[LeetCode]#1154. Day of the Year

Environment: Python 3.7

Key technique: datetime function, leap year

Given a string date representing a Gregorian calendar date formatted as YYYY-MM-DD, return the day number of the year.

Example 1:

Input: date = "2019-01-09"
Output: 9
Explanation: Given date is the 9th day of the year in 2019.

Example 2:

Input: date = "2019-02-10"
Output: 41

Example 3:

Input: date = "2003-03-01"
Output: 60

Example 4:

Input: date = "2004-03-01"
Output: 61

Constraints:

• date.length == 10
• date[4] == date[7] == '-', and all other date[i]'s are digits
• date represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.

Analysis:

if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)

Solution:

import datetime
class Solution:
    def dayOfYear(self, date):
        year_check=int(date[0:4])
        leap=0
        if year_check % 400 ==0:
            leap=1
        if year_check % 4 ==0 and year_check % 100 !=0:
            leap=1
        day1 = datetime.datetime(int(date[0:4]), int(date[5:7]) , int(date[8:10]))
        day2 = datetime.datetime(int(date[0:4]), int(12) , int(31))
        ans=365-(day2-day1).days+leap
        return ans

Submitted result:

Lesson learn:

This case let me know the rule of leap year.

Reference:

https://en.wikipedia.org/wiki/Leap_year