Numbering elements in vector - c++

Question

I have a vector of integers. For example: 26 58 32 47 . I need to replace them with their number in that sequence. In this case it would be: 4 1 3 2 . I tried this code:
int n = 1;
vector <int> vietos;
for (vector <int>::iterator i = vieta.begin(); i != vieta.end(); i++) {
for (vector <int>::iterator j = vieta.begin(); j != vieta.end(); j++) {
if (*i > *j)
n++;
}
vietos.push_back(n);
cout << n << " ";
n = 1;
}
Having numbers 23 25 38 28 26 28 (Note: In this case I number them in reverse order!) I get: 1 2 6 4 3 4 which is good except for two numbers are equal.
Maybe there is some way to number elements in vector using STL algorithms?

Answer 1

In my opinion the simplest way is to use std::reference_wrapper. The code will look simple and very clear.
Here is the program that demonstrates the approach.
Enjoy!:)
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
int main()
{
std::vector<int> v = { 23, 25, 38, 28, 26, 28 };
for ( int x : v ) std::cout << x << ' ';
std::cout << std::endl;
// Introducing a local block scope that the auxiliary vector would be automatically deleted
{
std::vector<std::reference_wrapper<int>> vr( v.begin(), v.end() );
std::stable_sort( vr.begin(), vr.end() );
for ( std::vector<std::reference_wrapper<int>>::size_type i = 0;
i < vr.size();
i++ )
{
vr[i].get() = i + 1;
}
}
for ( int x : v ) std::cout << x << ' ';
std::cout << std::endl;
return 0;
}
The output is
23 25 38 28 26 28
1 2 6 4 3 5
If you need to get the reverese order all you need is to add to the code functional object
std::greater<std::reference_wrapper<int>>()
in the call of std::stable_sort
For example
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
int main()
{
std::vector<int> v = { 23, 25, 38, 28, 26, 28 };
for ( int x : v ) std::cout << x << ' ';
std::cout << std::endl;
// Introducing a local block scope that the auxiliary vector would be automatically deleted
{
std::vector<std::reference_wrapper<int>> vr( v.begin(), v.end() );
std::stable_sort( vr.begin(), vr.end(),
std::greater<std::reference_wrapper<int>>() );
for ( std::vector<std::reference_wrapper<int>>::size_type i = 0;
i < vr.size();
i++ )
{
vr[i].get() = i + 1;
}
}
for ( int x : v ) std::cout << x << ' ';
std::cout << std::endl;
return 0;
}
The output is
23 25 38 28 26 28
6 5 1 2 4 3
Is not it the best solution is it? :)
EDIT: Maybe there is no sense to use std::stable_sort with the functional object. It will be enough to use for loop setting numbers in the reverse order. Something as
for ( std::vector<std::reference_wrapper<int>>::size_type i = 0;
i < vr.size();
i++ )
{
vr[vr.size() + i - 1].get() = i + 1;
}

Answer 2

In that case, you should increment when the value is greater than or equal to rather than just greater than, try this:
(*i >= *j)

Answer 3

If you were to replace the iterators with integers, i.e.: (I assume < isn't defined on iterators, but it could be)
for (int i = 0; i < vietos.size(); i++)
You could increment n as well when elements to the left are equal, i.e. when
vietos[i] > vietos[j] || (vietos[i] == vietos[j] && j < i).
Alternatively, you could create a vector<pair<int, int> >, with each pair containing the element and its index, then sort that, and iterate through the sorted vector, setting the index in the pair in the original vector to the index in the sorted vector.
This would give an O(n log n) running time, as opposed to the O(n²) of the above.
Pseudo-code:
vector arr
vector<pair> pairs
for i = 0 to n
pairs.insert(arr[i], i)
sort pairs
for i = 0 to n
arr[pairs[i].second] = i