We have introduced the linked list data structure in the previous post and discussed various types of linked lists. We have also covered the applications of linked list data structure and its pros and cons concerning arrays. This post will discuss various linked list implementation techniques in detail and construct a singly linked list in the C programming language.
Lets start by discussing the structure of a linked list node. Each node of a linked list contains a single data element and a pointer to the next node in the list.
1 2 3 4 5 6 | // A Linked List Node struct Node { int data; // integer data struct Node* next;// pointer to the next node }; |
Memory allocation of Linked List nodes
The nodes that will make up the lists body are allocated in the heap memory. We can allocate dynamic memory in C using the malloc[] or calloc[] function. malloc[] takes a single argument [the amount of memory to allocate in bytes]. In contrast, calloc[] needs two arguments [the total number of variables to allocate in memory and the size in bytes of a single variable]. malloc[] does not initialize the memory allocated, while calloc[] guarantees that all bytes of the allocated memory block have been initialized to 0. To deallocate the allocated memory, we can use free[].
1 2 3 4 5 6 7 8 9 10 11 12 | // Helper function in C to return new linked list node from the heap struct Node* newNode[int data] { // allocate a new node in a heap using `malloc[]` and set its data struct Node* node = [struct Node*]malloc[sizeof[struct Node]]; node->data = data; // set the `.next` pointer of the new node to point to null node->next = NULL; return node; } |
Constructing Linked List
This section covers various methods to construct a linked list.
1. Naive method
A naive solution is to construct individual linked list nodes first and rearrange their pointers later to build the list.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 | #include #include // Data structure to store a linked list node struct Node { int data; struct Node* next; }; // Helper function to return new linked list node from the heap struct Node* newNode[int data] { // allocate a new node in a heap and set its data struct Node* node = [struct Node*]malloc[sizeof[struct Node]]; node->data = data; // `.next` pointer of the new node points to nothing node->next = NULL; return node; } // Naive function for linked list implementation containing three nodes struct Node* constructList[] { // construct three linked list nodes struct Node* first = newNode[1]; struct Node* second = newNode[2]; struct Node* third = newNode[3]; // rearrange the pointers to construct a list struct Node* head = first; first->next = second; second->next = third; // return a pointer to the first node in the list return head; } // Helper function to print a linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // `head` points to the first node [also known as a head node] of a linked list struct Node* head = constructList[]; // print linked list printList[head]; return 0; } |
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2. Single Line
The above code can be rewritten in a single line by passing the next node as an argument to the newNode[] function:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 | #include #include // Data structure to store a linked list node struct Node { int data; struct Node* next; }; // Helper function to return new linked list node from the heap struct Node* newNode[int data, struct Node* nextNode] { // allocate a new node in a heap and set its data struct Node* node = [struct Node*]malloc[sizeof[struct Node]]; node->data = data; // set the `.next` pointer of the new node to point to the current // first node of the list. node->next = nextNode; return node; } // Naive function for linked list implementation containing three nodes struct Node* constructList[] { struct Node* head = newNode[1, newNode[2, newNode[3, NULL]]]; return head; } // Helper function to print a linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // `head` points to the first node [also known as a head node] of a linked list struct Node* head = constructList[]; // print linked list printList[head]; return 0; } |
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3. Generic Method
The above-discussed methods will become a pain if the total number of nodes required is huge in the linked list. We can construct a linked list easily using iteration if the keys are given in the form of an array or any other data structure [using its iterator]. Following is the implementation of the idea:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 | #include #include // Data structure to store a linked list node struct Node { int data; struct Node* next; }; // Helper function to return new linked list node from the heap struct Node* newNode[int data, struct Node* nextNode] { // allocate a new node in a heap and set its data struct Node* node = [struct Node*]malloc[sizeof[struct Node]]; node->data = data; // set the `.next` pointer of the new node to point to the current // first node of the list. node->next = nextNode; return node; } // Function for linked list implementation from a given set of keys struct Node* constructList[int keys[], int n] { struct Node *head = NULL, *node = NULL; // start from the end of the array for [int i = n - 1; i >= 0; i--] { node = newNode[keys[i], node]; head = node; } return head; } // Helper function to print a linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // input keys int keys[] = {1, 2, 3, 4}; int n = sizeof[keys]/sizeof[keys[0]]; // `head` points to the first node [also known as a head node] of a linked list struct Node* head = constructList[keys, n]; // print linked list printList[head]; return 0; } |
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4. Standard Solution
The standard function adds a single node to the head end of any list. This function is called push[] since we are adding the link to the head end, making a list look a bit like a stack. Alternately, it could be called InsertAtFront[].
