LeetCode--No.292--Nim Game-优快云博客

本文链接：https://blog.youkuaiyun.com/sophia_tone2w/article/details/52060503

本文探讨了一个经典的Nim游戏问题，玩家轮流从一堆石头中移除1到3个，拿走最后一个石头的人获胜。文章提供了判断先手是否能赢得游戏的简洁算法实现。

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.
思路：

古老的数学问题。每次可以取1-3个。那么我们可以控制的是，两个人取数的和为4的倍数。

因为是我先取，friend后取。所以如果是被4整除的话，则friend一定可以控制取到最后一个。
但不然的话，我则可以控制。第一次取，n%4即可。

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