How to do integer partitioning

The problem statement: given a number N, for example 5, print all the partitions of this number, meaning: the sum of all possible number that add up to N. In this example:

   1:          1+1+1+1+1

   2:          1+1+1+2

   3:          1+1+2+1

   4:          1+1+3

   5:          1+2+1+1

   6:          1+2+2

   7:          1+3+1

   8:          1+4

   9:          2+1+1+1

  10:          2+1+2

  11:          2+2+1

  12:          2+3

  13:          3+1+1

  14:          3+2

  15:          4+1

  16:          5

The answer is obviously recursive, and the complexity is, of course O(2^{n - 1}), exponential.

   1:  void doPartitions(int n, vector<int>* partition) {

   2:      if (n <= 0) { //the number cannot be partitioned further, so print

   3:          for (int i = 0; i < partition->size() - 1; i++)

   4:              cout << partition->at(i) << "+";

   5:          cout << partition->at(partition->size() - 1) << endl;

   6:          return;

   7:      }

   8:      int size = partition->size();

   9:      for (int i = 1; i <= n; i++) {

  10:          //add the current no to the partitions

  11:          //and generate partitions from the leftover number

  12:          partition->push_back(i);

  13:          doPartitions(n - i, partition);

  14:          while (partition->size() > size)

  15:              partition->pop_back(); //reset the partitions each iteration

  16:      }

  17:  }

What's left here is to find a way to print only the distinct solutions, but this is a simple exercise.