Question:

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?

 

Analysis:

给出一个整数，写一个函数判断它是否是3的n次幂。

注意：能否不用循环或递归解决这个问题？

 

Solution1: Using Loop.

//Solution1: Looppublic boolean isPowerOfThree(int n) {         if(n <= 0)             return false;         while(n > 1) {             if(n % 3 != 0)                 break;             else                  n /= 3;         }         return n == 1;}

 

Solution2: Using Recursion.

//Solution2: Recursion.public static boolean isPowerOfThree(int n) {         if(n <= 0)             return false;         if(n == 1)             return true;         if(n % 3 == 0)             return isPowerOfThree(n/3);         else return false;}

 

Solution3: Using Log Function. (logab = logcb / logca)

public boolean isPowerOfThree(int n) {        if(n <= 0)             return false;         double log = Math.log10(n) / Math.log10(3);         if(log - (int)log == 0)             return true;         else return false;    }