292. Nim游戏

你和你的朋友，两个人一起玩 Nim游戏：桌子上有一堆石头，每次你们轮流拿掉 1 - 3 块石头。 拿掉最后一块石头的人就是获胜者。你作为先手。

你们是聪明人，每一步都是最优解。 编写一个函数，来判断你是否可以在给定石头数量的情况下赢得游戏。

示例:

输入: 4
输出: false 
解释: 如果堆中有 4 块石头，那么你永远不会赢得比赛；
     因为无论你拿走 1 块、2 块 还是 3 块石头，最后一块石头总是会被你的朋友拿走。

 

Theorem: The first one who got the number that is multiple of 4 (i.e. n % 4 == 0) will lost, otherwise he/she will win.

Proof:

1.  

   the base case: when n = 4, as suggested by the hint from the problem, no matter which number that that first player, the second player would always be able to pick the remaining number.

2.  

   For 1* 4 < n < 2 * 4, (n = 5, 6, 7), the first player can reduce the initial number into 4 accordingly, which will leave the death number 4 to the second player. i.e. The numbers 5, 6, 7 are winning numbers for any player who got it first.

3.  

   Now to the beginning of the next cycle, n = 8, no matter which number that the first player picks, it would always leave the winning numbers (5, 6, 7) to the second player. Therefore, 8 % 4 == 0, again is a death number.

4.  

   Following the second case, for numbers between (2*4 = 8) and (3*4=12), which are 9, 10, 11, are winning numbers for the first player again, because the first player can always reduce the number into the death number 8.

Following the above theorem and proof, the solution could not be simpler:

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