如何找到两个std:set在C ++中? [英] how to find the intersection of two std:set in C++?
问题描述
我一直在尝试找到两个std :: set在C ++中的交集,但我一直收到一个错误。
I have been trying to find the intersection between two std::set in C++, but I keep getting an error.
我创建了一个小样本测试
I created a small sample test for this
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
int main() {
set<int> s1;
set<int> s2;
s1.insert(1);
s1.insert(2);
s1.insert(3);
s1.insert(4);
s2.insert(1);
s2.insert(6);
s2.insert(3);
s2.insert(0);
set_intersection(s1.begin(),s1.end(),s2.begin(),s2.end());
return 0;
}
后一个程序不会生成任何输出,使用以下值设置(我们称为s3):
The latter program does not generate any output, but I expect to have a new set (let's call it s3) with the following values:
s3 = [1,3]
而我得到的错误:
test.cpp: (std :: _ Rb_tree_const_iterator< int>,std :: _ Rb_tree_const_iterator< int>,std :: _ Rb_tree_const_iterator< int>,std()':
test.cpp:19:error:没有匹配的函数:: _ Rb_tree_const_iterator< int>)'
我理解这个错误,是 set_intersection
,它接受 Rb_tree_const_iterator< int>
作为参数。
What I understand out of this error, is that there's no definition in set_intersection
that accepts Rb_tree_const_iterator<int>
as parameter.
此外,我假设std :: set.begin()方法返回此类型的对象,
Furthermore, I suppose the std::set.begin() method returns an object of such type,
有没有更好的方法来找到两个std :: set在C ++的交集?最好是内置函数?
is there a better way to find the intersection of two std::set in C++? Preferably a built-in function?
非常感谢!
推荐答案
template <class InputIterator1, class InputIterator2, class OutputIterator>
OutputIterator set_intersection ( InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2,
OutputIterator result );
通过执行以下操作来修复此问题:
Fix this by doing something like
...;
set<int> intersect;
set_intersection(s1.begin(),s1.end(),s2.begin(),s2.end(),
std::inserter(intersect,intersect.begin()));
您需要一个 std :: insert
迭代器因为该集现在是空的。我们不能使用back_或front_inserter,因为set不支持这些操作。
You need a std::insert
iterator since the set is as of now empty. We cannot use back_ or front_inserter since set doesnt support those operations.
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