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231.power-of-two

Statement

Metadata
  • Link: 2 的幂
  • Difficulty: Easy
  • Tag: 位运算 递归 数学

给你一个整数 n,请你判断该整数是否是 2 的幂次方。如果是,返回 true ;否则,返回 false

如果存在一个整数 x 使得 n == 2x ,则认为 n 是 2 的幂次方。

 

示例 1:

输入:n = 1
输出:true
解释:20 = 1

示例 2:

输入:n = 16
输出:true
解释:24 = 16

示例 3:

输入:n = 3
输出:false

示例 4:

输入:n = 4
输出:true

示例 5:

输入:n = 5
输出:false

 

提示:

  • -231 <= n <= 231 - 1

 

进阶:你能够不使用循环/递归解决此问题吗?

Metadata
  • Link: Power of Two
  • Difficulty: Easy
  • Tag: Bit Manipulation Recursion Math

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 == 2x.

 

Example 1:

Input: n = 1
Output: true
Explanation: 20 = 1

Example 2:

Input: n = 16
Output: true
Explanation: 24 = 16

Example 3:

Input: n = 3
Output: false

 

Constraints:

  • -231 <= n <= 231 - 1

 

Follow up: Could you solve it without loops/recursion?

Solution

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        return (n & (n - 1)) == 0 if n != 0 else False

最后更新: January 15, 2023
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