Leetcode-292. Nim Game

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最新推荐文章于 2024-10-10 09:00:00 发布

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分类专栏： Leetcode 算法与数据结构程序员笔试面试文章标签： leetcode

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本文介绍了一种Nim游戏的胜负判断策略。在该游戏中，玩家轮流从一堆石头中移除1到3颗石头，移除最后一颗石头的玩家获胜。通过分析发现，当石头数量为4的倍数时无法取胜，否则总有策略可以赢得比赛。

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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.

Hint:

If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?

思路：

当石头个数为1，2，3时，我们可以上来就把石头取完，然后就赢了。

当石头个数为4时，我们取1个剩下3个，取2个剩下2个，取3个剩下1个，那么别人都可以一次取完，别人获胜。

当石头个数为n时，如果无论我们取1个还是2还是3个，只要这三种取法中有一种能取胜，我们就能赢。

所以就是递归解法：

class Solution {
public:
    bool canWinNim(int n) {
        if(n <= 3) return true;
        return !canWinNim(n-1) || !canWinNim(n-2) || !canWinNim(n-3);
    }
};

超时毫无疑问。

把上面的转换一下看看

!canWinNim(n-1) || !canWinNim(n-2) || !canWinNim(n-3) => !(canWinNim(n-1) && canWinNim(n-2) && canWinNim(n-3))

其实推倒一下，true和False是有规律出现的。当石头个数是4的整数倍时总是fasle.

class Solution {
public:
    bool canWinNim(int n) {
        if(n % 4 == 0) return false;
        else return true;
    }
};