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C Program For Finding Length Of A Linked List

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  • Last Updated : 27 Dec, 2021

Write a function to count the number of nodes in a given singly linked list.


For example, the function should return 5 for linked list 1->3->1->2->1.

Iterative Solution:

1) Initialize count as 0 
2) Initialize a node pointer, current = head.
3) Do following while current is not NULL
     a) current = current -> next
     b) count++;
4) Return count 

Following is the Iterative implementation of the above algorithm to find the count of nodes in a given singly linked list.


// Iterative C program to find length or count 
// of nodes in a linked list
// Link list node
struct Node
    int data;
    struct Node* next;
/* Given a reference (pointer to pointer) to 
   the head of a list and an int, push a new
   node on the front of the list. */
void push(struct Node** head_ref,  
          int new_data)
    // Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));
    // Put in the data 
    new_node->data = new_data;
    // Link the old list off the new node
    new_node->next = (*head_ref);
    // Move the head to point to the new node
    (*head_ref) = new_node;
// Counts no. of nodes in linked list
int getCount(struct Node* head)
    // Initialize count
    int count = 0;  
    // Initialize current
    struct Node* current = head;  
    while (current != NULL)
        current = current->next;
    return count;
// Driver code
int main()
    // Start with the empty list
    struct Node* head = NULL;
    // Use push() to construct list
    // 1->2->1->3->1 
    push(&head, 1);
    push(&head, 3);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);
    // Check the count function
    printf("count of nodes is %d"
    return 0;


count of nodes is 5


Please refer complete article on Find Length of a Linked List (Iterative and Recursive) for more details!

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