合并多个 STL 容器、删除重复元素的最佳方法? [英] Best way to merge multiple STL containers, removing duplicate elements?

问题描述

我有两个 STL 容器要合并，删除出现多次的任何元素.例如:

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 似乎仅适用于相邻元素，在我的情况下，容器可以按任何顺序排列.我想我可以做一些 std::set 诡计:

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());
}

是否有更好的方法，或者我是否遗漏了一些明显的出血?

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

推荐答案

对于无序列表，您的设置技巧可能是最好的技巧之一.每个插入应该是 O(log n)，需要 N 个插入，遍历将是 O(n)，给你 O(N*log n).另一种选择是在每个列表上单独运行 std::sort，然后使用 std::set_union，它会为您删除重复项.这也将是 O(n*log n)，因此如果您担心性能，则必须进行分析.如果不是，请选择对您更有意义的方式.

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.

set_union 仅在原始列表中没有重复项时才有效，否则您必须使用 sort、merge、unique 和 erase.大 O 性能仍然相同，但对性能分析的注意事项相同.

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;
}

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