Difference between revisions of "Haskell Quiz/FizzBuzz/Solution Ninju"

From HaskellWiki
Jump to navigation Jump to search
m
Line 22: Line 22:
   
 
<haskell>
  
<haskell>
−
module Main where
  
−  
 
main :: IO ()
  
main :: IO ()
−
main = mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100]
 +
main = mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100 :: Int]
 
where
  
where
  +
xor s t = if null s then t else s
 
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 :: Int)]
 +
join s t n = xor (s ++ t) (show n)
 
</haskell>
  
</haskell>

Revision as of 10:45, 8 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: 

main :: IO ()
main = mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100 :: Int]
  where
    xor s t = if null s then t else s
    loop n s = cycle $ replicate (n-1) [] ++ [s]
    join s t n = xor (s ++ t) (show n)
Retrieved from "https://wiki.haskell.org/index.php?title=Haskell_Quiz/FizzBuzz/Solution_Ninju&oldid=35182"