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?

4的幂次方 二进制里只有1个1

判断二进制里有没有1个1 用n & (n-1)== 0

 bool isPowerOfFour(int num) {

     if(num <= ) return false;

     if(num & (num - )) return false;

     if(num & 0x55555555) return true; // 再将不是 4 的 N 次方的数字去掉

     return false;

 }