# Difference between revisions of "99 questions/Solutions/9"

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< 99 questions | Solutions

(Added a new solution using break) |
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then (x:(head (pack xs))):(tail (pack xs)) |
then (x:(head (pack xs))):(tail (pack xs)) |
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else [x]:(pack xs) |
else [x]:(pack xs) |
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− | </haskell> |
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− | |||

− | |||

− | A simpler solution which is similar to the takeWhile/dropWhile solution, but in one step. |
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− | <haskell> |
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− | pack :: (Eq a) => [a] -> [[a]] |
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− | pack (x:xs) = (x:xs') : (pack ys) |
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− | where (xs',ys) = break (/=x) xs |
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− | pack [] = [] |
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</haskell> |
</haskell> |

## Revision as of 14:41, 24 April 2011

(**) Pack consecutive duplicates of list elements into sublists.

If a list contains repeated elements they should be placed in separate sublists.

```
pack (x:xs) = let (first,rest) = span (==x) xs
in (x:first) : pack rest
pack [] = []
```

This is implemented as `group`

in `Data.List`

.

A more verbose solution is

```
pack :: Eq a => [a] -> [[a]]
pack [] = []
pack (x:xs) = (x:first) : pack rest
where
getReps [] = ([], [])
getReps (y:ys)
| y == x = let (f,r) = getReps ys in (y:f, r)
| otherwise = ([], (y:ys))
(first,rest) = getReps xs
```

Similarly, using `splitAt`

and `findIndex`

:

```
pack :: Eq a => [a] -> [[a]]
pack [] = []
pack (x:xs) = (x:reps) : (pack rest)
where
(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs)
```

Another solution using `takeWhile`

and `dropWhile`

:

```
pack :: (Eq a) => [a] -> [[a]]
pack [] = []
pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)
```

Or we can use `foldr`

to implement this:

```
pack :: (Eq a) => [a] -> [[a]]
pack = foldr func []
where func x [] = [[x]]
func x (y:xs) =
if x == (head y) then ((x:y):xs) else ([x]:y:xs)
```

A simple solution:

```
pack :: (Eq a) => [a] -> [[a]]
pack [] = []
pack [x] = [[x]]
pack (x:xs) = if x `elem` (head (pack xs))
then (x:(head (pack xs))):(tail (pack xs))
else [x]:(pack xs)
```