LeetCode231. Power of Two

Given an integer, write a function to determine if it is a power of two.

Example 1:

Input: 1
Output: true 
Explanation: 20 = 1

Example 2:

Input: 16
Output: true
Explanation: 24 = 16

Example 3:

Input: 218
Output: false

---

思路：最开始自己想的很简单，就是一直除以2看是否最后等于1，或者调用python的bin()方法，看是不是二进制是不是1000...这种格式。两种思路代码如下：

class Solution:
    def isPowerOfTwo(self, n):
        """
        :type n: int
        :rtype: bool
        """
        while n>=2:
            n = n/2
        
        return n==1

class Solution:
    def isPowerOfTwo(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return n>0 and bin(n).count('1')==1

同时看到用&方法做的，很佩服。这种方法思路是当n二进制是1000..时,(n-1)二进制是0111...，这时候用&操作才能为0

class Solution:
    def isPowerOfTwo(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return n>0 and not(n & n-1)