Question

When implementintg a linked list, and inserting an element into the list, if index is 0, then

a) remove the middle element from the list.
b) create a new node to store the new element and locate where to insert it.
c) invoke addLast(e) to insert the element at the end of the list.
d) invoke addFirst(e) to insert the element at the beginning of the list

Answer 1

To insert an element at index 0 in a linked list, the correct operation is to invoke addFirst(e), which creates a new node and adds it to the beginning of the linked list.

Step-by-step explanation:

When implementing a linked list and inserting an element at the index 0, the correct action to take is to create a new node to store the new element and locate where to insert it. This option corresponds to:

1. Invoke addFirst(e) to insert the element at the beginning of the list.

This is because the index 0 corresponds to the first position in the list. By invoking addFirst(e), a new node is created and added at the start of the linked list, effectively placing the new element at the initial position. This operation will shift the former first node to become the second in the list.