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231. Power of Two

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: Bit Manipulation

According to the properties of bit manipulation, executing \(\texttt{n\&(n-1)}\) can eliminate the last bit \(1\) in the binary form of \(n\). Therefore, if \(n > 0\) and \(\texttt{n\&(n-1)}\) results in \(0\), then \(n\) is a power of \(2\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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

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

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

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

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

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impl Solution {
    pub fn is_power_of_two(n: i32) -> bool {
        n > 0 && (n & (n - 1)) == 0
    }
}

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

Solution 2: Lowbit

According to the definition of \(\text{lowbit}\), we know that \(\text{lowbit}(x) = x \& (-x)\), which can get the decimal number represented by the last bit \(1\) of \(n\). Therefore, if \(n > 0\) and \(\text{lowbit}(n)\) equals \(n\), then \(n\) is a power of \(2\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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

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

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

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

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

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impl Solution {
    pub fn is_power_of_two(n: i32) -> bool {
        n > 0 && n == (n & (-n))
    }
}

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

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