LeetCode 292. Nim Game-优快云博客

本文探讨了Nim游戏中的胜负策略，通过分析得出先手玩家在特定条件下必胜的结论。文章提供了简洁的代码实现，并解释了其背后的数学原理。

<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; background-color: rgb(255, 255, 255);">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.</span>

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.

分析：此题为简单的数字类题目，可以通过举例归纳出数字间的规律

题目是先手必赢的情况：每人拿1~3块石头，那么

如果石头数量是1~3，先手必赢；

如果是4，先手必输；

如果是5~7，可让后手面对数量为4的局面，先手必赢；

如果是8，先手会面临数量为4的局面，先手必输；

……

依次类推，当某一方面对石头数量为4的倍数的情况，必输，要满足题意，先手不能是4的倍数，必赢。

代码如下：

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