Reputation: 5544
I'm looking to make a recursive method iterative.
I have a list of Objects I want to iterate over, and then check their subobjects.
Recursive:
doFunction(Object)
while(iterator.hasNext())
{
//doStuff
doFunction(Object.subObjects);
}
I want to change it to something like this
doFunction(Object)
iIterator = hashSet.iterator();
while(Iterator.hasNext()
{
//doStuff
hashSet.addAll(Object.subObjects);
}
Sorry for the poor psuedo code, but basically I want to iterate over subobjects while appending new objects to the end of the list to check.
I could do this using a list, and do something like
while(list.size() > 0)
{
//doStuff
list.addAll(Object.subObjects);
}
But I would really like to not add duplicate subObjects. Of course I could just check whether list.contains(each subObject) before I added It.
But I would love to use a Set to accomplish that cleaner.
So Basically is there anyway to append to a set while Iterating over it, or is there an easier way to make a List act like a set rather than manually checking .contains()?
Any comments are appreciated.
Thanks
Upvotes: 2
Views: 7534
Reputation: 29569
I would use two data structures --- a queue (e.g. ArrayDeque
) for storing objects whose subobjects are to be visited, and a set (e.g. HashSet
) for storing all visited objects without duplication.
Set visited = new HashSet(); // all visited objects
Queue next = new ArrayDeque(); // objects whose subobjects are to be visited
// NOTE: At all times, the objects in "next" are contained in "visited"
// add the first object
visited.add(obj);
Object nextObject = obj;
while (nextObject != null)
{
// do stuff to nextObject
for (Object o : nextObject.subobjects)
{
boolean fresh = visited.add(o);
if (fresh)
{
next.add(o);
}
}
nextObject = next.poll(); // removes the next object to visit, null if empty
}
// Now, "visited" contains all the visited objects
NOTES:
ArrayDeque
is a space-efficient queue. It is implemented as a cyclic array, which means you use less space than a List
that keeps growing when you add elements.boolean fresh = visited.add(o)
" combines "boolean fresh = !visited.contains(o)
" and "if (fresh) visited.add(o)
".Upvotes: 5
Reputation: 12792
Why not create an additional set that contains the entire set of objects? You can use that for lookups.
Upvotes: 0
Reputation: 9574
Use more than one set and do it in "rounds":
/* very pseudo-code */
doFunction(Object o) {
Set processed = new HashSet();
Set toProcess = new HashSet();
Set processNext = new HashSet();
toProcess.add(o);
while (toProcess.size() > 0) {
for(it = toProcess.iterator(); it.hasNext();) {
Object o = it.next();
doStuff(o);
processNext.addAll(o.subObjects);
}
processed.addAll(toProcess);
toProcess = processNext;
toProcess.removeAll(processed);
processNext = new HashSet();
}
}
Upvotes: 0
Reputation: 5544
HashSet nextObjects = new HashSet();
HashSet currentObjects = new HashSet(firstObject.subObjects);
while(currentObjects.size() > 0)
{
Iterator iter = currentObjects.iterator();
while(iter.hasNext())
{
//doStuff
nextObjects.add(subobjects);
}
currentObjects = nextObjects;
nextObjects = new HashSet();
}
I think something like this will do what I want, I'm not concerned that the first Set contains duplicates, only that the subObjects may point to the same objects.
Upvotes: 0
Reputation: 3117
If you do not use a List, the iterator will throw an exception as soon as you read from it after modifying the set. I would recommend using a List and enforcing insertion limits, then using ListIterator as that will allow you to modify the list while iterating over it.
Upvotes: 0
Reputation: 87440
I think your problem is inherently a problem that needs to be solved via a List. If you think about it, your Set version of the solution is just converting the items into a List then operating on that.
Of course, List.contains() is a slow operation in comparison to Set.contains(), so it may be worth coming up with a hybrid if speed is a concern:
while(list.size() > 0)
{
//doStuff
for each subObject
{
if (!set.contains(subObject))
{
list.add(subObject);
set.add(subObject)
}
}
}
This solution is fast and also conceptually sound - the Set can be thought of as a list of all items seen, whereas the List is a queue of items to work on. It does take up more memory than using a List alone, though.
Upvotes: 1