[LeetCode]: 292: Nim Game

最新推荐文章于 2025-09-12 17:31:31 发布

转载最新推荐文章于 2025-09-12 17:31:31 发布 · 75 阅读

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原文链接：http://www.cnblogs.com/savageclc26/p/4876067.html

文章标签：

#数据结构与算法

本文介绍了一种Nim游戏胜负判断的方法。通过分析得知，在这种游戏中，当剩余石头数为4的倍数时，先手玩家无法赢得比赛。据此，文章提供了一个简洁的Java函数实现，用于快速判断在给定石头数量的情况下，先手玩家是否能获胜。

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.

分析：

当最后剩四个石头的时候，必输。所以，整个棋局的争四，以四个为循环

代码：

    public static boolean canWinNim(int n) {
        switch(n){
            case 0:
                return false;
            case 1:
                return true;
            case 2:
                return true;
            case 3:
                return true;
            default:
                if(n%4 == 0){
                    return false;
                }else{
                    return true;
                }
        }
    }