sort std::vector without losing index info

Question

I want to sort a std::vector using stored values without losing the index info. For example,

std::vector <int> vec;
vec.resize(3);
vec[0] = 20;
vec[1] = 10;
vec[2] = 6;
std::sort(vec.begin(), vec.end());
// Here I want to know the order of indices after sort operation which is 2, 1, 0

Answer 1

You want to save the permutation of your original vector, so you need another vector which builds the correct bijection from {0, ... , n - 1} to {0, ... , n - 1}:

vector<unsigned int> permutation( vec.size() );
for(unsigned int i = 0; i < vec.size(); ++i)
    permutation[i] = i;

We didn't permutate anything yet. Now you don't sort the second vector, instead you sort the permutation:

std::sort(permutation.begin(), permutation.end(), cmp);

If you use C++11, cmp can be a lambda:

[&vec](unsigned int a, unsigned int b) { return vec[a] < vec[b];}

If you use C++03 you'll need to use struct with bool operator()(unsigned int, unsigned int):

struct comparator{
   comparator(vector& v) : lookup(v){}
   bool operator()(unsigned int a, unsigned int b){
       return lookup[a] < lookup[b];
   }
   vector& lookup;
};

comparator cmp(vec);

The sorted vector can then be traversed with vec[permutation[i]].