LeetCode 292. Nim Game（取物游戏）

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本文探讨了一种名为Nim的游戏，玩家轮流从一堆石头中移除1到3个石头，取走最后一个石头的人获胜。文章提供了两种解决方案：一种是动态规划方法，但可能会超时；另一种是发现当石头总数能被4整除时必败的规律，从而快速判断胜败。

原题网址：https://leetcode.com/problems/nim-game/

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.

Hint:

If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?

方法一：此题最直接的想法是用动态规划，时间复杂度O(N)，会超时。

public class Solution {
    public boolean canWinNim(int n) {
        if (n<4) return true;
        boolean w1 = true, w2 = true, w3 = true;
        boolean win = false;
        for(int i=4; i<=n; i++) {
            win = !w1 | !w2 | !w3;
            w1 = w2;
            w2 = w3;
            w3 = win;
        }
        return win;
    }
}

方法二：分析发现当n能被4整除的时候必输，其他情况可以赢，时间复杂度O(1)

public class Solution {
    public boolean canWinNim(int n) {
        return !((n&0b11)==0);
    }
}