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 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 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
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!