Leetcode: Nim Game

最新推荐文章于 2022-10-01 23:34:25 发布

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最新推荐文章于 2022-10-01 23:34:25 发布

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本文探讨了游戏 Nim 的基本策略，并提供了一个简单的函数用于判断玩家是否能赢得游戏，给定石堆中石子的数量。通过分析游戏规则，作者提出了一种策略来确定在特定条件下是否能获胜。

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.

因为每个人都取不到4个，假设自己后走，要保证每轮自己和对方取得数量的和是4，这样就能确保每轮完后都有4的倍数个石头被取走。这样，如果我们先走的话，先把n除4的余数个石头拿走，这样不管怎样，到最后都会留4个下来，对方取1个你就取3个，对方取2个你就取2个，就必赢了。

如果一开始就是4的倍数，你就输了，因为对方可以用同样的策略

1 public class Solution {
2     public boolean canWinNim(int n) {
3         return n%4 != 0;
4     }
5 }