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

C Algorithm - Find Length of a Linked List both Iterative and Recursive

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

C++ Programming:

// 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)
{
    int count = 0;  // Initialize count
    struct node* current = head;  // Initialize current
    while (current != NULL)
    {
        count++;
        current = current->next;
    }
    return count;
}
 
/* Drier program to test count function*/
int main()
{
    /* Start with the empty list */
    struct node* head = NULL;
 
    /* Use push() to construct below 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;
}
[ad type=”banner”]

Output:

count of nodes is 5

Recursive Solution

int getCount(head)
1) If head is NULL, return 0.
2) Else return 1 + getCount(head->next)

C++ Programming:

// Recursive 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 the no. of occurences of a node
   (search_for) in a linked list (head)*/
int getCount(struct node* head)
{
    // Base case
    if (head == NULL)
        return 0;
 
    // count is 1 + count of remaining list
    return 1 + getCount(head->next);
}
 
/* Drier program to test count function*/
int main()
{
    /* Start with the empty list */
    struct node* head = NULL;
 
    /* Use push() to construct below 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;
}
[ad type=”banner”]

Output:

count of nodes is 5

Categorized in:

C++ programmingLinked ListSingly Linked List

Tagged in:

, , , , , , , , , , , ,