Java Code for Calculating Leap Year

Java Code for calculating Leap Year

The correct implementation is:

public static boolean isLeapYear(int year) {
  Calendar cal = Calendar.getInstance();
  cal.set(Calendar.YEAR, year);
  return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365;
}

But if you are going to reinvent this wheel then:

public static boolean isLeapYear(int year) {
  if (year % 4 != 0) {
    return false;
  } else if (year % 400 == 0) {
    return true;
  } else if (year % 100 == 0) {
    return false;
  } else {
    return true;
  }
}

Calculate leap year in Java

If you use the GregorianCalendar, you could do as below

Determines if the given year is a leap year. Returns true if the given
year is a leap year. To specify BC year numbers, 1 - year number must
be given. For example, year BC 4 is specified as -3.

GregorianCalendar cal = new GregorianCalendar();

if(cal.isLeapYear(year))
{
    System.out.print("Given year is leap year.");
}
else
{ 
    System.out.print("Given year is not leap year.");
}

Calculate age in java considering leap years

The correct way to do this is with the Period and LocalDate classes in the java.time package. However, it seems that you're trying to reinvent the calculation for yourself.

The way that I would recommend doing this is to write a class that lets you calculate a "day number" for a given date - that is, the number of days between the specified date, and some arbitrary date in the past. Then when you want to find the number of days between two specified dates, you can just use calculate the "day number" for both dates, and subtract them.

I have done that here, for a purely Gregorian calendar. This class is no good before the Gregorian cutover - I haven't tried to build a historically accurate Julian/Gregorian hybrid calendar, such as the JDK provides. And the arbitrary date in the past that it calculates day numbers from is 31 December, 2BC. This date, of course, isn't really part of the Gregorian calendar; but for our purposes here, it doesn't matter.

Since you're unlikely to encounter any dates before the Gregorian cutover, this class should be more than adequate for your purposes. I still recommend using the Period and LocalDate classes instead of this one, for production code. This is just here so you can see how to do the calculations.

public class GregorianDate {

    private final int day;
    private final int month;
    private final int year;

    private static final int[] DAYS_PER_MONTH = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    public GregorianDate(int day, int month, int year) {
        this.day = day;
        this.month = month;
        this.year = year;
    }

    public boolean isValid() {
        return month >= 1 && month <= 12 && day >= 1 && day <= daysThisMonth();
    }

    public static int daysBetween(GregorianDate from, GregorianDate to) {
        return to.dayNumber() - from.dayNumber();
    }

    public static int daysBetween(int fromDay, int fromMonth, int fromYear, int toDay, int toMonth, int toYear) {
        return daysBetween(new GregorianDate(fromDay, fromMonth, fromYear), new GregorianDate(toDay, toMonth, toYear));
    }

    private int daysThisMonth() {
        return (isLeapYear() && month == 2) ? 29 : DAYS_PER_MONTH[month];
    }

    private int dayNumber() {
        return year * 365 + leapYearsBefore() + daysInMonthsBefore() + day;
    }

    private boolean isLeapYear() {
        return ( year % 4 == 0 &&  year % 100 != 0 ) || year % 400 == 0;  
    }

    private int leapYearsBefore() {
        return year / 4 - year / 100 + year / 400;
    }

    private int daysInMonthsBefore() {
        switch(month) {
        case 1: 
            return 0;
        case 2:
            return 31;
        default:
            // Start with the number in January and February combined
            int toReturn = isLeapYear() ? 60 : 59; 
            for (int monthToConsider = 3; monthToConsider < month; monthToConsider++) {
                toReturn += DAYS_PER_MONTH[monthToConsider];
            }
            return toReturn;
        }
    }

}

Determine leap year using Java?

userInput % 4 == 0 && userInput % 100 ==0 is equivalent to userInput % 400 == 0

and userInput % 4 == 0 then it is definitely Leap year so need not to check any other condition.

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