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

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< 99 questions | Solutions

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+ | Another way to decode the simplified encoding (which in the opinion of this editor is a far more sensible one for Haskell): |
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+ | |||

+ | <haskell> |
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+ | decode :: Eq a => [(Int,a)] -> [a] |
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+ | decode xs = foldr f [] xs |
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+ | where |
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+ | f (1, x) r = x : r |
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+ | f (k, x) r = x : f (k-1, x) r |
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+ | </haskell> |
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+ | |||

+ | Or, to make it a good transformer for list fusion, |
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+ | |||

+ | <haskell> |
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+ | {-# INLINE decode #-} |
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+ | decode :: Eq a => [(Int,a)] -> [a] |
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+ | decode xs = build (\c n -> |
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+ | let |
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+ | f (1, x) r = x `c` r |
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+ | f (k, x) r = x `c` f (k-1, x) r |
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+ | in |
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+ | foldr f n xs) |
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+ | </haskell> |
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[[Category:Programming exercise spoilers]] |
[[Category:Programming exercise spoilers]] |

## Revision as of 20:16, 28 December 2014

(**) 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) []
```

Another way to decode the simplified encoding (which in the opinion of this editor is a far more sensible one for Haskell):

```
decode :: Eq a => [(Int,a)] -> [a]
decode xs = foldr f [] xs
where
f (1, x) r = x : r
f (k, x) r = x : f (k-1, x) r
```

Or, to make it a good transformer for list fusion,

```
{-# INLINE decode #-}
decode :: Eq a => [(Int,a)] -> [a]
decode xs = build (\c n ->
let
f (1, x) r = x `c` r
f (k, x) r = x `c` f (k-1, x) r
in
foldr f n xs)
```