Difference between revisions of "99 questions/Solutions/7"

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Line 14: Line 14:
 
flatten (List (x:xs)) = flatten x ++ flatten (List xs)
 
flatten (List (x:xs)) = flatten x ++ flatten (List xs)
 
flatten (List []) = []
 
flatten (List []) = []
  +
</haskell>
  +
or using things that act just like <hask>concatMap</hask>
  +
<haskell>
  +
flatten (Elem x) = return x
  +
flatten (List x) = flatten =<< x
  +
  +
flatten (Elem x) = [x]
  +
flatten (List x) = foldMap flatten x
 
</haskell>
 
</haskell>
 
<haskell>
 
<haskell>
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flatten3 (Elem x ) = [x]
 
flatten3 (Elem x ) = [x]
 
flatten3 (List xs) = foldr (++) [] $ map flatten3 xs
 
flatten3 (List xs) = foldr (++) [] $ map flatten3 xs
  +
</haskell>
  +
  +
or with an accumulator function:
  +
<haskell>
  +
flatten4 = reverse . rec []
  +
where
  +
rec acc (List []) = acc
  +
rec acc (Elem x) = x:acc
  +
rec acc (List (x:xs)) = rec (rec acc x) (List xs)
 
</haskell>
 
</haskell>
   
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Our NestedList datatype is either a single element of some type (Elem a), or a
 
Our NestedList datatype is either a single element of some type (Elem a), or a
 
list of NestedLists of the same type. (List [NestedList a]).
 
list of NestedLists of the same type. (List [NestedList a]).
  +
  +
[[Category:Programming exercise spoilers]]

Revision as of 13:29, 15 February 2015

(**) Flatten a nested list structure.

data NestedList a = Elem a | List [NestedList a]

flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List x) = concatMap flatten x

or without concatMap

flatten :: NestedList a -> [a]
flatten (Elem a   )   = [a]
flatten (List (x:xs)) = flatten x ++ flatten (List xs)
flatten (List [])     = []

or using things that act just like concatMap

flatten (Elem x) = return x
flatten (List x) = flatten =<< x

flatten (Elem x) = [x]
flatten (List x) = foldMap flatten x
flatten2 :: NestedList a -> [a]
flatten2 a = flt' a []
  where flt' (Elem x)      xs = x:xs
        flt' (List (x:ls)) xs = flt' x (flt' (List ls) xs)
        flt' (List [])     xs = xs

or with foldr

flatten3 :: NestedList a -> [a]
flatten3 (Elem x ) = [x]
flatten3 (List xs) =  foldr (++) [] $ map flatten3 xs

or with an accumulator function:

flatten4 = reverse . rec []
  where
  rec acc (List []) = acc
  rec acc (Elem x)  = x:acc
  rec acc (List (x:xs)) = rec (rec acc x) (List xs)

We have to define a new data type, because lists in Haskell are homogeneous. [1, [2, [3, 4], 5]] is a type error. Therefore, we must have a way of representing a list that may (or may not) be nested.

Our NestedList datatype is either a single element of some type (Elem a), or a list of NestedLists of the same type. (List [NestedList a]).