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 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.
// 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!