1 Description #

source: https://leetcode.com/problems/power-of-two/

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: 20 = 1

Example 2:

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

Example 3:

Input: n = 3
Output: false

Constraints:

• -2^31 <= n <= 2^31 - 1

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

2 Solution #

#include <bitset>
class Solution {
public:
  bool isPowerOfTwo(int n) {
    // Space complexity: O(33) => O(1)
    // Time complexity: O(1)
    return std::bitset<33>(n).count() == 1;
  }
};