Introduction
Linked list is one of the fundamental data structures, and can be used to implement other data structures. In alinked list there are different numbers of nodes. Each node is consists of two fields. The first field holds the value or data and the second field holds the reference to the next node or null if the linked list is empty.
Figure: Linked list
Linkedlist Node { data // The value or data stored in the node next // A reference to the next node, null for last node }
The singly-linked list is the easiest of the linked list, which has one link per node.
Pointer
To create linked list in C/C++ we must have a clear understanding about pointer. Now I will explain in brief what is pointer and how it works.
A pointer is a variable that contains the address of a variable. The question is why we need pointer? Or why it is so powerful? The answer is they have been part of the C/C++ language and so we have to use it. Using pointer we can pass argument to the functions. Generally we pass them by value as a copy. So we cannot change them. But if we pass argument using pointer, we can modify them. To understand about pointers, we must know how computer store variable and its value. Now, I will show it here in a very simple way.
Let us imagine that a computer memory is a long array and every array location has a distinct memory location.
int a = 50 // initialize variable a
Figure: Variable value store inside an array
It is like a house which has an address and this house has only one room. So the full address is-
Name of the house: a
Name of the person/value who live here is: 50
House Number: 4010
If we want to change the person/value of this house, the conventional way is, type this code line
a = 100 // new initialization
But using pointer we can directly go to the memory location of 'a' and change the person/value of this house without disturbing ‘a’. This is the main point about pointer.
Now the question is how we can use pointer. Type this code line:
int *b; // declare pointer b
We transfer the memory location of
a to b.b = &a; // the unary operator & gives the address of an object
Figure: Integer pointer
b store the address of the integer variable a
Now, we can change the value of
a without accessing a.*b = 100; // change the value of 'a' using pointer ‘b’ cout<<a; // show the output of 'a'
When you order the computer to access to access
*b, it reads the content inside b, which is actually the address ofa then it will follow the address and comes to the house of a and read a`s content which is 50.
Now the question is, if it is possible to read and change the content of
b without accessing b? The answer is affirmative. We can create a pointer of pointer.int **c; //declare a pointer to a pointer c = &b; //transfer the address of ‘b’ to ‘c’
So, we can change the value of
a without disturbing variable a and pointer b.**c = 200; // change the value of ‘a’ using pointer to a pointer ‘c’ cout<<a; // show the output of a
Now the total code is:
#include<iostream> using namespace std; int main() { int a = 50; // initialize integer variable a cout<<"The value of 'a': "<<a<<endl; // show the output of a int * b; // declare an integer pointer b b = &a; // transfer the address of 'a' to pointer 'b' *b = 100; // change the value of 'a' using pointer 'b' cout<<"The value of 'a' using *b: "<<a<<endl;// show the output of a int **c; // declare an integer pointer to pointer 'c' c = &b; // transfer the address of 'b' to pointer to pointer 'c' **c = 200; // change the value of 'a' using pointer to pointer 'c' cout<<"The value of 'a' using **c: "<<a<<endl;// show the output of a return 0; }
Output:
This program will give you the inside view of the pointer.
#include<iostream> using namespace std; int main() { int a = 50; // initialize integer variable a cout<<"Value of 'a' = "<<a<<endl; // show the output of a cout<<"Memory address of 'a': "<<&a<<endl; // show the address of a cout<<endl; int * b; // declare an integer pointer b b = &a; // transfer the address of 'a' to pointer 'b' cout<<"Value of Pointer 'b': "<<*b<<endl; // show the output of *b cout<<"Content of Pointer 'b': "<<b<<endl; // show the content of *b cout<<"Memory address of Pointer 'b': "<<&b<<endl; // show the address of *b cout<<endl; int **c; // declare an integer pointer to a pointer c = &b; // transfer the address of 'b' to 'c' cout<<"Value of Pointer 'c': "<<**c<<endl; // show the output of **c cout<<"Content of Pointer 'c': "<<c<<endl; // show the content of **c cout<<"Memory address of Pointer 'c': "<<&c<<endl; // show the address of **c cout<<endl; return 0; }
Output:
We can observe that the memory address of
a and the content of pointer b is same. The content of pointer c and the memory address of b is same.Linked list operation
Now we have a clear view about pointer. So we are ready for creating linked list.
