LeetCode.326.Power of Three

原题链接：Power of Three

题目内容：
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?

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

判断给定的整数是否是3的n次方数，题目希望使用非递归方式

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Python.01

普通递归（不推荐）

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

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Python.02

利用对数知识，loga(b) = logc(b) / logc(a)
理论可行，但是我LeetCode引用numpy失败。

from numpy import log10
class Solution(object):
    def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        res = log10(n) / log10(3)
        return res - int(res) == 0

C++

对数的C++版本

class Solution {
public:
    bool isPowerOfThree(int n) {
        double res = log10(n) / log10(3);
        return (res - int(res) == 0) ? true : false;  
    }
};

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Python.03

任何一个3的x次方一定能被范围内最大的3的x次方整除，int型最大的3的x次方数是1162261467 。略取巧

class Solution(object):
    def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return n > 0 and 1162261467 % n == 0