The document discusses different types of linked lists including singly linked lists, doubly linked lists, and circular linked lists. It provides examples of how to implement these linked lists in C by defining node structures containing data fields and pointer fields. The document also demonstrates how to perform operations on linked lists such as inserting nodes, deleting nodes, and finding the middle node.

1. L I N K E D L IST
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Dr. S. Lovelyn Rose
PSG College of Technology
Coimbatore, India

2. SINGLY LINKED LIST
DATA ADDRESS
Data 1 Data 2 . . . . . . Data n ADDRESS
DOUBLY LINKED LIST
ID Name Desig Address
Start
of Node2
ADDRESS Data 1 Data 2 . . . . . Data n ADDRESS
Start NULL Data 1 Address of
ID Name Desig Address of
Node 2
Node 3
Previous node Data part Next node
Address of Data 2 Address of
NodeID1 Name Desig NULL 3
Node
Address of Data 3
Node 2 NULL

3.  CIRCULAR LINKED LIST
SINGLY
Start
Data 1 Node 2 Data 2 Node 3 Data 3 Node1
Node 1 Node 2 Node 3
DOUBLY
Start
Node3 Data 1 Node 2 Node 1 Data 1 Node 3
Node 1 Node 2 Node 2 Data 1 Node1
Node 3

4. struct node
{
int x;
x c[10] next
char c[10];
struct node *next;
}*current;

5. create_node()
{
current=(struct node *)malloc(sizeof(struct node));
current->x=10; current->c=“try”; current->next=null;
return current;
}
Allocation x c[10] next
Assignment 10 try NULL

6. Create a linked list for maintaining the employee details
such as Ename,Eid,Edesig.
Steps:
1)Identify the node structure struct node
{
2)Allocate space for the node
int Eid;
3)Insert the node in the list char Ename[10],Edesig[10];
struct node *next;
} *current;
EID EName EDesig next

7. Number of nodes
Allocation
Insert_N_Nodes(N)
and
{ Assignment
Start=current=create_node();
Start,
current
011 ABI HR NULL

8. for(i=1;i<n-1;i++)
{
current->next=create_node();
current=current->next;
} }
Start,
current
011 ABI HR
i=1 012 Banu DBA NULL

9. start
Insert()
{ 10 c
c=create_node();
start=c;
20 c1
c1=create_node();
c->next=c1;
c2=create_node(); 30 c2
c1->next=c2;
}

10. Start
20 30
Insert_a_node(Start)
{ Start
current= create_node(); 10
current->next=Start; 20
Start=current;
} current 30
10
Note : Passing Start as a parameter implies the list already exists

11. Start
10 20 30 40
25
Steps:
1)Have a pointer (slow) to the node after which the new
node is to be inserted
2)Create the new node to be inserted and store the address
in ‘current’
3) Adjust the pointers of ‘slow’ and ‘current’

12.  Use two pointers namely, fast and slow
 Move the ‘slow’ pointer by one element in the list
 For every single movement of ‘slow’, move ‘fast’ by 2
elements
 When ‘fast’ points to the last element, ‘slow’ would be
pointing to the middle element
 This required moving through the list only once
 So the middle element is found in a time complexity of
O(n)

13. {
current=(struct node *)malloc(sizeof(struct node))
fast=slow=Start Start 10
do
fast slow
{
if(fast->next->next!=null) 20
fast=fast->next->next current
else break;
slow=slow->next 25 30
}while(fast->next!=null)
current->next=slow->next
slow->next=current 40
}
Note : fast and slow are below the node they point to

14. Insert_last(Start)
{ Start 10
current=Start;
while(current->next!=null)
20
{
current=current->next;
} temp
30
temp=create_node(); 25
current->next=temp;
} 40

15. delete_node(start,x)
{
temp=start;
temp1=temp;
while(temp!=null) Last node
{
if(temp->n==x) start
{ Only one node
if(temp==start && temp->next==null) 10
start=null temp

16. start
else if(temp==start)
{ First node
10 temp
start=start->next;
temp->next=null;
20
break;
}
start
else if(temp->next==null)
temp1
{
10
temp1->next=null;
temp
}
20
Last node

17. Start 50
else
{
temp1->next=temp->next; temp1 20
temp->next=null;
break;
temp 10
}
}
Any node
temp1=temp; 40
temp=temp->next;
}
}

18. Inserting N nodes
Insert_node(N) create_node()
{ {
c=(struct node*)malloc(sizeof(struct
start=c=create_node();
node))
for(i=0;i<N-1;i++) c->x;
{ c->previous=null;
c1=create_node(); c->next=null;
return c;
c->next=c1; }
c1->previous=c;
}
} 10 20
start c c1

19. delete_node(start,n)
{
start
temp=start
while(temp!=null) End of list 10
{ temp
if(temp->x==n)
{ single node
if(temp==start)
{
temp->next->previous=null;
start=temp->next
}

20. else if(temp->next==null) start
temp->previous->next=null;
else 20
Last node
{
temp->next->previous=temp->previous;
10
temp->previous->next=temp->next;
temp
}
middle node 30
}
else
temp=temp->next;
}
if(temp==null) print(“Element not found”);
}

21. http://datastructuresinterview.blogspot.in/
http://talkcoimbatore.blogspot.in/
http://simpletechnical.blogspot.in/