comments | difficulty | edit_url | rating | source | tags | |
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true |
Easy |
1227 |
Biweekly Contest 4 Q1 |
|
Given a year year
and a month month
, return the number of days of that month.
Example 1:
Input: year = 1992, month = 7 Output: 31
Example 2:
Input: year = 2000, month = 2 Output: 29
Example 3:
Input: year = 1900, month = 2 Output: 28
Constraints:
1583 <= year <= 2100
1 <= month <= 12
We can first determine whether the given year is a leap year. If the year can be divided by
February has
We can use an array
The time complexity is
class Solution:
def numberOfDays(self, year: int, month: int) -> int:
leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
days = [0, 31, 29 if leap else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
return days[month]
class Solution {
public int numberOfDays(int year, int month) {
boolean leap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
int[] days = new int[] {0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
return days[month];
}
}
class Solution {
public:
int numberOfDays(int year, int month) {
bool leap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
vector<int> days = {0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
return days[month];
}
};
func numberOfDays(year int, month int) int {
leap := (year%4 == 0 && year%100 != 0) || (year%400 == 0)
x := 28
if leap {
x = 29
}
days := []int{0, 31, x, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
return days[month]
}
function numberOfDays(year: number, month: number): number {
const leap = (year % 4 === 0 && year % 100 !== 0) || year % 400 === 0;
const days = [0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
return days[month];
}