# 99 questions/Solutions/7

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Markstoehr (Talk | contribs) m (Added in a new solution) |
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flt' (List []) xs = xs |
flt' (List []) xs = xs |
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</haskell> |
</haskell> |
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− | or with foldr1 |
+ | or with foldr |

<haskell> |
<haskell> |
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flatten3 :: NestedList a -> [a] |
flatten3 :: NestedList a -> [a] |
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− | flatten3 (Elem x) = [x] |
+ | flatten3 (Elem x ) = [x] |

− | flatten3 (List xs) = foldr1 (++) $ map flatten xs |
+ | flatten3 (List xs) = foldr (++) [] $ map flatten3 xs |

</haskell> |
</haskell> |
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## Revision as of 23:27, 25 December 2011

(**) 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 []) = []

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

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]).