向量相交

2. std::set_intersection() 三倍(等于您拥有的向量数量-1)。

时间复杂度分析:

• 4*O(nlogn)=O(nlogn)
• 线性2*(firstSecondInterSize+thirdSize)-1
• 线性2*(第一个第二个第三个中间尺寸+第四个尺寸)-1

完整代码示例:

#include <iostream>     // std::cout
#include <algorithm>    // std::set_intersection, std::sort
#include <vector>       // std::vector

int main () {
  std::vector<int> first = {5,10,15,20,25};
  std::vector<int> second = {50,40,30,20,10};
  std::vector<int> third = {10,20,3,4,0};
  std::vector<int> fourth = {4,20,10,3,6};
  std::vector<int> v(first.size() + second.size() + third.size() + fourth.size());
  std::vector<int>::iterator it;

  std::sort (first.begin(),first.end());
  std::sort (second.begin(),second.end());
  std::sort (third.begin(),third.end());
  std::sort (fourth.begin(),fourth.end());

  it=std::set_intersection (first.begin(), first.end(), second.begin(), second.end(), v.begin());
  it=std::set_intersection (v.begin(), v.end(), third.begin(), third.end(), v.begin());
  it=std::set_intersection (v.begin(), v.end(), fourth.begin(), fourth.end(), v.begin());
  v.resize(it-v.begin());

  std::cout << "The intersection has " << (v.size()) << " elements: ";
  for (it=v.begin(); it!=v.end(); ++it)
    std::cout << ' ' << *it;
  std::cout << '\n';

  return 0;
}

交叉点有2个元素:10 20

十字路口 ,不是 超集

如果可以对向量进行排序,直接的方法是使用 std::set_intersection .

你必须使用它三次,才能得到中间结果 A=交叉点(v1,v2) 和 B=交点(3,4) 在获得最终结果之前 .

使用此 linked answer

这个怎么样?您必须添加运算符<()在您的类中(和辅助运算符<())。从每个向量集合,然后计算这些集合的交集。再次-从结果集生成向量。

#include <iostream>
#include <vector>
#include <algorithm>
#include <iostream>
#include <set>

class CMyClass
 {
 public:
     CMyClass(int n){_n=n;}
     ~CMyClass(){}
     bool operator<(const CMyClass a) const{return _n <a._n;}
     friend std::ostream& operator<<(std::ostream& s,const CMyClass& a){s<<a._n;return s;}
 private:
     int _n;
 };
vector<CMyClass> find_intersect(vector<vector<CMyClass>>& vecs)
{
    vector<CMyClass> res;
    if (vecs.empty()) return res;
    set<CMyClass> S;
    for_each(vecs[0].begin(),vecs[0].end(),[&S](const CMyClass& e){S.insert(e);});
    for(auto &vec:vecs)
    {
        set<CMyClass> s,tmp;
        for_each(vec.begin(),vec.end(),[&s](const CMyClass& e){s.insert(e);});
        set_intersection(S.begin(),S.end(),s.begin(),s.end(),std::inserter(tmp,tmp.begin()));
        S=tmp;
        if (S.empty()) break;
    }
    for_each(S.begin(),S.end(),[&res](const CMyClass& e){res.push_back(e);});
    return res;
}

int main()
{
    vector<CMyClass> v1={1,2,3};
    vector<CMyClass> v2={1,2};
    vector<CMyClass> v3={1,4};
    vector<CMyClass> v4={1,5};
    vector<vector<CMyClass>> vectors;
    vectors.push_back(v1);
    vectors.push_back(v2);
    vectors.push_back(v3);
    vectors.push_back(v4);
    auto res = find_intersect(vectors);
    cout<<"[";
    for(auto &elem:res)
        cout<<elem<<",";
    cout<<"]"<<endl;
}

输出: [1,]