LeetCode 326. Power of Three

题目：
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?

思路：
判断一个数是否是3的幂，但是不建议用循环和递归。想到了用log函数，因为如果n是3的幂，那么log3(n)肯定是整数，将其result设成double型的，与转换成int型后的result对比，若相等则为3的幂；否则不是3的幂。但是因为没有3的底，所以要用换底公式将底换成10.

代码:

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

**注：
cmath中，log为以e为底，log10是以10为底，其它的都要用换底公式。**

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