'Easy way to determine leap year in ruby?
Is there an easy way to determine if a year is a leap year?
Solution 1:[1]
For your understanding:
def leap_year?(year)
if year % 4 == 0
if year % 100 == 0
if yearVar % 400 == 0
return true
end
return false
end
return true
end
false
end
This could be written as:
def leap_year?(year)
(year % 4 == 0) && !(year % 100 == 0) || (year % 400 == 0)
end
Solution 2:[2]
Try this:
is_leap_year = year % 4 == 0 && year % 100 != 0 || year % 400 == 0
Solution 3:[3]
Here is my answer for the exercism.io problem which asks the same question. You are explicitly told to ignore any standard library functions that may implement it as part of the exercise.
class Year
attr_reader :year
def initialize(year)
@year = year
end
def leap?
if @year.modulo(4).zero?
return true unless @year.modulo(100).zero? and not @year.modulo(400).zero?
end
false
end
end
Solution 4:[4]
def leap_year?(num)
if num%4 == 0 && num%100 != 0
true
elsif num%400 == 0
true
elsif num%4 == 0 && num%100 == 0 && num%400 != 0
false
elsif num%4 != 0
false
end
end
puts leap_year?(2000)
Solution 5:[5]
Any year that is evenly divisible by 4 is a leap year. However, there is still a small error that must be accounted for. To eliminate this error, the Gregorian calendar stipulates that a year that is evenly divisible by 100 (for example, 1900) is a leap year only if it is also evenly divisible by 400.
class Date
def self.leap?(year)
year % 4 == 0 && year % 100 != 0 || year % 400 == 0
end
end
Solution 6:[6]
This one takes a range:
(starting..ending).each do |year|
next if year % 4 != 0
next if year % 100 == 0 && year % 400 != 0
puts year
end
Source: Learn to Program by Chris Pine
Solution 7:[7]
Using the least amount possible of comparisons, you could do this:
- First/Longer version
def leap_year?(year)
# check first if year is divisible by 400
return true if year % 400 == 0
year % 4 == 0 && year % 100 != 0
end
- Shorter version
We can do the same check using short circuit OR (||):
def leap_year?(year)
year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)
end
Solution 8:[8]
The variable n takes year to test and prints true if it's a leap year and false if it's not.
n=gets.to_i
n%4==0 ? n%100==0 ? n%400 ?(puts true):(puts false) :(puts true) :(puts false)
Solution 9:[9]
Already a lot of answers, though here is a shorter version:
y%100 > 0 ? y%4 < 1 : y%400 < 1
If the year is not a century, then checks if it is a multiple of 4, otherwise checks if it is multiple of 400.
Solution 10:[10]
Here's my method:
def leap_year?(year)
Time.new(year, 12, 31).yday == 366
end
leap_year?(1999)
#=> false
leap_year?(2000)
#=> true
leap_year?(2100)
#=> false
leap_year?(2104)
#=> true
Sources
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Source: Stack Overflow