What is a pointer in a linked list?

Linked lists

Introduction

Linked lists are the best and simplest example of a dynamic data structure that uses pointers for its implementation. However, understanding pointers is crucial to understanding how linked lists work, so if you've skipped the pointers tutorial, you should go back and redo it. You must also be familiar with dynamic memory allocation and structures.

Table of Contents Show

Linked lists
Introduction
What is a linked list?
Iterating over a list
Adding an item to the end of the list
Adding an item to the beginning of the list (pushing to the list)
Removing the first item (popping from the list)
Removing the last item of the list
Removing a specific item
Linked List | Set 1 (Introduction)
What is a linked list?

Essentially, linked lists function as an array that can grow and shrink as needed, from any point in the array.

Linked lists have a few advantages over arrays:

Items can be added or removed from the middle of the list
There is no need to define an initial size

However, linked lists also have a few disadvantages:

There is no "random" access - it is impossible to reach the nth item in the array without first iterating over all items up until that item. This means we have to start from the beginning of the list and count how many times we advance in the list until we get to the desired item.
Dynamic memory allocation and pointers are required, which complicates the code and increases the risk of memory leaks and segment faults.
Linked lists have a much larger overhead over arrays, since linked list items are dynamically allocated (which is less efficient in memory usage) and each item in the list also must store an additional pointer.

What is a linked list?

A linked list is a set of dynamically allocated nodes, arranged in such a way that each node contains one value and one pointer. The pointer always points to the next member of the list. If the pointer is NULL, then it is the last node in the list.

A linked list is held using a local pointer variable which points to the first item of the list. If that pointer is also NULL, then the list is considered to be empty.

------------------------------ ------------------------------ | | | \ | | | | DATA | NEXT |--------------| DATA | NEXT | | | | / | | | ------------------------------ ------------------------------

Let's define a linked list node:

typedef struct node { int val; struct node * next; } node_t;

Notice that we are defining the struct in a recursive manner, which is possible in C. Let's name our node type node_t.

Now we can use the nodes. Let's create a local variable which points to the first item of the list (called head).

node_t * head = NULL; head = (node_t *) malloc(sizeof(node_t)); if (head == NULL) { return 1; } head->val = 1; head->next = NULL;

We've just created the first variable in the list. We must set the value, and the next item to be empty, if we want to finish populating the list. Notice that we should always check if malloc returned a NULL value or not.

To add a variable to the end of the list, we can just continue advancing to the next pointer:

node_t * head = NULL; head = (node_t *) malloc(sizeof(node_t)); head->val = 1; head->next = (node_t *) malloc(sizeof(node_t)); head->next->val = 2; head->next->next = NULL;

This can go on and on, but what we should actually do is advance to the last item of the list, until the next variable will be NULL.

Iterating over a list

Let's build a function that prints out all the items of a list. To do this, we need to use a current pointer that will keep track of the node we are currently printing. After printing the value of the node, we set the current pointer to the next node, and print again, until we've reached the end of the list (the next node is NULL).

void print_list(node_t * head) { node_t * current = head; while (current != NULL) { printf("%d\n", current->val); current = current->next; } }

Adding an item to the end of the list

To iterate over all the members of the linked list, we use a pointer called current. We set it to start from the head and then in each step, we advance the pointer to the next item in the list, until we reach the last item.

void push(node_t * head, int val) { node_t * current = head; while (current->next != NULL) { current = current->next; } /* now we can add a new variable */ current->next = (node_t *) malloc(sizeof(node_t)); current->next->val = val; current->next->next = NULL; }

The best use cases for linked lists are stacks and queues, which we will now implement:

Adding an item to the beginning of the list (pushing to the list)

To add to the beginning of the list, we will need to do the following:

Create a new item and set its value
Link the new item to point to the head of the list
Set the head of the list to be our new item

This will effectively create a new head to the list with a new value, and keep the rest of the list linked to it.

