Linked List

1. Singly Linked List

2. A node is a data structure that contains a
data item and one or more links. A link is
a reference to a node
An arrow of each node indicates that the reference of the next node. The
reference of the first node is stored at head. The null in the link part
of the last node indicates that no node follows the last node.
Each node in a linked list can be implemented as a class with two instance
variables data and link, placed inside a linked list class

3. class Node {
public:
int data;
Node* next;
// Default constructor
Node()
{
data = 0;
next = NULL;
}
// Parameterised Constructor
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
class Linkedlist {
Node* head;
};

4. Operations:
Traversal
nodeRef walks down the list, referencing each of
the list nodes in turn. The value of nodeRef is
null when the traversal is finished.
current = head
while current is not null:
print current.data
current = current.next

5. Line 1: Node newNode = new Node(5, null);
Line 2: newNode.next = head;
Line 3: head = newNode;
Line 1 creates a new node with data 5 and link null.
Line 2 makes the link part of the new node refers to the node
that has been pointed by head.
With line 3, now head refers to the new node.
Operations: Insertion at
Front

6. Operations: Insertion
after
First indicate where the new node will be
added. Let’s say we want to add a new node
after the node referred by p.
newNode.next = p.next;
p.next = newNode;

7. Operations: Insertion last
if (!head) {
head =
newNode;
return;
}
Node* current
= head;
while
(current->next) {
current =
current->next;
}
current->next
= newNode;
}

8. deleteFirst():
if head is NU
print "Li
else:
temp = he
head = he
delete te
Delete from beginning

9. Delete at middle
• Traverse to element before the element to
be deleted
• Change next pointers to exclude the node
from the chain
deleteMiddle(position):
if head is NULL:
print "List is empty"
else if position == 1:
deleteFirst()
else:
current = head
prev = NULL
for i from 1 to position -
prev = current
current = current.next
if current is NULL:
print "Position ou
bounds"
return
prev.next = current.next
delete current

10. Delete at End
Traverse to second last element
Change its next pointer to null
deleteLast():
if head is NULL:
print "List is
empty"
else if head.next is
NULL:
delete head
head = NULL
else:
current = head
while
current.next.next is not NULL
current =
current.next
delete
current.next
current.next =
NULL