Question

5. Given the following definition of class Node:
class Node { public int item; public Node next; public Node ( int i, Node n ) { item = i; next = n; }; }
and the following declarations:
Node list, p, q;
Assume the structure starts as below. Draw the structure created by executing the following statements. Each part is independent.

list 1 2 3 4 5  


[2] a) p = list; q = null; while ( p != null && p.item < 5 ) { p.item = p.item + 1; q = p.next; p = q.next; };







[2] b) p = list; while ( p != null ) { del(p); p = p.next; }; : private void del ( Node n ) { n = null; };








[6] c) Assume the declaration of the class Node as above. Write a method concat that concatenates (joins) list2 to the end of list1. That is, in the result the first node of list2 should follow the last node of list1.The method returns resulting list. No new nodes should be created. If list1 is empty the function should return list2 as the result.
private Node concat ( Node list1, Node list2 ) {

Answer 1

[2] a) ans.
list 1 2 3 4 5
p =list;
q=null;
while(p!= null && p.item<5){
p.item = p.item+1; 2 2 4 4 5
q = p.next; q=4
p = q.next; p=5
}
After execution of above code the updated list will be 2 2 4 4 5.
This is because after incrementing the first item (1) by 1, p will move to node 3 by skipping node 2 as p move two nodes (p.next.next) at a time.
similarly, it will modify 3 to 4 and then again moved to node 5 after skipping node 4.
Now since p.item = 5 and hence it will come out from the while loop as condition (p.item<5) will be false.
So final list will be 2 2 4 4 5.

[2] b) ans.
p = list;
while(p!=null){
del(p);
p=p.next;
}
private void del(Node n){
n=null;
}

After execution of the above code program will get crash as in the above code we are trying to access the address of next element after deleting the current node.
Once we will delete the node p, p will be assigned to null and hence we can not access the next node through p.

[6] c) ans.
private Node concat(Node list1,Node list2){
Node p,q;
p = list1;   // p pointing to start node of list1
q = list2;   // q pointing to start node of list2
/*If list1 is empty return list2*/
if(list1==null)
return list2;
/*If list2 is empty return list1*/
if(list2 == null)
return list1;
/*Both lists are not empty*/
while(p != null){  
p = p.next   //Moved p to end of list1
}
/*Run through the list2 and add each element of list2 to last of list1*/
while(q != null){
p.item = q.item;
p = p.next;
q = q.next;
}
return p;   //Return list1 after concating all element of list2.
}