(Added another solution that also indicates failure using Maybe (thanks to zygoloid from #haskell for the input regarding Maybe))
(Edit solution to align with problem statement and example)
|(5 intermediate revisions by 5 users not shown)|
Revision as of 02:05, 26 February 2014
(*) Remove the K'th element from a list.
removeAt :: Int -> [a] -> (a, [a]) removeAt k xs = case back of  -> error "removeAt: index too large" x:rest -> (x, front ++ rest) where (front, back) = splitAt (k - 1) xs
If the original list has fewer than k elements, the second list will be empty, and there will be no element to extract. Note that we treat 1 as the first element in the list.
removeAt n xs = (xs !! (n - 1), take (n - 1) xs ++ drop n xs)
Another solution that avoids throwing an error and using ++ operators. Treats 1 as the first element in the list.
removeAt :: Int -> [a] -> (Maybe a, [a]) removeAt _  = (Nothing, ) removeAt 1 (x:xs) = (Just x, xs) removeAt k (x:xs) = let (a, r) = removeAt (k - 1) xs in (a, x:r)
Another solution that also uses Maybe to indicate failure:
removeAt :: Int -> [a] -> (Maybe a, [a]) removeAt _  = (Nothing, ) removeAt 0 xs = (Nothing, xs) removeAt nr xs | nr > length xs = (Nothing, xs) | otherwise = (Just (xs !! nr), fst splitted ++ (tail . snd) splitted) where splitted = splitAt nr xs
And yet another solution (without error checking):
removeAt :: Int -> [a] -> (a, [a]) removeAt n xs = let (front, back) = splitAt n xs in (last front, init front ++ back)
Similar, point-free style:
removeAt n = (\(a, b) -> (head b, a ++ tail b)) . splitAt (n - 1)
A simple recursive solution:
removeAt 1 (x:xs) = (x, xs) removeAt n (x:xs) = (l, x:r) where (l, r) = removeAt (n - 1) xs