leetcode 693. Binary Number with Alternating Bits

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

Example 1:

Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101

Example 2:

Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.

Example 3:

Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.

Example 4:

Input: 10
Output: True
Explanation:
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