Java collections: compare items in a collection with each other and delete in a single loop

Question

Say I have a collection of some geolocations (in the format Country > Region [ > Town [ > District]] ), and I want to delete locations that overlap (for example, Europe > Germany overlaps Europe > Germany > Dresden and Europe > Germany > Hamburg ), so the last two should be deleted). I see that I need two instances of iterators to do something like this:

 final Iterator<Location> outerIterator = locations.newIterator(); while (outerIterator.hasNext()) { final Location outer = outerIterator.next(); final Iterator<Location> innerIterator = locations.newIterator(); while (innerIterator.hasNext()) { final Location inner = innerIterator.next(); if (!inner.equals(outer)) { if (inner.overlaps(outer)) outerIterator.remove(); else if (outer.overlaps(inner)) innerIterator.remove(); } } } 

But I can not get a new Iterator for the same collection. Is my algorithm incorrect or is there a way to do it right?

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The latest code, using tips provided by Carl Smotricz , is as follows:

 final Iterator<JobLocation> outerIterator = locations.iterator(); while (outerIterator.hasNext()) { final JobLocation outer = outerIterator.next(); final Iterator<JobLocation> innerIterator = locations.iterator(); while (innerIterator.hasNext()) { final JobLocation inner = innerIterator.next(); if (!inner.equals(outer) && inner.overlaps(outer)) { outerIterator.remove(); break; } } } 

Answer 1

If you delete an object from an external iterator, you need to exit the inner loop immediately. I'm not sure if this solves your problem completely, but it can help you a little further.

If there is still a problem, please display an error message and / or exception!

Answer 2

Are you sure you want to increase your outer element in the inner loop?

Answer 3

I think you should choose a different structure for your regions, for example, a tree in which each place has children that are contained in it. Thus, you can exclude all regions that overlap each other after a simple search. No nested iterations required.

This is easier if no location can intersect with two locations, which seems to be the case.