Linked list structure
typedef struct node { int data; // will store information node *next; // the reference to the next node };
First we create a structure “node”. It has two members and first is
int data which will store the information and second is node *next which will hold the address of the next node. Linked list structure is complete so now we will create linked list. We can insert data in the linked list from 'front' and at the same time from 'back’. Now we will examine how we can insert data from front in the linked list.1) Insert from front
At first initialize node type.
node *head = NULL; //empty linked list
Then we take the data input from the user and store in the
node info variable. Create a temporary node node *temp and allocate space for it.node *temp; //create a temporary node temp = (node*)malloc(sizeof(node)); //allocate space for node
Then place
info to temp->data. So the first field of the node *temp is filled. Now temp->next must become a part of the remaining linked list (although now linked list is empty but imagine that we have a 2 node linked listand head is pointed at the front) So temp->next must copy the address of the *head (Because we want insert at first) and we also want that *head will always point at front. So *head must copy the address of the node *temp.
Figure: Insert at first
temp->data = info; // store data(first field) temp->next=head; // store the address of the pointer head(second field) head = temp; // transfer the address of 'temp' to 'head'
2) Traverse
Now we want to see the information stored inside the linked list. We create node
*temp1. Transfer the address of*head to *temp1. So *temp1 is also pointed at the front of the linked list. Linked list has 3 nodes.
We can get the data from first node using
temp1->data. To get data from second node, we shift *temp1 to the second node. Now we can get the data from second node.while( temp1!=NULL ) { cout<< temp1->data<<" ";// show the data in the linked list temp1 = temp1->next; // tranfer the address of 'temp->next' to 'temp' }
Figure: Traverse
This process will run until the linked list’s next is NULL.
3) Insert from back
Insert data from back is very similar to the insert from front in the linked list. Here the extra job is to find the last node of the linked list.
node *temp1; // create a temporary node temp1=(node*)malloc(sizeof(node)); // allocate space for node temp1 = head; // transfer the address of 'head' to 'temp1' while(temp1->next!=NULL) // go to the last node temp1 = temp1->next;//tranfer the address of 'temp1->next' to 'temp1'
Now, Create a temporary node
node *temp and allocate space for it. Then place info to temp->data, so the first field of the node node *temp is filled. node *temp will be the last node of the linked list. For this reason,temp->next will be NULL. To create a connection between linked list and the new node, the last node of the existing linked list node *temp1`s second field temp1->next is pointed to node *temp.
Figure: Insert at last
node *temp; // create a temporary node temp = (node*)malloc(sizeof(node)); // allocate space for node temp->data = info; // store data(first field) temp->next = NULL; // second field will be null(last node) temp1->next = temp; // 'temp' node will be the last node
4) Insert after specified number of nodes
Insert data in the linked list after specified number of node is a little bit complicated. But the idea is simple. Suppose, we want to add a node after 2nd position. So, the new node must be in 3rd position. The first step is to go the specified number of node. Let, node *temp1 is pointed to the 2nd node now.
