Leetcode: 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:

1. 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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Naive Solution:
Simulate the process of the NIM game. Each time try to remove from 1 to 3 stones and go to the next round. 

Code (Java):

public class Solution {
    public boolean canWinNim(int n) {
        if (n <= 3) {
            return true;
        }
        
        return canWinNimHelper(n, 0);
    }
    
    private boolean canWinNimHelper(int n, int round) {
        if (round % 2 == 0 && n <= 3) {
            return true;
        } else if (round % 2 == 1 && n <= 3) {
            return false;
        }
        
        for (int i = 3; i >= 1; i--) {
            if (canWinNimHelper(n - i, round + 1)) {
                return true;
            }
        }
        
        return false;
    }
}

Analysis: 
It will get TLE error. There must be a clever way to solve this question. 

Better solution:
http://bookshadow.com/weblog/2015/10/12/leetcode-nim-game/

解题思路：

Nim游戏的解题关键是寻找“必胜态”。

根据题设条件：

当n∈[1,3]时，先手必胜。

当n == 4时，无论先手第一轮如何选取，下一轮都会转化为n∈[1,3]的情形，此时先手必负。

当n∈[5,7]时，先手必胜，先手分别通过取走[1,3]颗石头，可将状态转化为n == 4时的情形，此时后手必负。

当n == 8时，无论先手第一轮如何选取，下一轮都会转化为n∈[5,7]的情形，此时先手必负。

......

以此类推，可以得出结论：

当n % 4 != 0时，先手必胜；否则先手必负。

Code (Java):

public class Solution {
    public boolean canWinNim(int n) {
        return n % 4 > 0;
    }
}