What is the best way to sort a vector leaving the original one unaltered?

user_62811 0
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As the title says, I'm looking for a way to sort a vector without modifying the original one. My first idea is of course to create a copy of the vector before the sort, e.g.:

std::vector<int> not_in_place_sort(const std::vector<int>& original)
{
   auto copy = original;
   std::sort(copy.begin(), copy.end());
   return copy;
}

However, maybe there is a more efficient way to perform the sort using C++ standard algorithm (maybe a combination of sort and transform?)

sorting stl c++
6 Answers

Use partial_sort_copy. Here is an example:

vector<int> v{9,8,6,7,4,5,2,0,3,1};
vector<int> v_sorted(v.size());
partial_sort_copy(begin(v), end(v), begin(v_sorted), end(v_sorted));

Now, v remains untouched but v_sorted contains {0,1,2,3,4,5,6,7,8,9}.

user_8279270

You can create another vector to store the indices. Here is the code:

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main()
{
    vector<int> numbers = {50,30,20,10,40};
    vector<int> indexOfNumbers;
    
    for(int i = 0; i < numbers.size(); i++)
    {
        indexOfNumbers.push_back(i); 
    }
    // Now, indexOfNumbers = [0,1,2,3,4]

    std::sort(
        indexOfNumbers.begin(), indexOfNumbers.end(), 
        [numbers](int leftIndex, int rightIndex) 
        { 
            return numbers[leftIndex] < numbers[rightIndex]; // sort in ascending order
        }
    );
    // After sorting, indexOfNumbers = [3, 2, 1, 4, 0]

    // Access the sorted elements
    cout << "Accessing the sorted elements : ";
    for(int i = 0; i < numbers.size(); i++)
    {
        cout << numbers[indexOfNumbers[i]] << " ";
    }
    // prints numbers in sorted order i.e. [10,20,30,40,50]
   return 0;
}

Source: Made slight modification according to Tyrer's answer (https://stackoverflow.com/a/47537314)

user_95245650
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