从两个向量中删除公共实体？ [英] Remove the common entities from two vector?

问题描述

说我有向量< class1a>，向量< class1b> 如何从两者中删除公共实体
我已经定义==运算符class1对象class1a，class1b

say I have vector<class1a>,vector<class1b> how to remove the common entities from both of them I have defined ==operator for the class1 objects class1a,class1b

推荐答案

stl算法提供了几个函数来执行集合操作，特别是计算

The stl algorithms provide several functions to perform set operations, notably calculating the set symmetric difference, which is what you need.

这里是一个使用示例：

#include <algorithm>
#include <vector>

int main(int argc, char **argv) {

    std::vector<int> v1;
    v1.push_back(1);
    v1.push_back(2);
    v1.push_back(3);
    v1.push_back(4);
    v1.push_back(5);
    v1.push_back(6);

    std::vector<int> v2;
    v2.push_back(2);
    v2.push_back(4);
    v2.push_back(6);
    v2.push_back(8);

    // Ranges must be sorted!
    std::sort(v1.begin(), v1.end());
    std::sort(v2.begin(), v2.end());

    std::vector<int> res; // Will contain the symmetric difference
    std::set_symmetric_difference(v1.begin(), v1.end(), 
                                  v2.begin(), v2.end(), 
                                  std::back_inserter(res));

    // Copy result to the output
    std::copy(res.begin(), res.end(), std::ostream_iterator<int>(cout, " "));
    // Prints "1 3 5"

    return 0;
}

std :: set_symmetric_difference 采用两个范围（即两对OutputIterator）和一个InputIterator，它将把结果。它还返回一个迭代器到结果范围的结尾。

std::set_symmetric_difference takes two range (i.e. two pairs of OutputIterators) and an InputIterator where it will put the result. It also returns an iterator to the end of the result range.

---

EDIT

EDIT

我刚读到您对您问题的意见。如果你想修改两个原始向量，可以使用 std :: set_difference ：

I just read your comments on your question. If you want the two original vectors to be modified, you can use std::set_difference:

vector<int>::iterator endRange;
endRange = set_difference(v1.begin(), v1.end(), 
                          v2.begin(), v2.end(), 
                          v1.begin());
v1.erase(endRange, v1.end());

这里，我们将设置差异v1 - v2的结果放入v1。然而，我们不能做反之亦然，因为v1现在被修改。解决方案是计算v1和v2的交集，然后与这个交集的差：

Here, we put the result of the set difference v1 - v2 into v1. However, we can't do the vice-versa since v1 is now modified. The solution is to calculate the intersection of v1 and v2, and then the difference with this intersection:

vector<int> inter;
set_intersection(v1.begin(), v1.end(),
                 v2.begin(), v2.end(),
                 back_inserter(inter));
// inter is "2 4 6"

v1.erase(set_difference(v1.begin(), v1.end(),
                        inter.begin(), inter.end(),
                        v1.begin())
         v1.end());
// v1 is "1 3 5"

v2.erase(set_difference(v2.begin(), v2.end(),
                        inter.begin(), inter.end(),
                        v2.begin())
         v2.end());
// v2 is "8"

我想有更多的高性能解决方案，这一个是清楚的，真正地通过使用广为人知的stl算法来传达你的意图。

I guess there are much more performant solutions, but this one is clear, and really convey your intents by using widely known stl algorithms.

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