Is there a sorted container in the STL?
C++ do have sorted container e.g std::set and std::map
int main()
{
//ordered set
set<int> s;
s.insert(5);
s.insert(1);
s.insert(6);
s.insert(3);
s.insert(7);
s.insert(2);
cout << "Elements of set in sorted order: ";
for (auto it : s)
cout << it << " ";
return 0;
}
Output: Elements of set in sorted order: 1 2 3 5 6 7
int main()
{
// Ordered map
std::map<int, int> order;
// Mapping values to keys
order[5] = 10;
order[3] = 5;
order[20] = 100;
order[1] = 1;
// Iterating the map and printing ordered values
for (auto i = order.begin(); i != order.end(); i++) {
std::cout << i->first << " : " << i->second << '\n';
}
Output:
1 : 1
3 : 5
5 : 10
20 : 100
Yes, std::set
, std::multiset
, std::map
, and std::multimap
are all sorted using std::less
as the default comparison operation. The underlying data-structure used is typically a balanced binary search tree such as a red-black tree. So if you add an element to these data-structures and then iterate over the contained elements, the output will be in sorted order. The complexity of adding N elements to the data-structure will be O(N log N), or the same as sorting a vector of N elements using any common O(log N) complexity sort.
In your specific scenario, since you don't have key/value pairs, std::set
or std::multiset
is probably your best bet.
I'd like to expand on Jason's answer. I agree to Jason, that either std::set
or std::multiset
is the best choice for your specific scenario. I'd like to provide an example in order to help you to further narrow down the choice.
Let's assume that you have the following class Foo
:
class Foo {
public:
Foo(int v1, int v2) : val1(v1), val2(v2) {};
bool operator<(const Foo &foo) const { return val2 < foo.val2; }
int val1;
int val2;
};
Here, Foo
overloads the <
operator. This way, you don't need to specify an explicit comparator function. As a result, you can simply use a std::multiset
instead of a std::vector
in the following way. You just have to replace push_back()
by insert()
:
int main()
{
std::multiset<Foo> ms;
ms.insert(Foo(1, 6));
ms.insert(Foo(1, 5));
ms.insert(Foo(3, 4));
ms.insert(Foo(2, 4));
for (auto const &foo : ms)
std::cout << foo.val1 << " " << foo.val2 << std::endl;
return 0;
}
Output:
3 4
2 4
1 5
1 6
As you can see, the container is sorted by the member val2
of the class Foo
, based on the <
operator. However, if you use std::set
instead of a std::multiset
, then you will get a different output:
int main()
{
std::set<Foo> s;
s.insert(Foo(1, 6));
s.insert(Foo(1, 5));
s.insert(Foo(3, 4));
s.insert(Foo(2, 4));
for (auto const &foo : s)
std::cout << foo.val1 << " " << foo.val2 << std::endl;
return 0;
}
Output:
3 4
1 5
1 6
Here, the second Foo
object where val2
is 4 is missing, because a std::set
only allows for unique entries. Whether entries are unique is decided based on the provided <
operator. In this example, the <
operator compares the val2
members to each other. Therefore, two Foo
objects are equal, if their val2
members have the same value.
So, your choice depends on whether or not you want to store Foo
objects that may be equal based on the <
operator.
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