# H-99: Ninety-Nine Haskell Problems

These are Haskell translations of Ninety Nine Lisp Problems.

## Problem 1

(*) Find the last box of a list. Example: * (my-last '(a b c d)) (D)

This is "last" in Prelude, which is defined as:

```
last :: [a] -> a
last [x] = x
last (_:xs) = last xs
```

## Problem 2

(*) Find the last but one box of a list. Example: * (my-but-last '(a b c d)) (C D)

This can be done by dropping all but the last two elements of a list:

```
myButLast :: [a] -> [a]
myButLast list = drop ((length list) - 2) list
```

## Problem 3

(*) Find the K'th element of a list. The first element in the list is number 1. Example: * (element-at '(a b c d e) 3) C

This is (almost) the infix operator !! in Prelude, which is defined as:

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

Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:

```
elementAt :: [a] -> Int -> a
elementAt list i = list !! (i-1)
```

## Problem 4

(*) Find the number of elements of a list.

This is "length" in Prelude, which is defined as:

```
length :: [a] -> Int
length [] = 0
length (_:l) = 1 + length l
```

## Problem 5

(*) Reverse a list.

This is "reverse" in Prelude, which is defined as:

```
reverse :: [a] -> [a]
reverse = foldl (flip (:)) []
```

The standard definition is concise, but not very readable. Another way to define reverse is:

```
reverse :: [a] -> [a]
reverse [] = []
reverse (x:xs) = reverse xs ++ [x]
```

## Problem 6

(*) Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).

This is trivial, because we can use reverse:

```
isPalindrome :: (Eq a) => [a] -> Bool
isPalindrome xs = xs == (reverse xs)
```

## Problem 7

(**) Flatten a nested list structure. Transform a list, possibly holding lists as elements into a `flat' list by replacing each list with its elements (recursively). Example: * (my-flatten '(a (b (c d) e))) (A B C D E)

This is tricky, because lists in Haskell are homogeneous. [1, [2, [3, 4], 5]] is a type error. We have to devise some way of represent a list that may (or may not) be nested:

```
data NestedList a = Elem a | List [NestedList a]
flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List []) = []
flatten (List (x:xs)) = flatten x ++ flatten (List xs)
```

Our NestedList datatype is either a single element of some type (Elem a), or a list of NestedLists of the same type. (List [NestedList a]). Let's try it out in ghci:

*Main> flatten (Elem 5) [5] *Main> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]]) [1,2,3,4,5] *Main> flatten (List []) []