99 questions/Solutions/20

From HaskellWiki
< 99 questions‎ | Solutions
Revision as of 08:10, 29 July 2010 by Bbest86 (talk | contribs) (Adding a third solution.)
Jump to navigation Jump to search
The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

(*) 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 xs

Simply use the splitAt to split after k elements. If the original list has fewer than k+1 elements, the second list will be empty, and there will be no element to extract. Note that the Prolog and Lisp versions treat 1 as the first element in the list, and the Lisp version appends NIL elements to the end of the list if k is greater than the list length.

or

removeAt n xs = (xs!!n,take n xs ++ drop (n+1) 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)