Haskell Quiz/FizzBuzz/Solution Ninju
From HaskellWiki
(Difference between revisions)
| 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] | |
| - | + | ||
</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]
