Java 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.

Java

// Java program to count number of
// nodes in a linked list

// Linked list Node
class Node
{
    int data;
    Node next;
    Node(int d)  { data = d;  next = null; }
}

// Linked List class
class LinkedList
{
    // Head of list
    Node head;

    // Inserts a new Node at front
    // of the list.
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        // 3. Make next of new Node as head
        new_node.next = head;

        // 4. Move the head to point to new Node
        head = new_node;
    }

    // Returns count of nodes in linked list
    public int getCount()
    {
        Node temp = head;
        int count = 0;
        while (temp != null)
        {
            count++;
            temp = temp.next;
        }
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Start with the empty list
        LinkedList llist = new LinkedList();
        llist.push(1);
        llist.push(3);
        llist.push(1);
        llist.push(2);
        llist.push(1);

        System.out.println("Count of nodes is " +
                            llist.getCount());
    }
}

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. 

Java

// Recursive Java program to count number
// of nodes in a linked list

// Linked list Node
class Node
{
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}

// Linked List class
class LinkedList
{

    Node head;

    // Inserts a new Node at front of
    // the list.
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        // 3. Make next of new Node as head
        new_node.next = head;

        // 4. Move the head to point to new Node
        head = new_node;
    }

    // Returns count of nodes in linked list
    public int getCountRec(Node node)
    {
        // Base case
        if (node == null)
            return 0;

        // Count is this node plus rest
        // of the list
        return 1 + getCountRec(node.next);
    }

    // Wrapper over getCountRec()
    public int getCount()
    {
        return getCountRec(head);
    }

    // Driver code
    public static void main(String[] args)
    {
        // Start with the empty list
        LinkedList llist = new LinkedList();
        llist.push(1);
        llist.push(3);
        llist.push(1);
        llist.push(2);
        llist.push(1);

        System.out.println("Count of nodes is " +
                            llist.getCount());
    }
}

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!