Algorithm
Problem Name: 1154. Day of the 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
Constraints:
date.length == 10
date[4] == date[7] == '-'
, and all otherdate[i]
's are digitsdate
represents a calendar date between Jan 1st, 1900 and Dec 31th, 2019.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
List daysInMonth = Arrays.asList(31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
public int dayOfYear(String date) {
int year = Integer.parseInt(date.split("-")[0]);
int month = Integer.parseInt(date.split("-")[1]);
int day = Integer.parseInt(date.split("-")[2]);
int numOfDays = getNumOfDays(month, isLeapYear(year));
return numOfDays + day;
}
private boolean isLeapYear(int year) {
if (year % 4 == 0) {
if (year % 100 == 0) {
return year % 400 == 0;
}
else {
return true;
}
}
return false;
}
private int getNumOfDays(int month, boolean isLeapYear) {
int numOfDays = 0;
for (int i = 1; i < month; i++) {
if (i == 2) {
numOfDays += isLeapYear ? daysInMonth.get(i - 1) + 1 : daysInMonth.get(i - 1);
}
else {
numOfDays += daysInMonth.get(i - 1);
}
}
return numOfDays;
}
}
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#2 Code Example with Javascript Programming
Code -
Javascript Programming
const dayOfYear = function(date) {
const [year, month, day] = date.split('-').map(s => +s),
months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31],
isLeapYear = !(year % 4) && month > 2 && (!!(year % 100) || !(year % 400))
return months.splice(0, month - 1).reduce((a, b) => a + b, day + +isLeapYear)
}
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#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def dayOfYear(self, date: str) -> int:
cnt = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
y, m, d = map(int, date.split('-'))
days = sum(cnt[:m - 1]) + d
if m > 2:
if y % 400 == 0: days += 1
if y % 100 == 0: return days
if y % 4 == 0: days += 1
return days
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#4 Code Example with C# Programming
Code -
C# Programming
using System.Linq;
namespace LeetCode
{
public class _1154_DayOfTheYear
{
public int DayOfYear(string date)
{
var days = new int[] { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
var split = date.Split('-').Select(s => int.Parse(s)).ToArray();
var year = split[0];
var count = 0;
for (int i = 0; i < 12; i++)
{
if (split[1] == i + 1)
{
count += split[2];
break;
}
else
{
count += days[i];
}
}
if (split[1] > 2)
if ((year % 400 == 0) || (year % 100 != 0 && year % 4 == 0)) count++;
return count;
}
}
}
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