Difference between revisions of "99 questions/Solutions/20"
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m (One of removeAt examples were removing element at wrong index (off-by-one)) |
(Edit solution to align with problem statement and example) |
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[] -> error "removeAt: index too large" |
[] -> error "removeAt: index too large" |
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x:rest -> (x, front ++ rest) |
x:rest -> (x, front ++ rest) |
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| − | where (front, back) = splitAt k xs |
+ | where (front, back) = splitAt (k - 1) xs |
</haskell> |
</haskell> |
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| − | Simply use the <hask>splitAt</hask> to split after k elements. |
+ | Simply use the <hask>splitAt</hask> to split after k - 1 elements. |
| − | If the original list has fewer than k |
+ | 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 |
+ | Note that we treat 1 as the first element in the list. |
or |
or |
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| Line 59: | Line 59: | ||
where (l, r) = removeAt (n - 1) xs |
where (l, r) = removeAt (n - 1) xs |
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</haskell> |
</haskell> |
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| + | |||
| + | |||
| + | [[Category:Programming exercise spoilers]] |
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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
Simply use the splitAt to split after k - 1 elements.
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.
or
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