# 99 questions/Solutions/9

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< 99 questions | Solutions

Revision as of 13:48, 15 February 2015 by Henk-Jan van Tuyl (talk | contribs) (Added Category:Programming exercise spoilers)

(**) 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)
pack' [] = []
pack' [x] = [[x]]
pack' (x:xs)
| x == head h_p_xs = (x:h_p_xs):t_p_hs
| otherwise = [x]:p_xs
where p_xs@(h_p_xs:t_p_hs) = pack' xs
```

```
myPack [] = []
myPack (y:ys) = impl ys [[y]]
where
impl [] packed = packed
impl (x:xs) packed
| x == (head (last packed)) = impl xs ((init packed) ++ [x:(last packed)])
| otherwise = impl xs (packed ++ [[x]])
myPack' [] = []
myPack' (y:ys) = reverse $ impl ys [[y]]
where
impl [] packed = packed
impl (x:xs) p@(z:zs)
| x == (head z) = impl xs ((x:z):zs)
| otherwise = impl xs ([x]:p)
```