comments | difficulty | edit_url | rating | source | tags | ||
---|---|---|---|---|---|---|---|
true |
Easy |
1144 |
Weekly Contest 311 Q1 |
|
Given a positive integer n
, return the smallest positive integer that is a multiple of both 2
and n
.
Example 1:
Input: n = 5 Output: 10 Explanation: The smallest multiple of both 5 and 2 is 10.
Example 2:
Input: n = 6 Output: 6 Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.
Constraints:
1 <= n <= 150
If
The time complexity is
class Solution:
def smallestEvenMultiple(self, n: int) -> int:
return n if n % 2 == 0 else n * 2
class Solution {
public int smallestEvenMultiple(int n) {
return n % 2 == 0 ? n : n * 2;
}
}
class Solution {
public:
int smallestEvenMultiple(int n) {
return n % 2 == 0 ? n : n * 2;
}
};
func smallestEvenMultiple(n int) int {
if n%2 == 0 {
return n
}
return n * 2
}
function smallestEvenMultiple(n: number): number {
return n % 2 === 0 ? n : n * 2;
}
impl Solution {
pub fn smallest_even_multiple(n: i32) -> i32 {
if n % 2 == 0 {
return n;
}
n * 2
}
}
int smallestEvenMultiple(int n) {
return n % 2 == 0 ? n : n * 2;
}