Since we use a function to do this operation, we want to be able to modify the head variable. To do this, we must pass a pointer to the pointer variable (a double pointer) so we will be able to modify the pointer itself.

void push(node_t ** head, int val) { node_t * new_node; new_node = (node_t *) malloc(sizeof(node_t)); new_node->val = val; new_node->next = *head; *head = new_node; }

Removing the first item (popping from the list)

To pop a variable, we will need to reverse this action:

Take the next item that the head points to and save it
Free the head item
Set the head to be the next item that we've stored on the side

Here is the code:

int pop(node_t ** head) { int retval = -1; node_t * next_node = NULL; if (*head == NULL) { return -1; } next_node = (*head)->next; retval = (*head)->val; free(*head); *head = next_node; return retval; }

Removing the last item of the list

Removing the last item from a list is very similar to adding it to the end of the list, but with one big exception - since we have to change one item before the last item, we actually have to look two items ahead and see if the next item is the last one in the list:

int remove_last(node_t * head) { int retval = 0; /* if there is only one item in the list, remove it */ if (head->next == NULL) { retval = head->val; free(head); return retval; } /* get to the second to last node in the list */ node_t * current = head; while (current->next->next != NULL) { current = current->next; } /* now current points to the second to last item of the list, so let's remove current->next */ retval = current->next->val; free(current->next); current->next = NULL; return retval; }

Removing a specific item

To remove a specific item from the list, either by its index from the beginning of the list or by its value, we will need to go over all the items, continuously looking ahead to find out if we've reached the node before the item we wish to remove. This is because we need to change the location to where the previous node points to as well.

Here is the algorithm:

Iterate to the node before the node we wish to delete
Save the node we wish to delete in a temporary pointer
Set the previous node's next pointer to point to the node after the node we wish to delete
Delete the node using the temporary pointer

There are a few edge cases we need to take care of, so make sure you understand the code.

int remove_by_index(node_t ** head, int n) { int i = 0; int retval = -1; node_t * current = *head; node_t * temp_node = NULL; if (n == 0) { return pop(head); } for (i = 0; i < n-1; i++) { if (current->next == NULL) { return -1; } current = current->next; } if (current->next == NULL) { return -1; } temp_node = current->next; retval = temp_node->val; current->next = temp_node->next; free(temp_node); return retval; }

Linked List | Set 1 (Introduction)

Like arrays, Linked List is a linear data structure. Unlike arrays, linked list elements are not stored at a contiguous location; the elements are linked using pointers.

Why Linked List?
Arrays can be used to store linear data of similar types, but arrays have the following limitations.
1) The size of the arrays is fixed: So we must know the upper limit on the number of elements in advance. Also, generally, the allocated memory is equal to the upper limit irrespective of the usage.
2) Inserting a new element in an array of elements is expensive because the room has to be created for the new elements and to create room existing elements have to be shifted but in Linked list if we have the head node then we can traverse to any node through it and insert new node at the required position.
For example, in a system, if we maintain a sorted list of IDs in an array id[].
id[] = [1000, 1010, 1050, 2000, 2040].
And if we want to insert a new ID 1005, then to maintain the sorted order, we have to move all the elements after 1000 (excluding 1000).
Deletion is also expensive with arrays until unless some special techniques are used. For example, to delete 1010 in id[], everything after 1010 has to be moved due to this so much work is being done which affects the efficiency of the code.
Advantages over arrays
1) Dynamic size
2) Ease of insertion/deletion
Drawbacks:
1) Random access is not allowed. We have to access elements sequentially starting from the first node(head node). So we cannot do binary search with linked lists efficiently with its default implementation. Read about it here.
2) Extra memory space for a pointer is required with each element of the list.
3) Not cache friendly. Since array elements are contiguous locations, there is locality of reference which is not there in case of linked lists.
Representation:
A linked list is represented by a pointer to the first node of the linked list. The first node is called the head. If the linked list is empty, then the value of the head points to NULL.
Each node in a list consists of at least two parts:
1) data(we can store integer, strings or any type of data).
2) Pointer (Or Reference) to the next node(connects one node to another)
In C, we can represent a node using structures. Below is an example of a linked list node with integer data.
In Java or C#, LinkedList can be represented as a class and a Node as a separate class. The LinkedList class contains a reference of Node class type.

// A linked list node

struct Node {

int data;

struct Node* next;

};

C++

class Node {

public:

int data;

Node* next;

};

Java

class LinkedList {

Node head; // head of the list

/* Linked list Node*/

class Node {

int data;

Node next;

// Constructor to create a new node

// Next is by default initialized

// as null

Node(int d) { data = d; }

}
}

Python

# Node class

class Node:

# Function to initialize the node object

def __init__(self, data):

self.data = data # Assign data

self.next = None # Initialize

# next as null

# Linked List class

class LinkedList:

# Function to initialize the Linked

# List object

def __init__(self):

self.head = None

class LinkedList {

// The first node(head) of the linked list

// Will be an object of type Node (null by default)

Node head;

class Node {

int data;

Node next;

// Constructor to create a new node

Node(int d) { data = d; }

}
}

Javascript

<script>

var head; // head of the list
/* Linked list Node*/

class Node

{

// Constructor to create a new node

// Next is by default initialized

// as null

constructor(val) {

this.data = val;

this.next = null;

}

}
// This code is contributed by gauravrajput1
</script>

First Simple Linked List in C Let us create a simple linked list with 3 nodes.

