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;
  }
}

I suggest you put this code into a method and create a unit test.

public static boolean isLeapYear(int year) {
    assert year >= 1583; // not valid before this date.
    return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}

In the unit test

assertTrue(isLeapYear(2000));
assertTrue(isLeapYear(1904));
assertFalse(isLeapYear(1900));
assertFalse(isLeapYear(1901));

It's almost always wrong to have repetition in software. In any engineering discipline, form should follow function, and you have three branches for something which has two possible paths - it's either a leap year or not.

The mechanism which has the test on one line doesn't have that issue, but generally it would be better to separate the test into a function which takes an int representing a year and returns a boolean representing whether or not the year is a leap year. That way you can do something with it other that print to standard output on the console, and can more easily test it.

In code which is known to exceed its performance budget, it's usual to arrange the tests so that they are not redundant and perform the tests in an order which returns early. The wikipedia example does this - for most years you have to calculate modulo 400,100 and 4, but for a few you only need modulo 400 or 400 and 100. This is a small optimisation in terms of performance ( at best, only one in a hundred inputs are effected ), but it also means the code has less repetition, and there's less for the programmer to type in.

With the course of : TestMyCode Programming assignment evaluator, that one of the exercises was this kind of issue, I wrote this answer:

import java.util.Scanner;

public class LeapYear {

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        System.out.println("Type a year: ");

        int year = Integer.parseInt(reader.nextLine());

        if (year % 400 == 0 && year % 100 == 0 && year % 4 == 0) {
            System.out.println("The year is a leap year");
        } else
         if (year % 4 == 0 && year%100!=0 ) {
            System.out.println("The year is a leap year");
        } else 
        {
            System.out.println("The year is not a leap year");
        }

    }
}

easiest way ta make java leap year and more clear to understandenter code here

import  java.util.Scanner;

class que19{

public static void main(String[] args) {

    Scanner input=new Scanner(System.in);

    double a;

    System.out.println("enter the year here ");
    a=input.nextDouble();
    if ((a % 4 ==0 ) && (a%100!=0) || (a%400==0)) {
        System.out.println("leep year");

    }
    else {
        System.out.println("not a leap year");
    }
}

}

this answer is great but it won't work for years before Christ (using a proleptic Gregorian calendar). If you want it to work for B.C. years, then use the following adaptation:

public static boolean isLeapYear(final int year) {
    final Calendar cal = Calendar.getInstance();
    if (year<0) {
        cal.set(Calendar.ERA, GregorianCalendar.BC);
        cal.set(Calendar.YEAR, -year);
    } else
        cal.set(Calendar.YEAR, year);
    return cal.getActualMaximum(Calendar.DAY_OF_YEAR) > 365;
}

You can verify that for yourself by considering that year -5 (i.e. 4 BC) should be pronounced as a leap year assuming a proleptic Gregorian calendar. Same with year -1 (the year before 1 AD). The linked to answer does not handle that case whereas the above adapted code does.

Answer in Kotlin:

fun isLeapYear(year: Int): Boolean =
    when {
        year % 4 != 0 -> false
        year % 100 == 0 -> year % 400 == 0
        else -> true
    }

This is from Clean Code book by Martin R.

public static boolean isLeapYear(int year) {
   boolean fourth = year % 4 == 0;
   boolean hundredth = year % 100 == 0;
   boolean fourHundredth = year % 400 == 0;
   return fourth && (!hundredth || fourHundredth);
}