cout<<"ENTER THE NODE NUMBER:"; cin>>node_number; // take the node number from user node *temp1; // create a temporary node temp1 = (node*)malloc(sizeof(node)); // allocate space for node temp1 = head; for( int i = 1 ; i < node_number ; i++ ) { temp1 = temp1->next; // go to the next node if( temp1 == NULL ) { cout<<node_number<<" node is not exist"<< endl; break; } }
Now, Create a temporary node
node *temp and allocate space for it. Then place info to temp->next , so the first field of the node node *temp is filled.node *temp; // create a temporary node temp = (node*)malloc(sizeof(node)); // allocate space for node temp->data = info; // store data(first field)
To establish the connection between new node and the existing linked list, new node’s
next must pointed to the 2nd node’s next . The 2nd node’s (temp1) next must pointed to the new node(temp).temp->next = temp1->next; //transfer the address of temp1->next to temp->next temp1->next = temp; //transfer the address of temp to temp1->next
Figure: Insert after specified number of nodes
5) Delete from front
Delete a node from linked list is relatively easy. First, we create node
*temp. Transfer the address of *head to*temp. So *temp is pointed at the front of the linked list. We want to delete the first node. So transfer the address of temp->next to head so that it now pointed to the second node. Now free the space allocated for first node.node *temp; // create a temporary node temp = (node*)malloc(sizeof(node)); // allocate space for node temp = head; // transfer the address of 'head' to 'temp' head = temp->next; // transfer the address of 'temp->next' to 'head' free(temp);
Figure: Delete at first node
6) Delete from back
The last node`s
next of the linked list always pointed to NULL. So when we will delete the last node, the previous node of last nodenode * temp1 and *old_temp.// create a temporary node node *temp1; temp1 = (node*)malloc(sizeof(node)); // allocate space for node temp1 = head; //transfer the address of head to temp1 node *old_temp; // create a temporary node old_temp = (node*)malloc(sizeof(node)); // allocate space for node while(temp1->next!=NULL) // go to the last node { old_temp = temp1; // transfer the address of 'temp1' to 'old_temp' temp1 = temp1->next; // transfer the address of 'temp1->next' to 'temp1' }
Now
node *temp1 is now pointed at the last node and *old_temp is pointed at the previous node of the last node. Now rest of the work is very simple. Previous node of the last node old_temp will be NULL so it become the last node of the linked list. Free the space allocated for last lode.old_temp->next = NULL; // previous node of the last node is null free(temp1);
Figure: Delete at first last
7) Delete specified number of node
To delete a specified node in the linked list, we also require to find the specified node and previous node of the specified node. Create temporary
node * temp1, *old_temp and allocate space for it. Take the input from user to know the number of the node.node *temp1; // create a temporary node temp1 = (node*)malloc(sizeof(node)); // allocate space for node temp1 = head; // transfer the address of 'head' to 'temp1' node *old_temp; // create a temporary node old_temp = (node*)malloc(sizeof(node)); // allocate space for node old_temp = temp1; // transfer the address of 'temp1' to 'old_temp'
cout<<"ENTER THE NODE NUMBER:"; cin>>node_number; // take location
for( int i = 1 ; i < node_number ; i++ ) { old_temp = temp1; // store previous node temp1 = temp1->next; // store current node }
Now
node *temp1 is now pointed at the specified node and *old_temp is pointed at the previous node of the specified node. The previous node of the specified node must connect to the rest of the linked list so we transfer the address of temp1->next to old_temp->next. Now free the space allocated for the specified node.old_temp->next = temp1->next; // transfer the address of 'temp1->next' to 'old_temp->next' free(temp1);
8) Sort nodes
Linked list sorting is very simple. It is just like ordinary array sorting. First we create two temporary node
node *temp1, *temp2 and allocate space for it. Transfer the address of first node to temp1 and address of second node to temp2. Now check if temp1->data is greater than temp2->data. If yes then exchange the data. Similarly, we perform this checking for all the nodes.
node *temp1; // create a temporary node temp1 = (node*)malloc(sizeof(node)); // allocate space for node node *temp2; // create a temporary node temp2 = (node*)malloc(sizeof(node)); // allocate space for node int temp = 0; // store temporary data value for( temp1 = head ; temp1!=NULL ; temp1 = temp1->next ) { for( temp2 = temp1->next ; temp2!=NULL ; temp2 = temp2->next ) { if( temp1->data > temp2->data ) { temp = temp1->data; temp1->data = temp2->data; temp2->data = temp; } } }
Conclusion
From the above discussions, I hope that everybody understands what linked list is and how we can create it. Now we can easily modify linked list according to our program requirement and try to use it for some real tasks.
References
- Wikipedia, the free encyclopedia.
- Pointer in C, from P. Kanetkar.
