Welcome to Subscribe On Youtube
1118. Number of Days in a Month
Description
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
Solutions
Solution 1: Determine Leap Year
We can first determine whether the given year is a leap year. If the year can be divided by $4$ but not by $100$, or can be divided by $400$, then this year is a leap year.
February has $29$ days in a leap year and $28$ days in a common year.
We can use an array $days$ to store the number of days in each month of the current year, where $days[0]=0$, $days[i]$ represents the number of days in the $i$th month of the current year. Then the answer is $days[month]$.
The time complexity is $O(1)$, and the space complexity is $O(1)$.

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]; } };

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]

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]; }