【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.

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The solution is quite simple if you figure out the trick for playing this game. Focusing on the hint, we could know that 4 is a significant number. Why? It's because 4 is the least unit that you could pick up every round regardless of any number your enemy
post. If he post 3, you post 1; If he post 2, you post 2; If he post 1, you post 3. Thus, If the initial number is 13 and you will take the first move, you just need to take 1 for the first round. Then no matter what number he chose in any round, you just
choose (4-x) to complete until reach 13, which also means if the initial number is like 12 or 20, those could be divided by 4, you would definitely fail if your enemy is as clever as you.

I wrote this solution in JAVA for the record:

package testAndfun;

public class Nim {
public static void main(String args[]){
int n1 = 13;
int n2 = 20;
Nim nim = new Nim();
System.out.println("It's "+nim.canWinNim(n1)+" to win the game");
System.out.println("It's "+nim.canWinNim(n2)+" to win the game");
}

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