# Haskell/Solutions/Lists II

< Haskell | Solutions(Redirected from Haskell/Solutions/More about lists)

Note that, for our current purposes, it is indifferent whether to use `Int`

or `Integer`

, so don't worry if you have used one and the solutions below the other.

## Rebuilding listsEdit

1.

```
takeInt :: Int -> [a] -> [a]
takeInt 0 _ = []
takeInt _ [] = []
takeInt n (x:xs) = x : takeInt (n-1) xs
```

2.

```
dropInt :: Int -> [a] -> [a]
dropInt 0 list = list
dropInt _ [] = []
dropInt n (x:xs) = dropInt (n-1) xs
```

3.

```
sumInt :: [Int] -> Int
sumInt [] = 0
sumInt (x:xs) = x + sumInt xs
```

4.

```
-- "Direct" solution with pattern matching.
scanSum :: [Int] -> [Int]
scanSum [] = []
scanSum [x] = [x]
scanSum (x:y:xs) = x : scanSum ((x + y) : xs)
-- Alternatively, using a helper function with an accumulator argument:
scanSum :: [Int] -> [Int]
scanSum = scanSum' 0
where
-- The following type signature is entirely optional.
-- We have added it just for extra clarity.
scanSum' :: Int -> [Int] -> [Int]
scanSum' tot [] = []
scanSum' tot (x:xs) = tot' : scanSum' tot' xs
where
tot' = x + tot
-- Alternatively, using takeInt, dropInt and sumInt:
scanSum :: [Int] -> [Int]
scanSum [] = []
scanSum [x] = [x]
scanSum (x:xs) = x : scanSum ((x + sumInt (takeInt 1 xs)) : dropInt 1 xs)
```

5.

```
diffs :: [Int] -> [Int]
diffs [] = []
diffs (x:xs) = diffs' (x:xs) xs
where
diffs' :: [Int] -> [Int] -> [Int]
diffs' _ [] = []
diffs' [] _ = []
diffs' (x:xs) (y:ys) = (y-x) : diffs' xs ys
-- Alternatively, without the auxiliary function:
diffs :: [Int] -> [Int]
diffs [] = []
diffs [x] = []
diffs (x:y:xs) = (y-x) : diffs (y:xs)
```

## The `map`

functionEdit

1. A handful of variations for each function will be shown below, in a single block:

```
negateList, negateList2 :: [Int] -> [Int]
negateList = map negate
negateList2 xs = map negate xs
divisorsList, divisorsList2 :: [Int] -> [[Int]]
divisorsList = map divisors
divisorsList2 xs = map divisors xs
-- Note that there are even more possible ways of writing this one.
-- Remember that the dot operator composes functions: (g . f) x = g (f x)
negateDivisorsList, negateDivisorsList2, negateDivisorsList3, negateDivisorsList4 :: [Int] -> [[Int]]
negateDivisorsList = map (negateList . divisors)
negateDivisorsList2 = map negateList . divisorsList
negateDivisorsList3 list = map (negateList . divisors) list
negateDivisorsList4 list = map (map negate) (map divisors list)
```

2. One possible solution:

```
import Data.List
myRLEencoder :: String -> [(Int, Char)]
myRLEencoder s = map pairRLE (group s)
where
pairRLE xs = (length xs, head xs)
myRLEdecoder :: [(Int, Char)] -> String
myRLEdecoder l = concat (map expandRLE l)
where
expandRLE (n, x) = replicate n x
```

N.B.: the RLE example is inspired from a blog post by Don Stewart on the same subject. If you are curious, check Don's post for a neat solution which likely won't not be immediately understandable, as it uses some things we didn't see yet.

## Tips and tricksEdit

1. Both `scanSum (takeInt 10 [1..])`

and `takeInt 10 (scanSum [1..])`

have the same value. This is possible because Haskell is a lazy language, thus in both cases the result is only evaluated as needed.

2.

```
-- This is just like the Prelude function last.
lastOf :: [a] -> a
lastOf [] = error "Empty list"
lastOf [x] = x
lastOf (_:xs) = lastOf xs
-- This is just like the Prelude init.
dropLast :: [a] -> [a]
dropLast [] = error "Empty list"
dropLast [x] = []
dropLast (x:xs) = x : (dropLast xs)
```