LeetCode 326 Power of Three

Problem:

Given an integer, write a function to determine if it is a power of three.

Could you do it without using any loop / recursion?

Summary:

用非循环/递归的方法判断数n是否为3的整数幂。

Analysis:

1. 循环：将n逐次除以3，判断是否为3的整数幂。

 class Solution {
 public:
     bool isPowerOfThree(int n) {
         if (n <= ) {
             return false;
         }
         while (n > ) {
             if (n % ) {
                 return false;
             }

             n /= ;
         }

         return true;
     }
 };

2. 递归：若n为3的整数幂，n/3必为3的整数幂，以这个思想构建递归。

 class Solution {
 public:
     bool isPowerOfThree(int n) {
         if (n <= ) {
             return false;
         }

         if (n == ) {
             return true;
         }

         return (n %  == ) && isPowerOfThree(n / );
     }
 };

3. 非循环/递归：n若为3的整数幂，则 n = 3^log₃n = 3^{log₂n / log ₂3} 

 class Solution {
 public:
     bool isPowerOfThree(int n) {
         return (n > ) && (n == pow(, round(log(n) / log())));
     }
 };