[LeetCode] Challenge log 693

Posted on | Edited on |
693. Binary Number with Alterntive Bits

s1. n<<1 + n should be all 1s.

s2. similar to s1 but more pythonic

s3. check by bit

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

Example 1:

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

Example 2:

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

Example 3:

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

Example 4:

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Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.
Soulution:
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class Solution:
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        res = True
        mask = 0 
        while mask < n:
            mask = (mask << 1) + 1
        return (n >> 1) + n == mask
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class Solution:
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return bin((n >> 1) + n).count('0') == 1
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class Solution:
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        res = []
        while n > 0: 
            res.append(n % 2)
            n = n // 2
        if len(res) <= 1: return True
        for i in range(1, len(res)):
            if res[i] == res[i-1]:
                return False
        return True
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