99 questions/Solutions/17

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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 xs 0 = ([], xs)
split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l)