Linked List Implementation using dummy nodes

Question

I am working on a project where i have to union and intersect two sets. I am using Linked list for each set with dummy nodes. This is how i initialize my Sets LL class

public Set() {
 top = new Node(Integer.MIN_VALUE, new Node(Integer.MAX_VALUE, null) );
} //end Set

And this is how i insert items.

public void insert(int item) {
 Node prev = top;
 Node curr = top.next;

 while( curr.item < item ) {
  prev = curr;
  curr = curr.next;
 }
 prev.next = new Node( item, curr);
 size++;
} // insert

Now i am finding it hard to get a union or intersection of two sets. This is what i have in mind for intersection.

public Set intersection( Set setB ) {
 Set setC = new Set ();
 //loop over both sets and if both have same value add it otherwise 
 // get to the next node in both sets.

My question is, am i logically correct with the intersection pseudocode? My union pseudocode is laughable though. Can anyone please guide me though this problem?

Unfortunately I am not using that. My project is based on my own implementation of sets using LL @Abhijith — Saad, Feb 20 '17 at 19:00
But i do have remove duplicate values so it is kind of unique in that sense. @Abhijith — Saad, Feb 20 '17 at 19:14
I have added `insert` and `delete` so far but i need to add `union`, `intersection` and `difference` methods as well. @Abhijith — Saad, Feb 20 '17 at 19:25
Hey why did you remove your answer? It would have helped future readers, you know? @Abhijith — Saad, Feb 20 '17 at 23:06

score 0 · Accepted Answer · answered Feb 20 '17 at 19:22

0

Your idea will fail for this simple input:

1  2
2  3

Your idea:

//loop over both sets and if both have same value add it otherwise 
// get to the next node in both sets.

First iteration - we have 1 and 2, not same value, so we get to the next in both sets
Second iteration - we have 2 and 3, not same value, so get to next
End

What you actually need is:

On mismatch, advance only the list with lower element
On match, add to result and advance both (or to remove duplicates, keep advancing both as long as the same value is repeating)

For union, the idea is very similar:

On mismatch, add the lower one and advance the list that contained the lower element
On match, add and advance both (or keep advancing both as long as the same value is repeating)

answered Feb 20 '17 at 19:22

Jiri Tousek

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Can you please explain the second point? I got two if statements and they advance the list with lower element, after that i have else statement that advances the list and then add the element. @JiriTousek – Saad Feb 20 '17 at 19:33
Which one is the "second" one? Union? – Jiri Tousek Feb 21 '17 at 07:37
Both inputs are sorted and invariant we're maintaining is "all elements left in inputs are higher than all elements at output". Obviously true at the beginning. Then you always add and advance the lower input, or add one then advance both inputs, so it holds. This way you won't ever put duplicate values into output. – Jiri Tousek Feb 22 '17 at 08:06

Linked List Implementation using dummy nodes

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