题目：
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：本题与power of two,power of three略有不同，不能简单的套用power of three的方法。可以分析2、4的幂都有哪些特征，可以从2进制角度考虑，
1：1
4：100
8：1000
16：10000
32：100000
64：1000000
总结规律发现满足2或4的幂必须num&(num-1) == 0,其次每4位用0101判断是否为4的幂
代码：
class Solution {
public:
bool isPowerOfFour(int num) {
return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;
//0x55555555 is to get rid of those power of 2 but not power of 4
//so that the single 1 bit always appears at the odd position
}
};
