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[Solved] COMP2140 Lab3-Intersection of Linked Lists

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To implement intersection of sets implemented as simple linked lists with no dummy nodes. Intersection of Two Linked Lists File Lab03.java contains a nearly-complete application that creates pairs of sets of integers (implemented as simple linked lists) and then tests union and intersection methods on them. Union is already implemented, you will implement the intersection method only. If the set is {3,5,7}, then the linked list implementation looks like the following: top last This set implementation uses a simple linked with Furthermore, the list is ordered and does not allow duplicates. (Basically, class Set is a renamed class LinkedList.) Inside the Set class is a privateNode class with public instance members, item and next. Because the Node class is private inside the LinkedList class, no code outside the LinkedList class can access or know about Nodes. Because item and next are public, you can access the item and next of any node anywhere inside the LinkedList class. (This structure is not good object-oriented practice. However, it is very simple and will allow you to transfer your knowledge to non-object-oriented languages easily.) The method you must write: In the Set class, you must complete method intersection, which is passed two Sets: It returns another set containing all the elements that are in both this and otherSet, without changing this or otherSet. For example, if this is a linked list representing {1,3,5} and otherSet is a linked list representing {3,55}, then method intersection should return a linked list representing {3} (and this and otherSet should be unchanged). How does it work? A bit like merge in merge sort. Since the linked lists are ordered linked lists, you simply need to loop through this and otherSet at the same time, adding a new element to the (initially empty) result set whenever you see that this and otherSet have the same element. The intersection begins with pointers thisCurr and otherCurr at the start of the two sets: Since otherCurrs item is smaller than thisCurrs item, nothing is added to the intersection set and otherCurr is moved to the next node: Since thisCurrs item is equal to otherCurrs item, the item is added to the intersection set and both thisCurr and otherCurr are moved to the next node: Since thisCurrs item is less than otherCurrs item, nothing is added to the intersection set and thisCurr is moved to the next node: 2 Since thisCurrs item is equal to otherCurrs item, the item is added to the intersection set and both thisCurr and otherSet are moved to the next node: Since thisCurr is now null, there cannot be any more items that are common to both sets, so the method returns the intersection set (which now contains 3 and 7). So you need a loop that goes until one of thisCurr or otherCurr is null, moving the pointers as appropriate and only adding an item to the intersection set when thisCurrs and otherCurrs items are equal. (You can use private helper method addLast to add items to the intersection set, since the items you are adding are added in increasing order.) Reminder: Your method should have only ONE return statement. You should not have any break or continue statements.