LeetCode 326. Power of Three

https://leetcode.com/problems/power-of-three/description/

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?

• 数学题
• 第一种解法，简单直接，循环除以3直到最好余数为1。注意边界值。
• 第二种解法，int最大值在C/C++语言中与机器位长相关，32位下是2^32-1，Java中int是固定的32位，即2^32-1。那么可以算出来int下最大的3的幂为3^19=1162261467。只要能被它整除的必然是3的幂。
• https://leetcode.com/problems/power-of-three/solution/

//
//  main.cpp
//  LeetCode
//
//  Created by Hao on 2017/3/16.
//  Copyright © 2017年 Hao. All rights reserved.
//

#include <iostream>
using namespace std;

class Solution
{
public:
    bool isPowerOfThree(int n)
    {
        if (n < 1) {
            return false;
        }
        
        while (n % 3 == 0) {
            n /= 3;
        }
        
        return n == 1;
    }

    bool isPowerOfThree2(int n)
    {
        return n > 0 && 1162261467 % n == 0;
    }
};

int main(int argc, char* argv[])
{
    Solution    testSolution;
    
    auto nums = {0, 1, 2, 3, 4, 6, 9, 27, 1162261467};
    
    for (auto num : nums)
    {
        cout << num << " isPowerOfThree ? " << testSolution.isPowerOfThree(num) << endl;
        cout << num << " isPowerOfThree2 ? " << testSolution.isPowerOfThree2(num) << endl;
    }
    
    return 0;
}

0 isPowerOfThree ? 0
0 isPowerOfThree2 ? 0
1 isPowerOfThree ? 1
1 isPowerOfThree2 ? 1
2 isPowerOfThree ? 0
2 isPowerOfThree2 ? 0
3 isPowerOfThree ? 1
3 isPowerOfThree2 ? 1
4 isPowerOfThree ? 0
4 isPowerOfThree2 ? 0
6 isPowerOfThree ? 0
6 isPowerOfThree2 ? 0
9 isPowerOfThree ? 1
9 isPowerOfThree2 ? 1
27 isPowerOfThree ? 1
27 isPowerOfThree2 ? 1
1162261467 isPowerOfThree ? 1
1162261467 isPowerOfThree2 ? 1
Program ended with exit code: 0