Find Length of a Linked List (Iterative and Recursive) - GeeksforGeeks
Find Length of a Linked List (Iterative and Recursive)
- Difficulty Level : Basic
- Last Updated : 27 May, 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.
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
Iterative Solution
1) Initialize count as 02) Initialize a node pointer, current = head.3) Do following while current is not NULLa) current = current -> nextb) count++;4) Return count
Following are the Iterative implementations of the above algorithm to find the count of nodes in a given singly linked list.
- C++
- C
- Java
- Python
- C#
- Javascript
// Iterative C++ program to find length
// or count of nodes in a linked list
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node
{
public:
int data;
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(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node =new 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(Node* head)
{
int count = 0; // Initialize count
Node* current = head; // Initialize current
while (current != NULL)
{
count++;
current = current->next;
}
return count;
}
/* Driver program to test count function*/
int main()
{
/* Start with the empty list */
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 */
cout<<"count of nodes is "<< getCount(head);
return 0;
}
// This is code is contributed by rathbhupendra
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)
Following are the Recursive implementations of the above algorithm to find the count of nodes in a given singly linked list.
- C++
- Java
- Python
- C#
- Javascript
// Recursive C++ program to find length
// or count of nodes in a linked list
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node {
public:
int data;
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(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new 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;
}
/* Recursively count number of nodes in linked list */
int getCount(Node* head)
{
// Base Case
if (head == NULL) {
return 0;
}
// Count this node plus the rest of the list
else {
return 1 + getCount(head->next);
}
}
/* Driver program to test count function*/
int main()
{
/* Start with the empty list */
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 */
cout << "Count of nodes is " << getCount(head);
return 0;
}
// This is code is contributed by rajsanghavi9
Output:
Count of nodes is 5
