99 questions/Solutions/17
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Revision as of 20:44, 8 April 2013 by PiotrKukielka (talk | contribs) (Add one more example using arrows)
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(*) Split a list into two parts; the length of the first part is given.
Do not use any predefined predicates.
Solution using take
and drop
:
split xs n = (take n xs, drop n xs)
Or even simpler using splitAt
:
split = flip splitAt
But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:
split :: [a] -> Int -> ([a], [a])
split [] _ = ([], [])
split l@(x : xs) n | n > 0 = (x : ys, zs)
| otherwise = ([], l)
where (ys,zs) = split xs (n - 1)
The same solution as above written more cleanly:
split :: [a] -> Int -> ([a], [a])
split (x:xs) n | n > 0 = let (f,l) = split xs (n-1) in (x : f, l)
split xs _ = ([], xs)
Or (ab)using the "&&&" arrow operator for tuples:
split :: [a] -> Int -> ([a], [a])
split (x:xs) n | n > 0 = (:) x . fst &&& snd $ split xs (n - 1)
split xs _ = ([], xs)
A similar solution using foldl:
split :: [a] -> Int -> ([a], [a])
split [] _ = ([], [])
split list n
| n < 0 = (list, [])
| otherwise = (first output, second output)
where output = foldl (\acc e -> if third acc > 0 then (first acc ++ [e], second acc, third acc - 1) else (first acc, second acc ++ [e], third acc)) ([], [], n) list
Note that for the above code to work you must define your own first, second, and third functions for tuples containing three elements like so:
first :: (a, b, c) -> a
first (x, _, _) = x
second :: (a, b, c) -> b
second (_, y, _) = y
third :: (a, b, c) -> c
third (_, _, z) = z
Another foldl solution without defining tuple extractors:
split :: [a] -> Int -> ([a],[a])
split lst n = snd $ foldl helper (0,([],[])) lst
where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right))