两个单链表的无损递归相交

Question

I want to take two singly linked lists (this function is called from within one) and create a third singly linked list that contains all intersections between the two. So if p=[0,1,2,3] and q=[1,3,7,9] then out=[1,3], while leaving the old lists intact.

As you can see I need to declare "out" in two places. But if I hit the declaration by calling the function again, it naturally wipes what I previously wrote to it. I really can't figure out how to avoid it.

Singly linked lists can be generated with http://docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html . first is my header.

public List intersection(List l) {
    if(first.data == l.first.data) {
        List lTail = new List(l.first.next);
        List tail = new List(first.next);

        List out = new List(new Node(first.data, null)); //Bad idea #1
        // System.out.println(out);

        return tail.intersection(lTail);
    } else if (first.data > l.first.data && l.first.next != null) {
        List lTail = new List(l.first.next);
        return intersection(lTail);

    } else if (first.data < l.first.data && first.next != null) {
        List tail = new List(first.next);
        return tail.intersection(l);
    } else { //When both lists are at the end position
        List out = new List(new Node(0, null)); // Bad idea #2
        return out;
    }
}

Answer 1

List<T> p = new LinkedList<T>();
p.add...
...
List<T> q = new LinkedList<T>();
q.add...
...
List<T> intersection = new LinkedList<T>(p);
intersection.retainAll(q);

Now intersection contains only elements, which are in both lists, while lists theirselves remain untouched.