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:
Follow up: Could you solve it without loops/recursion?
Solutions#
Solution 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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| /**
* @param {number} n
* @return {boolean}
*/
var isPowerOfTwo = function (n) {
return n > 0 && (n & (n - 1)) == 0;
};
|
Solution 2#
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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 - 1)) === 0;
}
|
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| /**
* @param {number} n
* @return {boolean}
*/
var isPowerOfTwo = function (n) {
return n > 0 && n == (n & -n);
};
|
