Difference between revisions of "99 questions/Solutions/10"
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encode [] = [] |
encode [] = [] |
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encode (x:xs) = (length $ x : takeWhile (==x) xs, x) : encode (dropWhile (==x) xs) |
encode (x:xs) = (length $ x : takeWhile (==x) xs, x) : encode (dropWhile (==x) xs) |
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+ | </haskell> |
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+ | |||
+ | Or without higher order functions: |
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+ | |||
+ | <haskell> |
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+ | encode [] = [] |
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+ | encode (x:xs) = encode' 1 x xs where |
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+ | encode' n x [] = [(n, x)] |
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+ | encode' n x (y:ys) |
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+ | | x == y = encode' (n + 1) x ys |
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+ | | otherwise = (n, x) : encode' 1 y ys |
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</haskell> |
</haskell> |
Revision as of 05:05, 11 May 2011
(*) Run-length encoding of a list.
Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.
encode xs = map (\x -> (length x,head x)) (group xs)
which can also be expressed as a list comprehension:
[(length x, head x) | x <- group xs]
Or writing it Pointfree (Note that the type signature is essential here to avoid hitting the Monomorphism Restriction):
encode :: Eq a => [a] -> [(Int, a)]
encode = map (\x -> (length x, head x)) . group
Or (ab)using the "&&&" arrow operator for tuples:
encode :: Eq a => [a] -> [(Int, a)]
encode xs = map (length &&& head) $ group xs
Or with the help of foldr (pack is the resulting function from P09):
encode xs = (enc . pack) xs
where enc = foldr (\x acc -> (length x, head x) : acc) []
Or using takeWhile and dropWhile:
encode [] = []
encode (x:xs) = (length $ x : takeWhile (==x) xs, x) : encode (dropWhile (==x) xs)
Or without higher order functions:
encode [] = []
encode (x:xs) = encode' 1 x xs where
encode' n x [] = [(n, x)]
encode' n x (y:ys)
| x == y = encode' (n + 1) x ys
| otherwise = (n, x) : encode' 1 y ys