Question

Question 2 (3 points) Given the following singly linked list (a list with four nodes), what will be the value stored at the last node when the following operations are executed? head->1111-151->11011->1211 Node curr = head; while(curr next !=null) { curr.item = curritem-head. item; curr = cur next; اسرة Node newNode = new Node(8): curr.next = newNode: 23 O 12 OS O21 Question 3 (3 points) Given a recursive method as shown below, what is the return value for P(5) public static int P(int al 11 (n -- 1) return 2, else 12 -- 2) return 3; ela return PIA - 1) - Pin - 2); 13 11 O 10 Question 4 (3 points) Given a singly linked list with more than two nodes, to remove the second node from a linked list with only head reference, assume curr = head next, you do set head next to curr.next Set curr. next to head. next set head to CMIT. NEXT Set head. next to CHIT

Answer 1

Question 2)

The answer is option c(8).

Explanation:

The curr is assigned with the head.

The loop is used to traverse the entire list and last node 12 is reached.

When the last node 12 is reached, the curr.next becomes NULL. So the loop terminates.

Now a new node is created with value 8.

This is assigned to curr.next and made as the last node.

Hence the last node has the value 8.

Question 3)

The answer is option a(13).

Explanation:

The function uses recursion.

It is called when n=5.

The statement F(n-1)+F(n-2) is executed ie.., F(4)+F(3).

This again executes F(n-1)+F(n-2) i.e., F(3)+F(2)+F(2)+F(1)

This returns F(2)+F(1)+3+3+2
=3+2+3+3+2
=13.

Question 4)

The answer is option c.

Explanation:

The curr points to the second node now.

Set the head to curr.next. This makes the head point to the third node and eliminates the second node.