[leetcode]--292. Nim Game

Question 292

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.

中文解释与分析：
我是最先开始移动石头的, 如果n是4的倍数,那么无论我最开始移动1-3个,到我的时候是4的倍数,我必输;
如果n不是4的倍数,我移动之后都可以保证到对方手里时是4的倍数,我必赢.
所以核心的判断就是n是否是4的倍数。

实现源码：

package leetcode;import utils.InputUtil;import utils.LogUtil;public class Question292 {    public static boolean canWinNim(int n) {        /*if(n%4 == 0){            return false;        }else {            return true;        }*/        //return n%4==0? false:true;        return n%4!=0;    }    public static void main(String[] args) {        int n = InputUtil.inputInt(null);        if(canWinNim(n)){            LogUtil.log_debug("true");        }else{            LogUtil.log_debug("false");        }    }}

其实看看核心的函数就可以知道，这个问题是一行代码就可以解决的问题，短小精悍。return n%4!=0;