对向量进行排序而不改变原始向量的最佳方法是什么？

Question

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, eg:

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 ?)

Answer 1

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}.

Answer 2

Here is my favorite. Sort an index and not the original array/vector itself.

#include <algorithm>

int main() {

    int intarray[4] = { 2, 7, 3, 4 };//Array of values
    //or you can have vector of values as below
    //std::vector<int> intvec = { 2, 7, 3, 4 };//Vector of values
    int indexofarray[4] = { 0, 1, 2, 3 };//Array indices

    std::sort(indexofarray, indexofarray + 4, [intarray](int index_left, int index_right) { return intarray[index_left] < intarray[index_right]; });//Ascending order.
    //have intvec in place of intarray for vector.


}

After this, indexofarray[] elements would be 0, 2, 3, 1 , while intarray[] is unchanged.

Answer 3

As suggested in the comments pass the function argument by value std::vector<int> original :

#include <iostream>
#include <vector>
#include <algorithm>

std::vector<int> not_in_place_sort(std::vector<int> original) {
    std::sort(original.begin(), original.end());
    return original;
}

int main() {
    std::vector<int> v = { 8, 6, 7, 2, 3, 4, 1, 5, 9 };
    std::vector<int> v2 = not_in_place_sort(v); // pass the vector by value
    std::cout << "v1: " << '\n';
    for (auto el : v) {
        std::cout << el << ' ';
    }
    std::cout << "\nv2: " << '\n';
    for (auto el : v2) {
        std::cout << el << ' ';
    }
}

That will sort a copy of your original vector leaving the original intact. As pointed out below this might restrict some optimizations such as RVO but will call vector's move constructor in the return statement instead.

Answer 4

For the case where you are interested in proxy sorting (sorting an index list), you may want to implement a more flexible algorithm that allows you to deal with containers which do not support random access (such as std::list ). For example:

#include <algorithm>
#include <iostream>
#include <list>
#include <numeric>
#include <vector>

template <typename Container>
auto sorted_indices(const Container& c) {
  std::vector<typename Container::size_type> indices(c.size());
  std::iota(indices.begin(), indices.end(), 0);
  std::sort(indices.begin(), indices.end(), [&c](auto lhs, auto rhs) {
    return (*(std::next(c.begin(), lhs)) < *(std::next(c.begin(), rhs)));
  });
  return indices;
}

template <typename Container, typename Indices>
auto display_sorted(const Container& c, const Indices& indices) {
  std::cout << "sorted: ";
  for (auto&& index : indices) {
    std::cout << *(std::next(c.begin(), index)) << " ";
  }
  std::cout << std::endl;
}

template <typename Container>
auto display_sorted(const Container& c) {
  return display_sorted(c, sorted_indices(c));
}

template <typename Container>
auto display(const Container& c) {
  std::cout << "as provided: ";
  for (auto&& ci : c) std::cout << ci << " ";
  std::cout << std::endl;
}

int main() {
  // random access
  const std::vector<int> a{9, 5, 2, 3, 1, 6, 4};
  display(a);
  display_sorted(a);
  display(a);

  std::cout << "---\n";

  // no random access
  const std::list<int> b{9, 5, 2, 3, 1, 6, 4};
  display(b);
  display_sorted(b);
  display(b);
}

Sample run:

$ clang++ example.cpp -std=c++17 -Wall -Wextra
$ ./a.out
as provided: 9 5 2 3 1 6 4 
sorted: 1 2 3 4 5 6 9 
as provided: 9 5 2 3 1 6 4 
---
as provided: 9 5 2 3 1 6 4 
sorted: 1 2 3 4 5 6 9 
as provided: 9 5 2 3 1 6 4 

As you would expect, relying on proxy sorting could have important performance implications. For example: every time you want to traverse in order, you will possibly incur cache misses. In addition, the traversal will have the same complexity as the underlying container for random access: In the case of std::vector , std::next(v.begin(), n) is O(1) , but in the case of std::list , std::next(l.begin(), n) is O(n) .

Answer 5

For int's it doesn't make much difference if you're sorting an index or making a copy & sorting the copy; the data still needs to be initialized, and in the case of the indexes, this will involve a loop assigning values rather than faster memcpy routines; so may end up slower; in addition you're going to be jumping around memory lots more; so now the cache can't do its job nicely.

For larger objects I'd not sort the index, but use a vector of pointers. The copy of the pointers is cheap compared to copying the objects themselves; the containers are still obvious because they're containing pointers of your object; and the sort isn't attempting to reference another vector.

Answer 6

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 )