Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

Input: 5Output: TrueExplanation:The binary representation of 5 is: 101

Example 2:

Input: 7Output: FalseExplanation:The binary representation of 7 is: 111.

Example 3:

Input: 11Output: FalseExplanation:The binary representation of 11 is: 1011.

Example 4:

Input: 10Output: TrueExplanation:The binary representation of 10 is: 1010. 解法1:

class Solution(object):    def hasAlternatingBits(self, n):        """        :type n: int        :rtype: bool        """        # check every bit is diff from last bit        last_bit = n & 1        n = n >> 1                while n:                         bit = n & 1            if last_bit + bit != 1:                return False            last_bit = bit            n = n >> 1        return True

解法2: 移位再相加后，看是否为2的n次方

class Solution(object):    def hasAlternatingBits(self, n):        """        :type n: int        :rtype: bool        """        num = n + (n >> 1) + 1        return (num & (num-1)) == 0

解法3:直接使用bin函数

def hasAlternatingBits(self, n):        s = bin(n)        return '00' not in s and '11' not in s