231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂

231. Power of Two 

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

class Solution {
public:
    bool isPowerOfTwo(int n) {
        return n > 0 ? (n & (n-1)) == 0 : false;
    }
};

342. 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?

    bool isPowerOfFour(int num) {
        static int mask = 0b01010101010101010101010101010101;
        
        //edge case
        if (num<=0) return false;
        
        // there are multiple bits are 1
        if ((num & num-1) != 0) return false;
        
        // check which one bit is zero, if the place is 1 or 3 or 5 or 7 or 9...,
        // then it is the power of 4
        if ((num & mask) != 0) return true;
        return false;
    }