Список: реализовать объединение, пересечение и разность списков

-- | The 'union' function returns the list union of the two lists.
-- For example,
--
-- > "dog" `union` "cow" == "dogcw"
--
-- Duplicates, and elements of the first list, are removed from the
-- the second list, but if the first list contains duplicates, so will
-- the result.
-- It is a special case of 'unionBy', which allows the programmer to supply
-- their own equality test.
 
union                   :: (Eq a) => [a] -> [a] -> [a]
union                   = unionBy (==)
 
-- | The 'unionBy' function is the non-overloaded version of 'union'.
unionBy                 :: (a -> a -> Bool) -> [a] -> [a] -> [a]
unionBy eq xs ys        =  xs ++ foldl (flip (deleteBy eq)) (nubBy eq ys) xs

-- | The 'intersect' function takes the list intersection of two lists.
-- For example,
--
-- > [1,2,3,4] `intersect` [2,4,6,8] == [2,4]
--
-- If the first list contains duplicates, so will the result.
--
-- > [1,2,2,3,4] `intersect` [6,4,4,2] == [2,2,4]
--
-- It is a special case of 'intersectBy', which allows the programmer to
-- supply their own equality test.
 
intersect               :: (Eq a) => [a] -> [a] -> [a]
intersect               =  intersectBy (==)
 
-- | The 'intersectBy' function is the non-overloaded version of 'intersect'.
intersectBy             :: (a -> a -> Bool) -> [a] -> [a] -> [a]
intersectBy _  [] _     =  []
intersectBy _  _  []    =  []
intersectBy eq xs ys    =  [x | x <- xs, any (eq x) ys]

-- | The '\\' function is list difference ((non-associative).
-- In the result of @xs@ '\\' @ys@, the first occurrence of each element of
-- @ys@ in turn (if any) has been removed from @xs@.  Thus
--
-- > (xs ++ ys) \\ xs == ys.
--
-- It is a special case of 'deleteFirstsBy', which allows the programmer
-- to supply their own equality test.
 
(\\)                    :: (Eq a) => [a] -> [a] -> [a]
(\\)                    =  foldl (flip delete)

union  []ys = ys
union  (x|xs) ys
    |x'elem'ys=union xs ys
    |otherwise = x(union xs ys)

-- Содержится ли первый аргумент в списке, заданным вторым
 
memb :: (Eq a) => a -> [a] -> Bool
memb x [] = False
memb x (y:ys) = if (y == x) then True else memb x ys
 
-- Построить множество элементов (удалить повторяющиеся)
 
setof :: (Eq a) => [a] -> [a]
setof [] = []
setof (x:xs) = if memb x (setof xs) then setof xs else x:(setof xs)
 
-- Объединить два множества
 
unite :: (Eq a) => [a] -> [a] -> [a]
unite x y = setof (x++y)
 
-- Пересечение двух множеств
 
inter :: (Eq a) => [a] -> [a] -> [a]
inter x [] = []
inter [] y = []
inter (x:xs) y = setof (if memb x y then x:(inter xs y) else (inter xs y))
 
Main> inter [1,1,1,2,2] [2,3,3]
[2]
Main> unite "asdf" "qazwsx"
"dfqazwsx"