Difference between revisions of "Haskell Quiz/FizzBuzz/Solution Ninju"
< Haskell Quiz | FizzBuzz
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[[Category:Haskell Quiz solutions|FizzBuzz]] |
[[Category:Haskell Quiz solutions|FizzBuzz]] |
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| − | 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. |
+ | 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. |
<haskell> |
<haskell> |
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| n `mod` 3 == 0 = "Buzz" |
| n `mod` 3 == 0 = "Buzz" |
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| otherwise = show n |
| otherwise = show n |
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| − | </haskell> |
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| − | |||
| − | An alternate solution: |
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| − | |||
| − | <haskell> |
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| − | module Main where |
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| − | |||
| − | main :: IO () |
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| − | main = do |
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| − | mapM_ putStrLn $ zipWith3 join (loop 3 "Fizz") (loop 5 "Buzz") [1..100 :: Int] |
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| − | where |
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| − | loop n s = cycle $ replicate (n-1) "" ++ [s] |
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| − | join s t n = head $ filter (not . null) [s ++ t, show n] |
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</haskell> |
</haskell> |
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Latest revision as of 15:19, 6 February 2021
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.
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