就地C ++集交集 [英] In-place C++ set intersection

问题描述

在C ++中交叉两组的标准方法是执行以下操作：

The standard way of intersecting two sets in C++ is to do the following:

std::set<int> set_1;  // With some elements
std::set<int> set_2;  // With some other elements
std::set<int> the_intersection;  // Destination of intersect
std::set_intersection(set_1.begin(), set_1.end(), set_2.begin(), set_2.end(), std::inserter(the_intersection, the_intersection.end()));

我如何做一个就地集交集？也就是说，我想set_1有调用set_intersection的结果。显然，我只能做一个 set_1.swap（the_intersection），但这是一个很低的效率比交叉到位。

How would I go about doing an in-place set intersection? That is, I want set_1 to have the results of the call to set_intersection. Obviously, I can just do a set_1.swap(the_intersection), but this is a lot less efficient than intersecting in-place.

推荐答案

我想我有：

std::set<int>::iterator it1 = set_1.begin();
std::set<int>::iterator it2 = set_2.begin();
while ( (it1 != set_1.end()) && (it2 != set_2.end()) ) {
    if (*it1 < *it2) {
        set_1.erase(it1++);
    } else {
        if (*it2 < *it1) {
            ++it1;
        }
        ++it2;
    }
}
// Anything left in set_1 from here on did not appear in set_2,
// so we remove it.
set_1.erase(it1, set_1.end());

任何人都看到任何问题？似乎是O（n）对两个集合的大小。根据 cplusplus.com ，std :: set erase（position）被摊销常数，而擦除（第一，最后）是O（log n）。

Anyone see any problems? Seems to be O(n) on the size of the two sets. According to cplusplus.com, std::set erase(position) is amortized constant while erase(first,last) is O(log n).

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