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

(**) 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)
```

or making NestedList an instance of Foldable:

```import qualified Data.Foldable as F
instance F.Foldable NestedList where
foldMap f (Elem x)  = f x
foldMap f (List []) = mempty
foldMap f (List (x:xs)) = F.foldMap f x `mappend` F.foldMap f (List xs)

flatten5 :: NestedList a -> [a]
flatten5 = F.foldMap (\x -> [x])
```

or making use of -XDeriveFoldable:

```{-# LANGUAGE DeriveFoldable #-}

import Data.Foldable

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

flatten6 :: NestedList a -> [a]
flatten6 = toList
```

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]).