【LeetCode-Python】292. Nim Game

最新推荐文章于 2022-05-08 13:29:34 发布

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本文介绍了一种两人轮流从一堆石头中取石的游戏——Nim游戏，并提供了三种Python实现的方法来判断先手玩家是否能赢得比赛。通过观察规律发现，当初始石头数量为4的倍数时，先手玩家无法取胜。

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You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

简单解释：n个石头，你先拿，一次只能拿1个或者2个或者3个石头，拿走最后石头的人获胜，给定一个n值，确定你是否能够赢得这场比赛。

Solution1[Python]: 直接递归，LeetCode超时，未通过

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n < 4:
            return True
        elif n == 4:
            return False
        return ((not self.canWinNim(n-1)) or (not self.canWinNim(n-2)) or (not self.canWinNim(n-3)))

Solution2[Python]: 用一个list存储结果，类似动态规划的思想，逐步求解，从1到n，LeetCode提交失败

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n < 4:
            return True
        list = [1, 1, 1]
        for i in range(3, n):
            if sum(list) == 3:
                list.append(0)
            else:
                list.append(1)
            list = list[1:]
        return bool(list[-1])

Solution3[Python]: 上面两个方法太傻了，找规律就可以了，TTTFTTTFTTTFTTTF。。。一句话

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return (False if n%4 == 0 else True )

或者

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return bool(n%4)