LeetCode 292. Nim Game 题解（C++）

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题目描述

• 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.

示例

• 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.

思路

超时解法

• 看到这道题的第一想法是使用递归进行解答，在确定了stone数为1,2,3,5,6,7时都能赢得游戏，stone数为4时游戏会输后，在递归函数的入口进行了判断
• 接下来，分别对取1,2,3块石头的情况进行递归，所得的bool值是对方能不能取胜的结果，所以需要进行取反
• 使用上述算法submit的结果是在n=35时超时

最佳解法

• 其实这道题是有规律的，在n分别为1,2,3时胜利，n等于4时失败，当n分别为5,6,7时，此时你只需取1,2,3块stone，则对方取时n都为4，此时对方失败；同理，当n等于8时，无论你取1,2,3块stone，对方都可以在7,6,5块stone里取，按照上面的推理，此时对方胜利…接下去可推得只要n为4的倍数，对方就取得胜利，否则我方获得胜利，所以这道题只需要一行代码便可以解决。

代码

超时解法

class Solution {public:    bool canWinNim(int n)     {        if (n <= 3 || n == 5 || n == 6 || n == 7)        {            return true;        }        if (n == 4)        {            return false;        }        bool takeOne = canWinNim(n - 1);        bool takeTwo = canWinNim(n - 2);        bool takeThree = canWinNim(n - 3);        return (!takeOne) || (!takeTwo) || (!takeThree);    }};

最佳解法

class Solution {public:    bool canWinNim(int n)     {        return (n % 4 != 0);    }};