Consider the following snippet:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 | #include #include // Data Structure to store a linked list node struct Node { int data; struct Node* next; }; void push[struct Node* head, int data] { // allocate a new node in a heap and set its data struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; // set the `.next` pointer of the new node to point to the current // first node [head node] of the list. newNode->next = head; // change the head pointer to point to the new node, so it is // now the first node in the list. head = newNode; // No, this line does not work! [Why?] } // Function for linked list implementation from a given set of keys struct Node* constructList[int keys[], int n] { struct Node* head = NULL; // start from the end of the array for [int i = n - 1; i >= 0; i--] { push[head, keys[i]];// try to push a key at front doesn't work } return head; } // Helper function to print given linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // input keys int keys[] = {1, 2, 3, 4}; int n = sizeof[keys]/sizeof[keys[0]]; // head points to first node [also called as head node] of linked list struct Node* head = constructList[keys, n]; // print linked list printList[head]; return 0; } |
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The above code does not work as changes to local parameters are never reflected in the callers memory. The traditional method to allow a function to change its callers memory is to pass a pointer to the callers memory instead of a copy. So, in C, to change an int in the caller, pass an int* instead. In general, to change X, we pass X*. So, in this case, the value we want to change is struct Node*, so we pass a struct Node** instead, i.e., the type of the head pointer is pointer to a struct node and to change that pointer, we need to pass a pointer to it, which will be a pointer to a pointer to a struct node.
Correct push[] code:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 | /* push[]: Takes a list and a data value, creates a new link with the given data and pushes it onto the list's front. Its head pointer does not pass in the list. Instead, the list is passed in as a "reference" pointer to the head pointer this allows us to modify the caller's memory. The parameter has the word "ref" in it as a reminder that this is a "reference". [struct Node**] pointer to the head pointer instead of an ordinary [struct Node*] copy of the head pointer. */ void push[struct Node** headRef, int data] { // allocate a new node in a heap and set its data struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; // set the `.next` pointer of the new node to point to the current // first node of the list. newNode->next = *headRef; // '*' to dereferences back to the real head // change the head pointer to point to the new node, so it is // now the first node in the list. *headRef = newNode; } // Function for linked list implementation from a given set of keys struct Node* constructList[int keys[], int n] { struct Node* head = NULL; // start from the end of the array for [int i = n - 1; i >= 0; i--] { push[&head, keys[i]]; } return head; } |
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5. Make head pointer global
This approach is not recommended as global variables are usually considered bad practice precisely because of their non-locality: a global variable can potentially be modified from anywhere [unless they reside in protected memory or are otherwise, rendered read-only], and any part of the program may depend on it. Therefore, a global variable has unlimited potential for creating mutual dependencies, and adding mutual dependencies increases complexity. Global variables also make it challenging to integrate modules because others may use the same global names unless names are reserved by agreement or by naming convention.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 | #include #include // A Linked List Node struct Node { int data; struct Node* next; }; // Global head pointer struct Node* head = NULL; // Takes a list and a data value, creates a new link with the given // data and pushes it onto the list's front void push[int data] { // allocate a new node in a heap and set its data struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; // set the `.next` pointer of the new node to point to the current // head node of the list. newNode->next = head; // change the head pointer to point to the new node, so it is // now the first node in the list. head = newNode; } // Function for linked list implementation from a given set of keys void constructList[int keys[], int n] { // start from the end of the array for [int i = n - 1; i >= 0; i--] { push[keys[i]]; } } // Helper function to print the global linked list `head` void printList[] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // input keys int keys[] = {1, 2, 3, 4}; int n = sizeof[keys]/sizeof[keys[0]]; // `head` points to the first node [also known as a head node] of a linked list constructList[keys, n]; // print linked list printList[]; return 0; } |
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6. Return head from the push[] function
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 | #include #include // Data structure to store a linked list node struct Node { int data; struct Node* next; }; /* Takes a list and a data value, creates a new link with the given data and pushes it onto the list's front */ struct Node* push[struct Node* head, int data] { // allocate a new node in a heap and set its data struct Node* newNode = [struct Node*]malloc[sizeof[struct Node]]; newNode->data = data; // set the `.next` pointer of the new node to point to the current // first node of the list. newNode->next = head; // return the new node, so it becomes the first node in the list return newNode; } // Function for linked list implementation from a given set of keys struct Node* constructList[int keys[], int n] { struct Node* head = NULL; // start from the end of the array for [int i = n - 1; i >= 0; i--] { head = push[head, keys[i]];// update head here } return head; } // Helper function to print a linked list void printList[struct Node* head] { struct Node* ptr = head; while [ptr] { printf["%d -> ", ptr->data]; ptr = ptr->next; } printf["NULL"]; } int main[void] { // input keys int keys[] = {1, 2, 3, 4}; int n = sizeof[keys]/sizeof[keys[0]]; // `head` points to the first node [also known as a head node] of a linked list struct Node* head = constructList[keys, n]; // print linked list printList[head]; return 0; } |
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Exercise: Modify the push[] function to add nodes to the tail of the list.
[Hint Locate the last node in the list, and then changing its .next field from NULL to point to the new node, or maintain a tail pointer along with a head pointer to perform insertion in constant time.]
Continue Reading:
Linked List Insertion at Tail | C, Java, and Python Implementation
Also See:
Linked List Implementation in C++
Linked List Implementation in Java
Linked List Implementation in Python
References: //cslibrary.stanford.edu/103/LinkedListBasics.pdf