Question

231. Power of Two

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Description

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?

Solutions

Solution 1

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

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

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

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;
};

Solution 2

class Solution:
  def isPowerOfTwo(self, n: int) -> bool:
    return n > 0 and n == n & (-n)

class Solution {
  public boolean isPowerOfTwo(int n) {
    return n > 0 && n == (n & (-n));
  }
}

class Solution {
public:
  bool isPowerOfTwo(int n) {
    return n > 0 && n == (n & (-n));
  }
};

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

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 & -n);
};