Question

New programmer here, I am trying to understand and break down this code below for a remove method, sorted linked list. I have added comments below for what my understand is and what i do not understand. can someone shed some light on the things which are unclear?

thanks in advance.

/* 1  */ public void remove(E e) throws NotFoundException{
/* 2  */     Node<E> p; //declares node p
/* 3  */     // chunk below determines where to start traversing based on element value. should traverse from head if new element < pos value
/* 4  */     if(pos == head || pos.compareTo(e) >= 0 ){ //I do not understand 2nd equality..why?
/* 5  */         p = head; //traverse list from head
/* 6  */     }else{
/* 7  */         //traverse list from pos
/* 8  */         p = pos;
/* 9  */     }
/* 10 */     for( ;p.next!=null && p.next.compareTo(e)<0; p = p.next); //nothing to initialize?
/* 11 */     //e not found in the list
/* 12 */     if(p.next == null || p.next.compareTo(e) > 0){
/* 13 */         throw new NotFoundException();
/* 14 */     }
/* 15 */     if(p.next == pos){
/* 16 */         //if node to be deleted is pos, update pos to head
/* 17 */         pos = head;
/* 18 */     }
/* 19 */     p.next = p.next.next; //delete node
/* 20 */ }

Answer 1

4. if(pos == head || pos.compareTo(e) >= 0 ){ //I do not understand 2nd equality..why? 
5. p = head; //traverse list from head
6. }else{
7. //traverse list from pos 
8. p = pos;  
9. }

First, here is a documentation for compareTo
The 2nd equality checks if pos points to a node that comes after “e”. if it’s true then u must traverse the list from its head because e comes before pos. Else, e comes after pos so you can traverse the list from pos. This is true because the list is sorted.

10. for( ;p.next!=null && p.next.compareTo(e)<0; p = p.next); //nothing to initialize?
11. //e not found in the list
12. if(p.next == null || p.next.compareTo(e) > 0){
13. throw new NotFoundException();
14. }

Here you start scanning the list from the position that was chosen and if you get to a node that is null (the end of the list) or a node that is greater than “e” then you know that “e” is not found in the list (because the list is sorted), so you throw an exception

Line 10: you don’t have to initialize anything here because you already initialized p above.