/**
* Given a master list and the new sub list, replace the items in the master list with the
* matching items from the new sub list. This process works even if the length of the new sublist
* is different.
*
* <p>For example, givn:
*
* <pre>
* replace A by A':
* M=[A,B,C], S=[A'] => [A',B,C]
* M=[A,B,A,B,C], S=[A',A'] => [A',B,A',B,C]
*
* when list length is different:
* M=[A,A,B,C], S=[] => [B,C]
* M=[A,B,C], S=[A',A'] => [A',A',B,C]
* M=[B,C], S=[A',A'] => [B,C,A',A']
* </pre>
*/
private static List<Child> stitchList(
List<Child> list, String name, List<? extends Child> newSubList) {
List<Child> removed = new LinkedList<Child>();
// to preserve order, try to put new itesm where old items are found.
// if the new list is longer than the current list, we put all the extra
// after the last item in the sequence. That is,
// given [A,A,B,C] and [A',A',A'], we'll update the list to [A',A',A',B,C]
// The 'last' variable remembers the insertion position.
int last = list.size();
ListIterator<Child> itr = list.listIterator();
ListIterator<? extends Child> jtr = newSubList.listIterator();
while (itr.hasNext()) {
Child child = itr.next();
if (child.name.equals(name)) {
if (jtr.hasNext()) {
itr.set(jtr.next()); // replace
last = itr.nextIndex();
removed.add(child);
} else {
itr.remove(); // remove
removed.add(child);
}
}
}
// new list is longer than the current one
if (jtr.hasNext()) list.addAll(last, newSubList.subList(jtr.nextIndex(), newSubList.size()));
return removed;
}
