Difference between revisions of "99 questions/Solutions/20"
m (One of removeAt examples were removing element at wrong index (off-by-one))
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removeAt n xs = (xs!!n,take n xs ++ drop
removeAt n xs = (xs !! n , take n xs ++ drop n xs)
Revision as of 19:55, 8 April 2013
(*) 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.
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