LeetCode 题解(277) ：Flip Game II

题目：

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.

题解：

递归或者 Game Theory。

C++版：

class Solution {public:    bool canWin(string s) {        int cur = 0, maxL = 0;        vector<int> subGames;        for(int i = 0; i < s.length(); i++) {            if(s[i] == '+') {                cur++;                if(cur >= 2 && (i == s.length() - 1 || s[i + 1] == '-')) {                    subGames.push_back(cur);                    maxL = max(maxL, cur);                    cur = 0;                }            } else                cur = 0;        }                vector<int> eachGame(maxL + 1, 0);        for(int i = 2; i <= maxL; i++) {            unordered_set<int> nums;            for(int j = 0; j < i / 2; j++) {                int k = i - j - 2;                nums.insert(eachGame[j] ^ eachGame[k]);            }            eachGame[i] = firstMissingNumber(nums);        }                int result = 0;        for(auto i : subGames)            result ^= eachGame[i];        return result != 0;    }        int firstMissingNumber(unordered_set<int>& s) {        int l = s.size();        for(int i = 0; i < l; i++)            if(s.count(i) == 0)                return i;        return l;    }};

Python版：

class Solution(object):    def canWin(self, s):        """        :type s: str        :rtype: bool        """        if len(s) < 2:            return False        for i in range(len(s) - 1):            if s[i] == '+' and s[i+1] == '+' and not self.canWin(s[:i] + '-' + s[i+2:]):                return True        return False