I'm a C programmer and trying to get better at C++. I want to implement a permutation function (without using the STL algorithms). I came up with the following algorithm (out of my C way of thinking), but
a) it crashes for k > 2 (I suppose because the element that the iterator
points to, gets deleted, is inserted back and then incremented).
b) erase/insert operation seem unnecessary.
How would the C++ experts amongst you implement it?
template <class T>
class Ordering {
public:
Ordering(int n);
int combination(int k);
int permutation(int k);
private:
set<T> elements;
vector<T> order;
}
template <class T>
int Ordering<T>::permutation (int k) {
if (k > elements.size()) {
return 0;
}
if (k == 0) {
printOrder();
return 1;
}
int count = 0;
for (typename set<T>::iterator it = elements.begin();
it != elements.end();
it++
)
{
order[k-1] = *it;
elements.erase(*it);
count += permutation(k-1);
elements.insert(*it);
}
return count;
}
The problem is in your iteration over the elements
set. You try to increment an iterator which you have removed. That cannot work.
If you insist in using this approach, you must store the successor of it
, before calling set::erase
. That means you have to move the incrementation part of your for loop into the loop.
Like this:
for (typename set<T>::iterator it = elements.begin();
it != elements.end();
/* nothing here */
)
{
order[k-1] = *it;
typename set<T>::iterator next = it;
++next;
elements.erase(*it);
count += permutation(k-1);
elements.insert(order[k-1]);
it = next;
}
Edit: One possible way of temporarily "removing" objects from your set would be to have a std::set<std::pair<T,bool>>
and simply write it->second = false
and afterwards it->second = true
. Then, while iterating, you can skip entries where the second value is false
. This adds a bit of an overhead since you have to do a lot more work while descending. But inserting+removing elements adds a logarithmic overhead every time, which is probably worse.
If you used a (custom) linked list (perhaps you can even get std::list
to do that) you could very inexpensively remove and re-insert objects.
T
is) - bitmask 2012-04-04 01:39