Difference between revisions of "Haskell Quiz/FizzBuzz/Solution Ninju"
< Haskell Quiz | FizzBuzz
Jump to navigation
Jump to search
Line 26: | Line 26: | ||
main :: IO () |
main :: IO () |
||
main = do |
main = do |
||
− | mapM_ putStrLn $ |
+ | mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100 :: Int] |
where |
where |
||
loop n s = cycle $ replicate (n-1) "" ++ [s] |
loop n s = cycle $ replicate (n-1) "" ++ [s] |
||
− | + | join s t n = head $ filter (not . null) [s ++ t, show n] |
|
− | xor s t = if null s then t else s |
||
</haskell> |
</haskell> |
Revision as of 22:32, 6 July 2010
I think this is probably what I'd do in the interview situation - i.e. the first and most obvious thing that comes to mind. (Alex Watt)
module Main where
main :: IO ()
main = printAll $ map fizzBuzz [1..100]
where
printAll [] = return ()
printAll (x:xs) = putStrLn x >> printAll xs
fizzBuzz :: Integer -> String
fizzBuzz n | n `mod` 15 == 0 = "FizzBuzz"
| n `mod` 5 == 0 = "Fizz"
| n `mod` 3 == 0 = "Buzz"
| otherwise = show n
An alternate solution:
module Main where
main :: IO ()
main = do
mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100 :: Int]
where
loop n s = cycle $ replicate (n-1) "" ++ [s]
join s t n = head $ filter (not . null) [s ++ t, show n]