LeetCode 342. Power of Four

https://leetcode.com/problems/power-of-four/

Given an integer (signed 32 bits), write a function to check whether it is a power of 4.

Example:
Given num = 16, return true. Given num = 5, return false.

Follow up: Could you solve it without loops/recursion?

• 数学题+位运算。
• 因为是signed 32 bits，所以先保证num > 0。
• 然后2的幂的特征(num & (num - 1)) == 0，e.g. 0x1000 & 0x0111。
• 再对2的幂中筛选不是4的幂的。4的幂的二进制中1总是出现在奇数位，e.g. 4^1(0x0100), 4^2(0x00010000), 4^3(0x01000000)，其他则是在偶数位，e.g. 2^1(0x0010), 2^3(0x1000), 2^5(0x00100000)。所以(num & 0x55555555) != 0，这里0x55555555正是signed 32 bits integer。
• 之前没注意位运算运算符&优先级低于==，需要加上括号。
• C语言运算符_百度百科
  • http://baike.baidu.com/link?url=7D3QzeLlfI9pELy4OqJyyGE-WwRZhZ_mCI8ZSI6YdQHrldIIly3WnCTGnjzbypatAAodGjGFTUrTGJxkIbWFBq#2
• 还有一种解法（4^n - 1）都可以被3整除，所以可以判断(num > 0) && ((num & (num - 1)) == 0) && ((num - 1) % 3 == 0)
  • https://helloacm.com/how-to-check-if-integer-is-power-of-four-cc/

#include <iostream>
using namespace std;

class Solution {
public:
    bool isPowerOfFour(int num) {
        return ((num > 0) && ((num & (num - 1)) == 0) && ((num & 0x55555555) != 0));
    }
};

int main ()
{
    Solution testSolution;
    bool result;
    
    result = testSolution.isPowerOfFour(16);    
    cout << result << endl;
        
    char ch;
    cin >> ch;
    
    return 0;
}