List with unique elements

Question

I need a container, where:

when I add a new element that does not exist yet, it is added to the top of the list
when I add an element that already exists, it is not added and I I get its index in the list
once the element is inserted, it always has the same index and it can be accessed using this index

std::set alone is insufficient, because I cannot access the elements with [index]. std::list neither, because it does not store unique only elements.

I used a mixed solution with list and map but maybe there is some standard, generic template for that?

I don't want to use boost. Invoking list::unique after every insertion is no solution.

How about rolling your own? Have you tried? Sounds like a very thin wrapper over `list`... — user703016, Apr 20 '12 at 10:12
@Soohjun: Implementing this with `list` would not be good; the initial collision detection would be O(n), as would looking things up by index. — Oliver Charlesworth, Apr 20 '12 at 10:12
Yup, I have my own class, that maintains list together with map (for uniqueness of the items) but I thought that maybe I missed some standard template for that — Jakub M., Apr 20 '12 at 10:13
I know you don't want to use Boost, but this is effectively what Boost.bimap does! — Oliver Charlesworth, Apr 20 '12 at 10:13
**Why** is using `list::unique` no solution, if its done automatically? — Sebastian, Apr 20 '12 at 10:13
@JakubM. have you seen this similar question: [How to make elements of vector unique?](http://stackoverflow.com/questions/1453333/how-to-make-elements-of-vector-unique-remove-non-adjacent-duplicates)? — Sebastian, Apr 20 '12 at 10:15
@OliCharlesworth right, i forgot, so it's no solution at all. Thanks for the hint. — Sebastian, Apr 20 '12 at 10:17
List and set sounds best. Why use a map storing both key and value when key and value are the same? — Grimm The Opiner, Apr 20 '12 at 10:18
@user1158692: There are effectively 2 keys (the element itself, and the associated integer index that's its assigned). — Oliver Charlesworth, Apr 20 '12 at 10:19
@JakubM.: "maybe there is some standard, generic template" - there's nothing standard, but Boost.MultiIndex is close to a standard. Obviously that's not an option if you don't want to use Boost for some reason; and for simple cases like yours it's probably easier to roll your own. — Mike Seymour, Apr 20 '12 at 10:59

score 4 · Answer 1 · answered Apr 20 '12 at 10:25

If you're using just a std::list (or std::vector, for that matter), you're not going to get around a linear search if you don't want to avoid duplicated, but you want to keep the original order. A simple std::vector based solution might be:

int
createIndex( std::vector<T>& references, T const& newValue )
{
    int results = std::find( references.begin(), references.end(), newValue )
                                    - references.begin();
    if ( results == references.size() ) {
        references.push_back( newValue );
    }
    return results;
}

Alternatively, you can use std::map:

int
createIndex( std::map<T, int>& references, T const& newValue )
{
    st::map<T, int>::iterator results = references.find( newValue );
    if ( results == references.end() ) {
        results = references.insert(
                    std::make_pair( newValue, references.size() ) ).first;
    }
    return results->second;
}

(This supposes that T supports <. If not, you'll have to establish an ordering critera. Or use unordered_map and define a hash code for it.)

List with unique elements

1 Answers1