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?

解法1，经典的数学解法：

class Solution(object):    def isPowerOfFour(self, num):        """        :type num: int        :rtype: bool        """        if num <= 0: return False        n = int(round(math.log(num, 4)))        return 4**n == num

解法2，迭代：

class Solution(object):    def isPowerOfFour(self, num):        """        :type num: int        :rtype: bool        """        if num <= 0: return False        while num >= 4:            if num % 4 != 0:                return False            num = num / 4            return num == 1

 解法3，最牛叉，

class Solution(object):    def isPowerOfFour(self, num):        """        :type num: int        :rtype: bool        """        return num > 0 and (num & (num - 1)) == 0 and (num - 1) % 3 == 0

 因为，4^n - 1 = C(n,1)*3 + C(n,2)*3^2 + C(n,3)*3^3 +.........+ C(n,n)*3^n

i.e (4^n - 1) = 3 * [ C(n,1) + C(n,2)*3 + C(n,3)*3^2 +.........+ C(n,n)*3^(n-1) ]

This implies that (4^n - 1) is multiple of 3.

类似解法：

return n & (n-1) == 0 and n & 0xAAAAAAAA == 0

或者是：

class Solution(object):    def isPowerOfFour(self, n):        """        :type num: int        :rtype: bool        """                 #1, 100, 10000, 1000000, 100000000, ....        #1, 100 | 10000 | 1000000 | 100000000, ... = 0101 0101 0101 0101 0101 0101 0101 0101        return n > 0 and n & (n-1) == 0 and (n & 0x55555555 != 0)

 因为n & n-1 == 0 就可以确定只有1个1， so 只要保证1的位置在1，3，5，7，。。。。这些位置上就行。