Question

我需要一个集装箱,在那里:

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

单单是不足的,因为我不能使用<代码>[index]的内容。 <代码>td:list 既不储存独一无二的内容。

我采用了一种混合解决办法,即<条码>、<条码>和<条码>>>,但可能有一些通用的标准模板?

我不想利用动力。在每一次插入之后援引<代码>:unique是没有解决办法的。

Answer 1

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;
}

或者,你可以使用<代码>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.)

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