Python实现"3的幂"的两种方法

给定一个整数，写一个函数判断它是否是3的幂

Example 1:

Input: 27
Output: true

Example 2:

Input: 0
Output: false

Example 3:

Input: 9
Output: true

Example 4:

Input: 45
Output: false

进阶：

你能不适用循环或者迭代完成这个题么？

1：累除3

循环

def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n == 0:
            return False
        while n % 3 == 0:
            n //= 3
        return n == 1

递归

def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n == 0:
            return False
        if n == 1:
            return True
        if n % 3 == 0:
            return self.isPowerOfThree(n // 3)
        else:
            return False

2：round()+math.log()方法（参考他人）

def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n <= 0:
            return False
        return 3 ** round(math.log(n, 3)) == n

类似方法（参考他人）

def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n <= 0:
            return False
        sum = 3 ** 100
        return sum % n == 0

算法题来自：https://leetcode-cn.com/problems/power-of-three/description/