Set union algorithm using vector in C++

Question

I'm only using std::vector in this problem, and I can guarantee no duplicates in each vector (but there isn't any order in each vector). How do I union the vectors I have?

Example:

If I have following vectors...

1
1
3 2
5
5 4
2
4
4 2

After the union I should have only two vectors left:

1
2 3 4 5

Again I'm only using vector, std::set isn't allowed.

Answer 1

You can use std::set_union algorithm.

int first[] = {5,10,15,20,25};
  int second[] = {50,40,30,20,10};
  std::vector<int> v(10);                      // 0  0  0  0  0  0  0  0  0  0
  std::vector<int>::iterator it;

  std::sort (first,first+5);     //  5 10 15 20 25
  std::sort (second,second+5);   // 10 20 30 40 50

  it=std::set_union (first, first+5, second, second+5, v.begin());
                                               // 5 10 15 20 25 30 40 50  0  0
  v.resize(it-v.begin());                      // 5 10 15 20 25 30 40 50

Refer :http://www.cplusplus.com/reference/algorithm/set_union/

Answer 2

Here is my code:

template<class T> bool vectorExist (vector<T> c, T item)
{
    return (std::find(c.begin(), c.end(), item) != c.end());
}

template<class T> vector<T> vectorUnion (vector<T> a, vector<T> b)
{
    vector<T> c;

    std::sort(a.begin(), a.end());
    std::sort(b.begin(), b.end());

    auto i = a.begin();
    auto j = b.begin();

    while (i != a.end() || j != b.end())
    {
        if (j == b.end() || *i < *j)
        {
            if(!exist(c,*i)) c.insert(*i);
            i++;
        }
        else
        {
            if(!exist(c,*j)) c.insert(*j)
            j++;
        }
    }

    return c;
}

Answer 3

Sort the vectors, then merge them like in mergesort, but don't insert duplicates.

vector<int> a, b, c;
sort( a.begin(), a.end());
sort( b.begin(), b.end());
int i = 0, j = 0;
while( i < a.size() && j < b.size())
if( a[ i ] == b[ j ] )
{
   c.push_back( a[ i ] );
   ++i, ++j;
}
else if( a[ i ] < b[ j ] )
   c.push_back( a[ i++ ] );
else 
   c.push_back( b[ j++ ] );

while( i < a.size()) c.push_back( a[ i++ ] );
while( j < b.size()) c.push_back( b[ j++ ] );