Question

I have two STL containers that I want to merge, removing any elements that appear more than once. For example:

typedef std::list<int> container; container c1; container c2;  c1.push_back(1); c1.push_back(2); c1.push_back(3);  c2.push_back(2); c2.push_back(3); c2.push_back(4);  container c3 = unique_merge(c1, c2); // c3 now contains the following 4 elements: //   1, 2, 3, 4 

std::unique seems to be for adjacent elements only, and in my case the containers could be in any order. I could do some std::set trickery I guess:

container unique_merge(const container& c1, const container& c2) {     std::set<container::value_type> s;     BOOST_FOREACH(const container::value_type& val, c1)         s.insert(val);     BOOST_FOREACH(const container::value_type& val, c2)         s.insert(val);     return container(s.begin(), s.end()); } 

Is there a better way or have I missed something bleeding obvious?

Answer 1

For an unordered lists, your set trick is probably one of the best. It each insert should be O(log n), with N inserts required, and traversing will be O(n), giving you O(N*log n). The other option is to run std::sort on each list individually and then walk through them in parallel using std::set_union, which removes duplicates for you. This will also be O(n*log n), so if you’re worried about performance, you’ll have to profile. If you’re not, do whichever makes more sense to you.

Edit: set_union will only work if there are no duplicates in the original lists, otherwise you’ll have to go with sort, merge, unique and erase. The big O performance is still the same, with the same caveats about profiling.

template <typename container> container unique_merge(container c1, container c2) {     std::sort(c1.begin(), c1.end());     std::sort(c2.begin(), c2.end());     container mergeTarget;     std::merge(c1.begin(), c1.end(), c2.begin(), c2.end(),          std::insert_iterator(mergeTarget, mergeTarget.end())     );     std::erase(         std::unique(mergeTarget.begin(), mergeTarget.end()),          mergeTarget.end()     );      return mergeTarget; }