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?

思路：

首先是2的幂，然后二进制表示类似于4⁰ = 1，4¹=100，4²=10000。1总是出现在奇数位置上，所以可以用num & 0x55555555==num来判断。

bool isPowerOfFour(int num) {        return (num > 0) && ((num & (num - 1)) == 0) && ((num & 0x55555555) == num);    }