C Program For Finding Length Of A Linked List

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.

C

// Iterative C program to find length or count
// of nodes in a linked list
#include<stdio.h>
#include<stdlib.h>

// 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)
    {
        count++;
        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",
            getCount(head));
    return 0;
}

Output: 

count of nodes is 5

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