描述
Given an integer n, return true if it is a power of two. Otherwise, return false.
An integer n is a power of two, if there exists an integer x such that n == 2^x.
Example 1:
Input: n = 1
Output: true
Explanation: 2^0 = 1
Example 2:
Input: n = 16 Output: true Explanation: 2^4 = 16
Example 3:
Input: n = 3
Output: false
Example 4:
Input: n = 4
Output: true
Example 5:
Input: n = 5
Output: false
Note:
-2^31 <= n <= 2^31 - 1
解析
根据题意，就是判断 n 是不是 2 的 x 次方的值，用递归的方法进行判断。如果 n 小于等于 0 ，直接返回 False ，如果 n 为 1 ，直接返回 True ，如果 n%2==0 那么递归对 n//2 进行上述类似的操作。其他情况直接返回 False 。
解答
class Solution(object):
def isPowerOfTwo(self, n):
"""
:type n: int
:rtype: bool
"""
if n<=0:
return False
elif n == 1:
return True
elif n % 2 == 0 :
return self.isPowerOfTwo(n//2)
else:
return False
运行结果
Runtime: 16 ms, faster than 83.61% of Python online submissions for Power of Two.
Memory Usage: 13.5 MB, less than 34.60% of Python online submissions for Power of Two.
解析
不用递归，直接写逻辑代码。
解答
class Solution(object):
def isPowerOfTwo(self, n):
"""
:type n: int
:rtype: bool
"""
if n<=0:
return False
if n==1:
return True
while n%2==0:
n//=2
return n==1
运行结果
Runtime: 16 ms, faster than 83.61% of Python online submissions for Power of Two.
Memory Usage: 13.5 MB, less than 34.60% of Python online submissions for Power of Two.
原题链接：leetcode.com/problems/po…
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