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).
PythonJavaC++GoTypeScriptRustJavaScript
class Solution: def isPowerOfTwo(self, n: int) -> bool: return n > 0 and (n & (n - 1)) == 0(code-box)
Solution 2: Lowbit
According to the definition of lowbit, we know that lowbit(x) = x & (-x), which can get the decimal number represented by the last bit 1 of n. Therefore, if n > 0 and lowbit(n) equals n, then n is a power of 2.
The time complexity is O(1), and the space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScript
class Solution: def isPowerOfTwo(self, n: int) -> bool: return n > 0 and n == n & (-n)(code-box)
