Haskell Quiz/FizzBuzz/Solution Ninju: Difference between revisions
< Haskell Quiz | FizzBuzz
No edit summary |
No edit summary |
||
| Line 25: | Line 25: | ||
main :: IO () | main :: IO () | ||
main = | main = mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100] | ||
where | where | ||
loop n s = cycle $ replicate (n-1) | loop n s = cycle $ replicate (n-1) [] ++ [s] | ||
join s t n = head $ filter (not . null) [s ++ t, show n] | join s t n = head $ filter (not . null) [s ++ t, show (n :: Int)] | ||
</haskell> | </haskell> | ||
Revision as of 06:09, 7 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 nAn alternate solution:
module Main where
main :: IO ()
main = mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100]
where
loop n s = cycle $ replicate (n-1) [] ++ [s]
join s t n = head $ filter (not . null) [s ++ t, show (n :: Int)]