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342. Power of Four

Description

Given an integer n, return true if it is a power of four. Otherwise, return false.

An integer n is a power of four, if there exists an integer x such that n == 4x.

 

Example 1:

Input: n = 16
Output: true

Example 2:

Input: n = 5
Output: false

Example 3:

Input: n = 1
Output: true

 

Constraints:

• -231 <= n <= 231 - 1

 

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

Solutions

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

• class Solution {
      public boolean isPowerOfFour(int n) {
          return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
      }
  }
  

• class Solution {
  public:
      bool isPowerOfFour(int n) {
          return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
      }
  };
  

• class Solution:
      def isPowerOfFour(self, n: int) -> bool:
          return n > 0 and (n & (n - 1)) == 0 and (n & 0xAAAAAAAA) == 0
  
  

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

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

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

• public class Solution {
      public bool IsPowerOfFour(int n) {
          return n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa) == 0;
      }
  }
  
  

• impl Solution {
      pub fn is_power_of_four(n: i32) -> bool {
          n > 0 && (n & (n - 1)) == 0 && (n & 0xaaaaaaaa_u32 as i32) == 0
      }
  }
  
  

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