Leap year
You are encouraged to solve this task according to the task description, using any language you may know.
Determine whether a given year is a leap year.
See Also
ActionScript
<lang actionscript>public function isLeapYear(year:int):Boolean {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
ALGOL 68
<lang algol68>MODE YEAR = INT, MONTH = INT, DAY = INT;
PROC year days = (YEAR year)DAY: # Ignore 1752 CE for the moment #
( month days(year, 2) = 28 | 365 | 366 );
PROC month days = (YEAR year, MONTH month) DAY:
( month | 31,
28 + ABS(year MOD 4 NE 0 OR year MOD 400 EQ 0),
31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
PROC is leap year = (YEAR year)BOOL: year days(year)=366;
test:(
[]INT test cases = (1900, 1996, 1997, 2000);
FOR i TO UPB test cases DO
YEAR year = test cases[i];
printf(($g(0)" is "b("","not ")"a leap year."l$, year, is leap year(year)))
OD
)</lang> Output:
1900 is not a leap year. 1996 is not a leap year. 1997 is a leap year. 2000 is a leap year.
J
<lang j>isLeap=: 0 -/@:= 4 100 400 |/ ]</lang>
Example use:
<lang> isLeap 1900 1996 1997 2000 0 1 0 1</lang>
Java
<lang java>public static boolean isLeapYear(int year){
if(year % 100 == 0) return year % 400 == 0; return year % 4 == 0;
}</lang>
JavaScript
<lang javascript>function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
PHP
<lang php><?php function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
With date('L'):
<lang php><?php function isLeapYear(year) {
return (date('L', mktime(0, 0, 0, 2, 1, year)) === '1')
}</lang>
Python
<lang python>import calendar calendar.isleap(year) </lang>
or <lang python>def is_leap_year(year):
if year % 100 == 0:
return year % 400 == 0
return year % 4 == 0
</lang>
Asking for forgiveness instead of permission:
<lang python>import datetime
def is_leap_year(year):
try:
datetime.date(year, 2, 29)
except ValueError:
return False
return True
</lang>