Int determines whether a power of 3 - Leetcode (326)

leetcode链接：326. Power of Three

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?

General universal solution:

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class Solution { public: bool isPowerOfThree(int n) { while (n >= 3 ){ if(n % 3 != 0) return false; n /= 3; } return n>0 && n != 2; } };

By the above, but the topic that best try acyclic or recursive solution.

I feel such a method is a little tricky. For example, to find the intlargest factor in the range of 3, so that any multiple of 3 nmay be divisible thereof.

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class Solution { public: bool isPowerOfThree(int n) { // Method 2 const int maxint = 0x7fffffff; //假设3^k 是int范围内最大的3的幂 int k = int(log(maxint) / log(3)); int max_power_3 = pow(3,k); return n>0 && max_power_3 % n ==0; } };

或者将上述代码压缩到一行：

1	return n> 0 && int( pow( 3, int( log( 0x7fffffff) / log( 3)))) % n == 0;

pow函数在头文件math.h中。

同类题：

1. 231. Power of Two
2. 326. Power of Three
3. 342. Power of Four

原文:大专栏  判断int是否为3的幂—— Leetcode(326)