If you cannot use another list, you could solve your problem by keeping a count of the number of elements you processed via the iterator and compare that to the original size of the list: all new element will be at the end of the list, so you can end your loop when you have reached the original size.
LinkedList<Schedule> queue = new LinkedList<Schedule>(schedules);
int origSize = queue.size();
int currCount = 0;
ListIterator<Schedule> iterator = queue.listIterator();
while (iterator.hasNext()) {
++currCount;
if (currCount >= origSize) {
break; // reached the end of the original collection
}
Schedule schedule = iterator.next();
if(condition)
iterator.add(new Schedule());
}
You could also use an extra list to keep track of the new elements and add that to the original list after the processing is over:
LinkedList<Schedule> queue = new LinkedList<Schedule>(schedules);
LinkedList<Schedule> addQueue = new LinkedList<Schedule>();
ListIterator<Schedule> iterator = queue.listIterator();
while (iterator.hasNext()) {
Schedule schedule = iterator.next();
if(condition)
addQueue.add(new Schedule());
}
queue.addAll(addQueue);
Also, note that iterator.add()
Inserts the specified element into the list (optional operation). The element is inserted immediately before the next element that would be returned by next, if any, and after the next element that would be returned by previous, if any. (If the list contains no elements, the new element becomes the sole element on the list.) The new element is inserted before the implicit cursor: a subsequent call to next would be unaffected, and a subsequent call to previous would return the new element. (This call increases by one the value that would be returned by a call to nextIndex or previousIndex.)
so if you have more than one elements in the list, it will not add the new ones to the end, but between the current one and the one returned by next()
. If you indeed want to place the new elements at the end of the list, use queue.add(...)
In general, it is not advisable to modify a collection while traversing it via an iterator, so I suggest you use the second approach (collect the extra elements in a separate list and add them to the original at the end)