LeetCode 231. Power of Two

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

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

• 简单数学题，位运算
• 2的幂在2进制中只有1个0，所以可以利用n & (n -1) == 0来判断是不是2的幂

//
//  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 isPowerOfTwo(int n)
    {
        return ((n > 0) && ((n & (n - 1)) == 0));
    }
};

int main(int argc, char* argv[])
{
    Solution    testSolution;
    
    auto nums = {0, 1, 2, 3, 4, 8, 16, 1000, 1024};
    
    for (auto num : nums)
    {
        cout << num << " isPowerOfTwo ? " << testSolution.isPowerOfTwo(num) << endl;
    }
    
    return 0;
}

0 isPowerOfTwo ? 0
1 isPowerOfTwo ? 1
2 isPowerOfTwo ? 1
3 isPowerOfTwo ? 0
4 isPowerOfTwo ? 1
8 isPowerOfTwo ? 1
16 isPowerOfTwo ? 1
1000 isPowerOfTwo ? 0
1024 isPowerOfTwo ? 1
Program ended with exit code: 0