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Sorting the map by value

Suppose I have std::map<std::string,int>

, this card stores an ID along with the amount owed. I want to know if there is a way to get the 5 highest (int) values ​​from the map. I know I can scroll the map and do normal sorting, but do you have an algorithm that can help me with this? What's the most efficient way?

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Only when storing them in a different location when inserting into a card and saving it. A map is just a map .... Getting a maximum 5 would be o (N) on the map. if you control them on insert you can do it in o (1)



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Create a card std::map<int, std::string>

and insert all data from the old card into this card. Then the 5 keys from the end of the new card have the greatest value.



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you could create a vector of iterators on the map:

#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

int main()
{
    // make some data
    using data_map_type = map<string, int>;
    data_map_type data_map = {
        { "a", 10 },
        { "b", 5 },
        { "c", 3 },
        { "d", 3 },
        { "e", 4 },
        { "f", 1 },
        { "g", 2 },
        { "h", 5 },
        { "i", 0 },
        { "j", 2 },
        { "k", 1 },
    };

    // build reverse index (initially unordered)
    std::vector<data_map_type::const_iterator> second_index;
    for(auto it = begin(data_map) ; it != end(data_map) ; ++it)
        second_index.push_back(it);

    // order the secondary index by descending debt amount
    sort(begin(second_index),
         end(second_index),
         [](const data_map_type::const_iterator& l,
            const data_map_type::const_iterator& r)
         {
             return l->second > r->second;
         });

    // emit the top 5 (or fewer)
    const auto limit = std::min(second_index.size(), size_t(5));
    cout << "top " << limit << " entries:" << endl;
    for(size_t i = 0 ; i < limit ; ++i)
    {
        cout << "key=" << second_index[i]->first << ", value=" << second_index[i]->second << endl;
    }

    return 0;
}
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