leetcode/solution/0200-0299/0231.Power of Two

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0200-0299/0231.Power%20of%20Two/README_EN.md	Bit Manipulation Recursion Math

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

Python3

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

Java

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

C++

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

Go

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

TypeScript

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

JavaScript

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

Solution 2

Python3

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

Java

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

C++

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

Go

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

TypeScript

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

JavaScript

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