Difference between revisions of "99 questions/Solutions/12"

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m (Improve code layout.)
(Fix function name)
Line 31: Line 31:
  
 
<haskell>
 
<haskell>
decode :: [ListItem a] -> [a]
+
decodeModified :: [ListItem a] -> [a]
decode = concatMap (uncurry replicate . toTuple)
+
decodeModified = concatMap (uncurry replicate . toTuple)
 
</haskell>
 
</haskell>
  
 
a naïve solution with <hask>foldl</hask>:
 
a naïve solution with <hask>foldl</hask>:
 
<haskell>
 
<haskell>
decode :: [ListItem a]-> [a]
+
decodeModified :: [ListItem a]-> [a]
decode = foldl (\x y -> x ++ decodeHelper y) []  
+
decodeModified = foldl (\x y -> x ++ decodeHelper y) []
 
     where
 
     where
 
         decodeHelper :: ListItem a -> [a]
 
         decodeHelper :: ListItem a -> [a]
Line 48: Line 48:
  
 
<haskell>
 
<haskell>
decode :: [ListItem a] -> [a]
+
decodeModified :: [ListItem a] -> [a]
decode = foldl (\acc e -> case e of Single x -> acc ++ [x]; Multiple n x -> acc ++ replicate n x) []
+
decodeModified = foldl (\acc e -> case e of Single x -> acc ++ [x]; Multiple n x -> acc ++ replicate n x) []
 
</haskell>
 
</haskell>

Revision as of 21:01, 8 April 2013

(**) Decode a run-length encoded list.

Given a run-length code list generated as specified in problem 11. Construct its uncompressed version.

decodeModified :: [ListItem a] -> [a]
decodeModified = concatMap decodeHelper
    where
      decodeHelper (Single x)     = [x]
      decodeHelper (Multiple n x) = replicate n x

We only need to map single instances of an element to a list containing only one element and multiple ones to a list containing the specified number of elements and concatenate these lists.

A solution for the simpler encoding from problem 10 can be given as:

decode :: [(Int, a)] -> [a]
decode = concatMap (uncurry replicate)

This can be easily extended given a helper function:

toTuple :: ListItem a -> (Int, a)
toTuple (Single x)     = (1, x)
toTuple (Multiple n x) = (n, x)

as:

decodeModified :: [ListItem a] -> [a]
decodeModified = concatMap (uncurry replicate . toTuple)

a naïve solution with foldl:

decodeModified :: [ListItem a]-> [a]
decodeModified = foldl (\x y -> x ++ decodeHelper y) []
    where
        decodeHelper :: ListItem a -> [a]
        decodeHelper (Single x)     = [x]
        decodeHelper (Multiple n x) = replicate n x

foldl can also be used to solve this problem:

decodeModified :: [ListItem a] -> [a]
decodeModified = foldl (\acc e -> case e of Single x -> acc ++ [x]; Multiple n x -> acc ++ replicate n x) []