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 = 0; i < str.length() - 1; i++) { if(str[i] == '+' && str[i + 1] == '+') { str[i] = str[i + 1] = '-'; vs.push_back(str); str[i] = str[i + 1] = '+'; } } 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 = 0; i < str.length() - 1; i++) { if(str[i] == '+' && str[i + 1] == '+') { str[i] = str[i + 1] = '-'; int win = !canwin(str); str[i] = str[i + 1] = '+'; if(win) return true; } } return false; }