[LeetCode]292. 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.

思路：深搜的话时间肯定会爆掉，然后如果用动规的话，可以用递推式dp[i]=!(dp[i-1]&&dp[i-2]&&dp[i-3])，dp里边是自己在i的时候是否可以赢，这个其实是基于两个动规数组，自己和Nim的，结果最后发现，其实两个人的数组是一样的。。然后就用一个就好，然后用动规是AC不了的，有大数据导致超空间，然后就发现。。其实自己按着这个递推公式推几步就发现，只有n为4的倍数的时候，自己是输的，其他时候都是赢的。。然后就解出来了。。

public class Solution {        public boolean canWinNim(int n) {        if(n%4==0){            return false;        }else{            return true;        }    }    /*    public boolean canWinNim(int n) {        if(n<4){            return true;        }else if(n==4){            return false;        }        boolean dp[]=new boolean[n+1];        dp[1]=true;        dp[2]=true;        dp[3]=true;        dp[4]=false;        for(int i=5;i<n+1;i++){            dp[i]=!(dp[i-1]&&dp[i-2]&&dp[i-3]);        }        return dp[n];    }    */}