Comparing two lists for unique items in each

问题

I have two collections (they happen to be arrays, but it doesn't really matter, I think): L and R. They are both sorted and now I want to compare them. I want to end up with two collections: one for each input array containing the items which were not in the other.

I could just take the first item from L and then search R and, if there isn't a match, add it to my "unique" collection (Lu). But that's extremely inefficient, and I am expecting to have some very large collections to process in the near future.

I though about possibly "playing hopscotch":

• Step 1: Take two lists, L and R, and compare the head of each list ( l :: L and r :: R):

  • Branch 1: if l < r, then add l to Lu and recurse, passing in L and r :: R

  • Branch 2: if l > r, then add r to Ru and recurse, passing in l :: L and R

  • Branch 3: if l = r, then recurse, passing in L and R

• Step 2: return Lu and Ru

I can write this function, but before I put in the effort I was wondering if a function already exists which can do this for me. It seems like a not-to-uncommon scenario, and I'd always rather use an existing solution to rolling my own.

(Also, if there's a more recognizable name for this algorithm, I'd like to know what it's called.)

回答1:

(I wrote the question above about 2 hours ago. Since then, I found the answer on my own. The following is what I discovered.)

In set theory, the "list" of items in L but not in R is known as "the relative complement of R in L", also known as "set-theoretic difference of L and R"

(See Wikipedia's Complement (set theory) article)

F#, being a mathematical language, has this concept baked right in to it's Core library. First, you need to build your collections as sets:

// example arrays:
let arr1 = [| 1; 2; 3 |]
let arr2 = [| 2; 3; 4 |]

// build the L and R sets
let L = set arr1
let R = set arr2

Now you can call the "difference" function and quickly get the relative complement for each array:

let Lu = Set.difference L R |> Set.toArray
let Ru = Set.difference R L |> Set.toArray

> val Lu : int [] = [|1|]
> val Ru : int [] = [|4|]

There's also a shorter syntax. The Set type has overloaded the minus operator. Set.difference just subtracts the second parameter from the first, so you can actually just use the following:

let Lu = L - R |> Set.toArray
let Ru = R - L |> Set.toArray

> val Lu : int [] = [|1|]
> val Ru : int [] = [|4|]

来源：https://stackoverflow.com/questions/21389733/comparing-two-lists-for-unique-items-in-each