从映射中删除键不包含在另一个向量中的元素

Remove elements from map with keys not containing in another vector

本文关键字：另一个 向量 元素 映射 删除 包含      更新时间：2023-10-16

给定一个 stl 容器，比如 std：：map：

std::map<int, CustomClass> some_container;
std::vector<int> real_objects;

如何正确有效地从地图中删除some_container键不在矢量real_objects元素？地图是此类任务的最佳容器吗？

我的第一直觉是批量擦除非真实对象块：

// Assuming real_objets is sorted (otherwise sort it)
auto first = some_container.begin();
for(int i : real_objects) {
  // end of the chunk to erase is the place where i could be
  // inserted without breaking the order
  auto last = some_container.lower_bound(i);
  some_container.erase(first, last);
  // if i is a key in the container, skip that element
  if(last != some_container.end() && last->first == i) {
    ++last;
  }
  first = last;
}
// and in the end, kill stuff that comes after real_objects.back()
some_container.erase(first, some_container.end());

这具有运行时复杂度O（n * log（m）），其中n是real_objects.size()，m是some_container.size()，这意味着如果some_container远大于real_objects，它的性能最佳。否则，由于可以在线性时间内循环访问std::map，因此您可以按顺序步调遍历两者并删除差异：

// again, sort real_objects.
if(!some_container.empty()) { // else there's nothing to do
  auto ir  = real_objects.begin();
  auto ire = std::lower_bound(real_objects.begin(),
                              real_objects.end  (),
                              some_container.rbegin()->first);
  auto is  = some_container.begin();
  for(;;) {
    // find the next place where the real_objects and the keys of some_container don't match
    auto p = std::mismatch(ir, ire, is, [](int i, std::pair<int, int> const &p) { return i == p.first; });
    if(p.first  == real_objects  .end() ||
       p.second == some_container.end())
    {
      // upon reaching the end, remove all that comes after the
      // end of real_objects (if anything)
      some_container.erase(p.second, some_container.end());
      break;
    } else if(*p.first < p.second->first) {
      // if there's something in real_objects that's not in
      // some_container, skip it
      ++p.first;
    } else {
      // otherwise there's something in some_container that's
      // not in real_objects, so delete it.
      p.second = some_container.erase(p.second);
    }
    ir = p.first;
    is = p.second;
  }
}

这具有运行时复杂度 O（max（n， m）），因此如果 some_container 和 real_objects几乎匹配，它应该表现良好。

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