LeetCode Nim Game

最新推荐文章于 2025-08-15 16:44:13 发布

转载最新推荐文章于 2025-08-15 16:44:13 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/4903343.html

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#数据结构与算法

本文介绍了一种快速判断Nim游戏胜负的方法。通过分析游戏规则，得出结论：当石堆中石头数量为4的倍数时，先手玩家无法获胜；其他情况下，先手玩家可以获胜。此算法的时间复杂度和空间复杂度均为O(1)。

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原题链接在这里：https://leetcode.com/problems/nim-game/

题目：

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.

题解：

列出前几个数，发现只要是四的倍数就return false, 其他都是return true.

Time Complexity: O(1). Space: O(1).

AC Java:

1 public class Solution {
2     public boolean canWinNim(int n) {
3         if(n < 0){
4             throw new IllegalArgumentException("input number can't be smaller than zero.");
5         }
6         return n % 4 != 0;
7     }
8 }