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 <= 0) {
            return false;
        }
        while (n > 1) {
            if (n % 3) {
                return false;
            }
            
            n /= 3;
        }
        
        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 <= 0) {
            return false;
        }
        
        if (n == 1) {
            return true;
        }
        
        return (n % 3 == 0) && isPowerOfThree(n / 3);
    }
};

 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 > 0) && (n == pow(3, round(log(n) / log(3))));
    }
};