C++

// A simple CPP program to introduce
// a linked list
#include <bits/stdc++.h>

using namespace std;

class Node {

public:

int data;

Node* next;
};
// Program to create a simple linked
// list with 3 nodes

int main()
{

Node* head = NULL;

Node* second = NULL;

Node* third = NULL;

// allocate 3 nodes in the heap

head = new Node();

second = new Node();

third = new Node();

/* Three blocks have been allocated dynamically.

We have pointers to these three blocks as head,

second and third

head second third

| | |

| | |

+---+-----+ +----+----+ +----+----+

| # | # | | # | # | | # | # |

+---+-----+ +----+----+ +----+----+

# represents any random value. 
Data is random because we haven’t assigned 
anything yet */

head->data = 1; // assign data in first node

head->next = second; // Link first node with

// the second node

/* data has been assigned to the data part of first

block (block pointed by the head). And next

pointer of the first block points to second.

So they both are linked.

head second third

| | |

| | |

+---+---+ +----+----+ +-----+----+

| 1 | o----->| # | # | | # | # |

+---+---+ +----+----+ +-----+----+
*/

// assign data to second node

second->data = 2;

// Link second node with the third node

second->next = third;

/* data has been assigned to the data part of the second

block (block pointed by second). And next

pointer of the second block points to the third

block. So all three blocks are linked.

head second third

| | |

| | |

+---+---+ +---+---+ +----+----+

| 1 | o----->| 2 | o-----> | # | # |

+---+---+ +---+---+ +----+----+ */

third->data = 3; // assign data to third node

third->next = NULL;

/* data has been assigned to the data part of the third

block (block pointed by third). And next pointer

of the third block is made NULL to indicate

that the linked list is terminated here.

We have the linked list ready.

head

|

|

+---+---+ +---+---+ +----+------+

| 1 | o----->| 2 | o-----> | 3 | NULL |

+---+---+ +---+---+ +----+------+

Note that only the head is sufficient to represent

the whole list. We can traverse the complete

list by following the next pointers. */

return 0;
}
// This code is contributed by rathbhupendra

// A simple C program to introduce
// a linked list
#include <stdio.h>
#include <stdlib.h>

struct Node {

int data;

struct Node* next;
};
// Program to create a simple linked
// list with 3 nodes

int main()
{

struct Node* head = NULL;

struct Node* second = NULL;

struct Node* third = NULL;

// allocate 3 nodes in the heap

head = (struct Node*)malloc(sizeof(struct Node));

second = (struct Node*)malloc(sizeof(struct Node));

third = (struct Node*)malloc(sizeof(struct Node));

/* Three blocks have been allocated dynamically.

We have pointers to these three blocks as head,

second and third

head second third

| | |

| | |

+---+-----+ +----+----+ +----+----+

| # | # | | # | # | | # | # |

+---+-----+ +----+----+ +----+----+

# represents any random value.

Data is random because we haven’t assigned

anything yet */

head->data = 1; // assign data in first node

head->next = second; // Link first node with

// the second node

/* data has been assigned to the data part of the first

block (block pointed by the head). And next

pointer of first block points to second.

So they both are linked.

head second third

| | |

| | |

+---+---+ +----+----+ +-----+----+

| 1 | o----->| # | # | | # | # |

+---+---+ +----+----+ +-----+----+

*/

// assign data to second node

second->data = 2;

// Link second node with the third node

second->next = third;

/* data has been assigned to the data part of the second

block (block pointed by second). And next

pointer of the second block points to the third

block. So all three blocks are linked.

head second third

| | |

| | |

+---+---+ +---+---+ +----+----+

| 1 | o----->| 2 | o-----> | # | # |

+---+---+ +---+---+ +----+----+ */

third->data = 3; // assign data to third node

third->next = NULL;

/* data has been assigned to data part of third

block (block pointed by third). And next pointer

of the third block is made NULL to indicate

that the linked list is terminated here.

