Question

I have 3 maps:

map<string, vector<int> > map1
map<string, vector<int> > map2
map<string, vector<int> > map3

I need to create a third map with all the strings existing in both map1 and map2 with their corresponding vectors. Thing is, even if the strings are same, their vectors can be different and i have to append all the vectors from both the common strings into 1 vector.
This is what I’m trying but I’m kind of lost:

for(map<string, vector<int> >::iterator it1=map1.begin(); it1 != map1.end(); it1++)
{
    string name = (*it1).first;
    for(map<string, vector<int> >::iterator it2=map2.begin(); it2 != map2.end(); it2++)
    {
        string name2 = (*it2).first;
        if (name == name2)
        {
            map3[name2] = (*it2).second;

        }
    }
}

Thanks so much for your help!!

Answer 1

Since the maps are ordered you can do this in linear time. This is one possible way to do the task:

#include <map>
#include <string>
#include <vector>

typedef std::vector<int> vec_t;
typedef std::map<std::string, vec_t> map_t;
typedef map_t::const_iterator iter_t;

int main()
{
    map_t map1, map2, map3;

    iter_t it1 = map1.begin();
    iter_t it2 = map2.begin();

    while (it1 != map1.end() && it2 != map2.end()) {
        if (it1->first < it2->first) {
            ++it1;
        } else if (it2->first < it1->first) {
            ++it2;
        } else { // equal keys
            vec_t temp(it1->second);
            temp.insert(temp.end(), it2->second.begin(), it2->second.end());
            map3[it1->first].swap(temp); // swap contents to new map entry
            ++it1;
            ++it2;
        }
    }
}

Note that this assumes the map key ordering is the default less<Key>.