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 nodestruct 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 of 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 listint 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 codeint 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
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Find Length of a Linked List (Iterative and Recursive) for more details!
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