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Question

Formatted question description: https://leetcode.ca/all/231.html

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?

Algorithm

Observe the characteristics of the binary notation of the power of 2:

1   2   4       8       16 　　....

1   10  100     1000    10000　...

The power of 2 has only one 1, and the rest is 0s, so our problem-solving idea is to determine whether the lowest bit is 1 each time, then shift to the right, and finally count the number of 1s to determine whether it is Power of 2

Code

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 
  public class Power_of_Two {
  
  
      class Solution {
          public boolean isPowerOfTwo(int n) {
              int cnt = 0;
              while (n > 0) {
                  cnt += (n & 1);
                  n >>= 1;
              }
              return cnt == 1;
  
          }
      }
  }
  
  ############
  
  class Solution {
      public boolean isPowerOfTwo(int n) {
          return n > 0 && (n & (n - 1)) == 0;
      }
  }
  

• // OJ: https://leetcode.com/problems/power-of-two/
  // Time: O(1)
  // Space: O(1)
  class Solution {
  public:
      bool isPowerOfTwo(int n) {
          if (n <= 0) return false;
          while ((n & 1) == 0) n >>= 1;
          return n == 1;
      }
  };
  

• class Solution:
      def isPowerOfTwo(self, n: int) -> bool:
          return n > 0 and (n & (n - 1)) == 0
  
  ############
  
  class Solution(object):
    def isPowerOfTwo(self, n):
      """
      :type n: int
      :rtype: bool
      """
      return n != 0 and (n & -n) == n
  
  

• func isPowerOfTwo(n int) bool {
  	return n > 0 && (n&(n-1)) == 0
  }
  

• function isPowerOfTwo(n: number): boolean {
      return n > 0 && (n & (n - 1)) == 0;
  }
  
  

• /**
   * @param {number} n
   * @return {boolean}
   */
  var isPowerOfTwo = function (n) {
      return n > 0 && (n & (n - 1)) == 0;
  };
  
  

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