律，可以直接从当前状态判断是否必胜，经过尝试，难以直接判断；所以采用一般解法，当两边都是没有失误的高手状态下，有以下规律：

　　　　1、终结点是必败点（P点）；

　　　　2、从任何必胜点（N点）操作，至少有一种方法可以进入必败点（P点）

　　　　3、无论如何操作， 从必败点（P点）都只能进入必胜点（N点）.

　　这里用到的就是1，2，3规律，对手无点可操作，则以失败终结；自己必胜，则至少一种方法进入下一轮必败点；若不能必胜，则找不到方法进入下一轮必败点；若自己必败，则无论用什么方法下一轮都是必胜点。

代码：

bool canwin(string str) {
    for(int i = ; i < str.length() - ; i++) {
        if(str[i] == '+' && str[i + ] == '+') {
            str[i] = str[i + ] = '-';
            int win = !canwin(str);
            str[i] = str[i + ] = '+';
            if(win)
                return true;
        }
    }
    return false;
}