Constructs a sorted range beginning consisting of elements that are found in both sorted ranges [first1, last1) and [first2, last2).
If some element is found m times in [first1, last1) and n times in [first2, last2), the first x number of elements will be copied from the first range to the destination range where x is the minimum of m and n.
The order of equivalent elements is preserved.
The resulting range cannot overlap with either of the input ranges.
set_intersection(first1,last1,first2,last2,result)
where, first1, last1: Input iterators to the initial and final positions of the first sorted sequence.The range used is [first1,last1), which contains all the elements between first1 and last1, including the element pointed by first1 but not the element pointed by last1.
first2, last2: Input iterators to the initial and final positions of the second sorted sequence.The range used is [first2,last2), which contains all the elements between first2 and last2, including the element pointed by first2 but not the element pointed by last2.
result: Output iterator to the initial position of the range where the resulting sequence is stored.
Return value: Returns an iterator past the end of the constructed range.
#include <iostream>
#include <vector>
#include <algorithm> //set_intersection, sort
#include <iterator> //back_inserter
using namespace std;
int main()
{
vector<int> v1{1,2,3,4,5,6,7,8};
vector<int> v2{5,7,9,10};
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
vector<int> v_intersection;
set_intersection(v1.begin(), v1.end(),v2.begin(), v2.end(),back_inserter(v_intersection));
for(int n : v_intersection) cout << n << ' ';
}
Output:
5 7