Question

The add(index, e) method inserts an element into the list at the specified index. It can be implemented as follows:

public void add(int index, E e) {

     if (index == 0) addFirst(e); // Insert first

     else if (index >= size) addLast(e);// Insert last

     else { // Insert in the middle

     Node<E> current = head;

         

     for (int i = 1; i < index; i++)

          current = current.next;

     Node<E> temp = current.next;

     current.next = new Node<E>(e);

     (current.next).next = temp;

     size++;

     }

}

Using the above code snippet, your author states that “There are three cases when inserting an element into the list”.

Describe the three cases when inserting an element into the list (Need more than one sentence to describe each case)

1. ________________

2. ________________

3. ________________

Answer 1

ANSWER

While inserting a node to a list, there are 3 possible scenarios:

• Insert at the beginning
• Insert at the end
• Insert at middle (Specified position)

SCENARIO – 1: Insert at the beginning:

• If the index passed to the function is 0, then that means, the passed element is to be inserted at the beginning of the list. This can be done by,
  • Create a temp node using the element passed
    • Node <E> temp = new Node<E>(e)
  • Point the next of temp node to head
    • temp.next = head
  • make the temp node as the new head
    • temp = head
  • Increment the size of list
  • Now, the passed node is inserted at the beginning

SCENARIO – 2: Insert at the end:

• If the passed index is greater than or equal to the size of the list (Probably equal to the size of the list), the passed element should be inserted at the end
  • Create a temp node using the element passed
    • Node <E> temp = new Node<E>(e)
  • Create a current element and make it point to head
    • Node <E> current = head
  • Iterate the current element “SIZE OF THE LIST TIMES” by doing,
    • current = current.next
  • After iterating the list, size times, point the next of current to temp node
    • current.next = temp
  • Increment the size of list
  • Now, the element passed is inserted at the end of the list

SCENARIO – 3: Insert at the middle (Specified index):

• If the passed index is not 0, not equal to size of list, then the element has to be inserted somewhere in the middle of the list. To do this,
  • Create a current node and point it to head of the list
    • Node <E> current = head
  • Iterate the list “index” times starting from head
    • Current = current.next;
  • Once desired position is reached, create a temp node to hold the rest of the elements of the list from current.next ‘s position
    • Node <E> temp = current.next
  • Now point the current.next to the new node
    • current.next = new Node<E>(e)
  • Now that new element is added at current.next, add the rest of the list fter new added node by pointing current.next.next to temp (which temporarily holds the rest of elements)
    • (current.next).next = temp
  • Increment the size of list
  • Now the element passed is added to the desired position