Re: Merging Linked Lists

From:
"Daniel Pitts" <googlegroupie@coloraura.com>
Newsgroups:
comp.lang.java.programmer
Date:
21 Dec 2006 12:33:00 -0800
Message-ID:
<1166733180.555915.254910@n67g2000cwd.googlegroups.com>
djthomp wrote:

On Dec 21, 11:02 am, bugbear <bugbear@trim_papermule.co.uk_trim> wrote:

Damo wrote:

Hi,

I (will) have anythng up to 6 Linked Lists of strings. I want to merge
them and remove duplicate entries at the same time. So that I end up
with one Linked List with every node containing a distinct string. I
dont really want(need) to sort the list. Does anyone know how I would
go about doing this efficiently.
Any advice would be much appreciated.1) bear in mind what other have said about not

using lists at all.

2) If you don't mind some temporary storage,

LinkedList merge(List multipleLists) {
     LinkedList newList = new LinkedList();
     Set newListSet = new HashSet();

     for(Iterator li = multipleLists.iterator(); li.hasNext();) {
         List l = (List)li.next();
         for(Iterator i = l.iterator(); i.hasNext();) {
             Object o = i.next();
             if(!newListSet.contains(o)) {
                 newListSet.add(o);
                 newList.add(o);
             }
         }
     }
     return newList;

}may serve (untested code)

I'm assuming your multiple lists are held in a list...
I think that's O(N).

It also kind of retains the order of the input lists

    BugBear


It can be simpler if you take advantage of set's addAll method:

LinkedList merge(List multipleLists) {
    Set newListSet = new HashSet();

    for(Iterator li = multipleLists.iterator(); li.hasNext();)
        newListSet.addAll((List)li.next());

    return new LinkedList(newListSet);
}


Or, using a more type safe, Java 1.5 approach:

public static <E> List<E> merge(
   Iterable<? extends Collection<? extends E>> collections) {
    final Set<E> set = new LinkedHashSet<E>();
    for (Collection<? extends E> collection: collections) {
      set.addAll(collection);
    }
    return new LinkedList<E>(set);
}

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