leetcode-292-easy

NimGame

You are playing the following Nim Game with your friend:

Initially, there is a heap of stones on the table.
You and your friend will alternate taking turns, and you go first.
On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
The one who removes the last stone is the winner.
Given n, the number of stones in the heap, return true if you can win the game assuming both you and your friend play optimally, otherwise return false.

Example 1:

Input: n = 4
Output: false
Explanation: These are the possible outcomes:
1. You remove 1 stone. Your friend removes 3 stones, including the last stone. Your friend wins.
2. You remove 2 stones. Your friend removes 2 stones, including the last stone. Your friend wins.
3. You remove 3 stones. Your friend removes the last stone. Your friend wins.
In all outcomes, your friend wins.
Example 2:

Input: n = 1
Output: true
Example 3:

Input: n = 2
Output: true
Constraints:

1 <= n <= 231 - 1

思路一：有下面思考

• 对于 1,2,3 个石头，先出手必赢
• 如果是 4,先出手的人无论怎么操作都是必输
• 对于 5,6,7，先出手的人只要让下一个人拿 4，他无论怎么操作都是必输
• 对于 8，先出手的人无论怎么操作，下一个人拿到的都是 5,6,7，他可以反制我们
  所以有这样的规律拿到 n % 4 == 0 人必输

public boolean canWinNim(int n) {
    if (n % 4 == 0) return false;
    else return true;
}

原文地址：http://www.cnblogs.com/iyiluo/p/16829648.html