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 <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 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(Node* head)
{
    // Initialize count
    int count = 0;

    // Initialize current
    Node* current = head;
    while (current != NULL)
    {
        count++;
        current = current->next;
    }
    return count;
}

// Driver code
int main()
{
    // Start with the empty list
    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
    cout << "count of nodes is " <<
             getCount(head);
    return 0;
}

// This is code is contributed by rathbhupendra

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.

Recursive Solution:

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

Following is the Recursive implementation of the above algorithm to find the count of nodes in a given singly linked list.

C++

// 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 of 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 code
int main()
{
    // Start with the empty list
    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
    cout << "Count of nodes is " <<
             getCount(head);
    return 0;
}
// This is code is contributed by rajsanghavi9

Output:

Count of nodes is 5

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive stack where n represents the length of the given linked list.

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