99 questions/Solutions/18
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(Difference between revisions)
m (Updated my solution to a more paranoid one) |
(added a cleaner version of the final solution) |
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| Line 48: | Line 48: | ||
chunk = fst $ splitAt (k - i') chop -- Remove the part after k | chunk = fst $ splitAt (k - i') chop -- Remove the part after k | ||
i' = i - 1 | i' = i - 1 | ||
| + | </haskell> | ||
| + | A little cleaner, using the previous problem's split (a.k.a. <hask>splitAt</hask>): | ||
| + | <haskell> | ||
| + | slice xs (i+1) k = snd (split (fst (split xs k)) i) | ||
</haskell> | </haskell> | ||
Revision as of 14:47, 17 September 2010
(**) Extract a slice from a list.
Given two indices, i and k, the slice is the list containing the elements between the i'th and k'th element of the original list (both limits included). Start counting the elements with 1.
slice xs (i+1) k = take (k-i) $ drop i xs
The same solution as above, but the more paranoid (maybe too paranoid?) version of it (uses guards and Maybe):
slice :: [a] -> Int -> Int -> Maybe [a] slice [] _ _ = Just [] slice xs k n | k == n = Just [] | k > n || k > length xs || n > length xs || k < 0 || n < 0 = Nothing | k == 0 = Just (take n xs) | otherwise = Just (drop (k-1) $ take n xs)
Or, an iterative solution:
slice :: [a]->Int->Int->[a] slice lst 1 m = slice' lst m [] where slice' :: [a]->Int->[a]->[a] slice' _ 0 acc = reverse acc slice' (x:xs) n acc = slice' xs (n - 1) (x:acc) slice (x:xs) n m = slice xs (n - 1) (m - 1)
Or:
slice :: [a] -> Int -> Int -> [a] slice (x:xs) i k | i > 1 = slice xs (i - 1) (k - 1) | k < 1 = [] | otherwise = x:slice xs (i - 1) (k - 1)
splitAt
take
drop
slice :: [a] -> Int -> Int -> [a] slice xs i k = chunk where chop = snd $ splitAt i' xs -- Get the piece starting at i chunk = fst $ splitAt (k - i') chop -- Remove the part after k i' = i - 1
splitAt
slice xs (i+1) k = snd (split (fst (split xs k)) i)
