A leap year is a year containing one additional day (366 days a year). Review the leap year algorithm :
if year is divisible by 400 then
is_leap_year
else if year is divisible by 100 then
not_leap_year
else if year is divisible by 4 then
is_leap_year
else
not_leap_year
P.S Algorithm from wikipedia leap year.
1. Java Leap Year Example
Java example to determine if the given year is a leap year.
DateTimeExample.java
package com.mkyong.utils;
public class DateTimeExample {
public static void main(String[] args) {
DateTimeExample obj = new DateTimeExample();
System.out.println("1993 is a leap year : " + obj.isLeapYear(1993));
System.out.println("1996 is a leap year : " + obj.isLeapYear(1996));
System.out.println("2012 is a leap year : " + obj.isLeapYear(2012));
}
public boolean isLeapYear(int year) {
if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0))) {
return true;
} else {
return false;
}
}
}
Output
1993 is a leap year : false
1996 is a leap year : true
2012 is a leap year : true
2. GregorianCalendar Example
Alternatively, you can use GregorianCalendar.isLeapYear() API.
import java.util.GregorianCalendar;
//...
public boolean isLeapYear(int year) {
GregorianCalendar cal = (
GregorianCalendar) GregorianCalendar.getInstance();
return cal.isLeapYear(year);
}
Wikipedia algorithm is inefficient. Instead, use:
“return ((year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0));”. See: http://stackoverflow.com/a/11595914/733805)
public boolean isLeapYear(int year) {
return ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0)));
}
It should doing the same, however it is a nice article.
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