342. Power of Four

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             }};