We have the linked list ready.

head

|

|

+---+---+ +---+---+ +----+------+

| 1 | o----->| 2 | o-----> | 3 | NULL |

+---+---+ +---+---+ +----+------+

Note that only head is sufficient to represent

the whole list. We can traverse the complete

list by following next pointers. */

return 0;
}

Java

// A simple Java program to introduce a linked list

class LinkedList {

Node head; // head of list

/* Linked list Node. This inner class is made static so that

main() can access it */

static class Node {

int data;

Node next;

Node(int d)

{

data = d;

next = null;

} // Constructor

}

/* method to create a simple linked list with 3 nodes*/

public static void main(String[] args)

{

/* Start with the empty list. */

LinkedList llist = new LinkedList();

llist.head = new Node(1);

Node second = new Node(2);

Node third = new Node(3);

/* Three nodes have been allocated dynamically.

We have references to these three blocks as head,

second and third

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | null | | 2 | null | | 3 | null |

+----+------+ +----+------+ +----+------+ */

llist.head.next = second; // Link first node with the second node

/* Now next of the first Node refers to the second. So they

both are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | null | | 3 | null |

+----+------+ +----+------+ +----+------+ */

second.next = third; // Link second node with the third node

/* Now next of the second Node refers to third. So all three

nodes are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | o-------->| 3 | null |

+----+------+ +----+------+ +----+------+ */

}
}

Python

# A simple Python program to introduce a linked list
# Node class

class Node:

# Function to initialise the node object

def __init__(self, data):

self.data = data # Assign data

self.next = None # Initialize next as null
# Linked List class contains a Node object

class LinkedList:

# Function to initialize head

def __init__(self):

self.head = None
# Code execution starts here

if __name__=='__main__':

# Start with the empty list

llist = LinkedList()

llist.head = Node(1)

second = Node(2)

third = Node(3)

'''

Three nodes have been created.

We have references to these three blocks as head,

second and third

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | None | | 2 | None | | 3 | None |

+----+------+ +----+------+ +----+------+

'''

llist.head.next = second; # Link first node with second

'''

Now next of first Node refers to second. So they

both are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | null | | 3 | null |

+----+------+ +----+------+ +----+------+

'''

second.next = third; # Link second node with the third node

'''

Now next of second Node refers to third. So all three

nodes are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | o-------->| 3 | null |

+----+------+ +----+------+ +----+------+

'''

// A simple C# program to introduce a linked list

using System;

public class LinkedList {

Node head; // head of list

/* Linked list Node. This inner class is made static so that

main() can access it */

public class Node {

public int data;

public Node next;

public Node(int d)

{

data = d;

next = null;

} // Constructor

}

/* method to create a simple linked list with 3 nodes*/

public static void Main(String[] args)

{

/* Start with the empty list. */

LinkedList llist = new LinkedList();

llist.head = new Node(1);

Node second = new Node(2);

Node third = new Node(3);

/* Three nodes have been allocated dynamically.

We have references to these three blocks as head,

second and third

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | null | | 2 | null | | 3 | null |

+----+------+ +----+------+ +----+------+ */

llist.head.next = second; // Link first node with the second node

/* Now next of first Node refers to second. So they

both are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | null | | 3 | null |

+----+------+ +----+------+ +----+------+ */

second.next = third; // Link second node with the third node

/* Now next of the second Node refers to third. So all three

nodes are linked.

llist.head second third

| | |

| | |

+----+------+ +----+------+ +----+------+

| 1 | o-------->| 2 | o-------->| 3 | null |

+----+------+ +----+------+ +----+------+ */

}
}
// This code has been contributed by 29AjayKumar

Linked List Traversal
In the previous program, we have created a simple linked list with three nodes. Let us traverse the created list and print the data of each node. For traversal, let us write a general-purpose function printList() that prints any given list.

What is a linked list?

A linked list is a common data structure made of a chain of nodes in which each node contains a value and a pointer to the next node in the chain.

The head pointer points to the first node, and the last element of the list points to null. When the list is empty, the head pointer points to null.

svg viewer

Linked lists can dynamically increase in size and it is easy to insert and delete from a linked list because unlike arrays, we only need to change the pointers of the previous element and the next element to insert or delete an element.

Linked lists are typically used to create file systems, adjacency lists, and hash tables.