Given an integer, write a function to determine if it is a power of three.
Follow up:
Could you do it without using any loop / recursion?
不用递归和循环,这里有一个讨巧的办法,因为输入的数据是int类型,int类型最大的3的次幂是1162261467所以解决办法是:
class Solution(object): def isPowerOfThree(self, n): """ :type n: int :rtype: bool """ return n > 0 and 1162261467 % n == 0
原文：http://www.cnblogs.com/feiqiangs/p/5791647.html