将链表添加到链表时，更好的重复数据删除方法？

Question

I'm trying to add some dedupe logic to a LinkedList. The duplicates come from new LinkedLists getting added to the master LinkedList.:

masterList.addAll(0, newList)

The tricky part for me is that, each new list is getting added to the beginning of the master list (as the above code shows). And the dedupe needs to keep the elements that are added later. For instance:

masterList = [3, 4]
newList = [5, 4]
masterList.addAll(0, newList)

Now masterList = [5, 4, 3, 4] . And the "4" at the end of masterList should be removed as dupe while the later-added "4" needs to remain. So the dedupe result should be masterList = [5, 4, 3] .

My current solution is to dedupe AFTER the "add" is done:

protected List<String> dedupeIds(List<String> masterList) {
    // HashSet to store seen values
    HashSet<String> set = new HashSet<>();
    for (Iterator<String> iter = masterList.iterator(); iter.hasNext();) {
      String doc_id = iter.next();
      // put the doc id in Set hs, if cannot add as key, it means dupe
      if (!set.add(doc_id)) {
        iter.remove();
      }
    }
    return masterList;
  }

The current solution works, but I wonder if there is a way to dedupe during "add"?

Answer 1

As Chrylis suggested, perhaps consider a different collection type. LinkedHashSet would be a good choice for storing unique elements (it's a Set ) while also preserving order.

final Set<Integer> a = new LinkedHashSet<>(Arrays.asList(3, 4));
System.out.println(Arrays.toString(a.toArray())); // Prints [3, 4]

final Set<Integer> b = new LinkedHashSet<>(Arrays.asList(5, 4));
System.out.println(Arrays.toString(b.toArray())); // Prints [5, 4]

b.addAll(a);
System.out.println(Arrays.toString(b.toArray())); // Prints [5, 4, 3]