leetcode Nim Game题解

本文探讨了Nim游戏的胜负策略,通过分析得出当石子数不是4的倍数时,先手玩家将赢得游戏。文章提供了一个简洁的Java代码实现,用于判断任意数量的石子下,先手玩家是否能获胜。

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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.

Example:

Input: 4
Output: false 
Explanation: 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.

中文理解:玩游戏,假设你和伙伴都足够聪明,有一堆石子,每次没人只能移除1-3个,给定石子数,问最终的赢家。

解题思路:分析可得当石子数不是4的倍数是先手赢。

代码(java):

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

 

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