一种低效率的方式是

  intersect'xs ys = xs \\（xs \\ ys）
 

例如

  [1,2,2,2,3,4] \\\ [6,4,4， 2,2] == [1,2,3] 
 [1,2,2,2,3,4] \\ [1,2,3] == [2,2,4] 
 

和

  [6,4,4,2,2] \\ [1,2,2,2,3,4] == [6,4] 
 [6,4,4， 2,2] \\ [6,4] == [4,2,2] 
 

从 http://hackage.haskell.org/package/base-4.7。 0.1 / docs / Data-List.html ：

在 xs \\ ys $ c $的结果中c>，首先出现 ys 的每个元素（如果有）已从 xs 中删除​​。因此



 （xs ++ ys）\\ xs == ys。 
 

I am new to Haskell and this has to be quite simple, but I have been searching the net for an hour without finding a convenient answer.

What I want is a function that returns an 'intersection' of two lists: a list of the elements that exists in both lists, taking account of duplicates.

I thought the function intersect would be what I wanted but, as stated in the docs, if the first list contains duplicates, so will the result. E.g.:

[1,2,2,2,3,4] `intersect` [6,4,4,2,2] == [2,2,2,4]

This is not what I want, since [2,2,2,4] isn't a part of [6,4,4,2,2] because there are only two 2's in that list. My desired result is:

[1,2,2,2,3,4] `intersect` [6,4,4,2,2] == [2,2,4]

How could this be accomplished?

One inefficient way is

intersect' xs ys = xs \\ (xs \\ ys)

For example

[1,2,2,2,3,4] \\ [6,4,4,2,2]   == [1,2,3]
[1,2,2,2,3,4] \\ [1,2,3]       == [2,2,4]

And

[6,4,4,2,2]   \\ [1,2,2,2,3,4] == [6,4]
[6,4,4,2,2]   \\ [6,4]         == [4,2,2]

From http://hackage.haskell.org/package/base-4.7.0.1/docs/Data-List.html :

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.