Java Program To Find The Sum Of Last N Nodes Of The Given Linked List
Last Updated :
23 Jul, 2025
Given a linked list and a number n. Find the sum of the last n nodes of the linked list.
Constraints: 0 <= n <= number of nodes in the linked list.
Examples:
Input: 10->6->8->4->12, n = 2
Output: 16
Sum of last two nodes:
12 + 4 = 16
Input: 15->7->9->5->16->14, n = 4
Output: 44
Method 1: (Recursive approach using system call stack)
Recursively traverse the linked list up to the end. Now during the return from the function calls, add up the last n nodes. The sum can be accumulated in some variable passed by reference to the function or to some global variable.
Java
// Java implementation to find the sum of
// last 'n' nodes of the Linked List
import java.util.*;
class GFG{
// A Linked list node
static class Node
{
int data;
Node next;
};
static Node head;
static int n, sum;
// Function to insert a node at the
// beginning of the linked list
static 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 to the
// new node
new_node.next = head_ref;
// Move the head to point to the
// new node
head_ref = new_node;
head = head_ref;
}
// Function to recursively find the sum of last
// 'n' nodes of the given linked list
static void sumOfLastN_Nodes(Node head)
{
// if head = NULL
if (head == null)
return;
// Recursively traverse the remaining
// nodes
sumOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0)
{
// Accumulate sum
sum = sum + head.data;
// Reduce node count 'n' by 1
--n;
}
}
// Utility function to find the sum of
// last 'n' nodes
static int sumOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
sum = 0;
// Find the sum of last 'n' nodes
sumOfLastN_Nodes(head);
// Required sum
return sum;
}
// Driver Code
public static void main(String[] args)
{
head = null;
// Create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
n = 2;
System.out.print("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by 29AjayKumar
Output:
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), if system call stack is being considered.
Method 2: (Iterative approach using user-defined stack)
It is an iterative procedure to the recursive approach explained in Method 1 of this post. Traverses the nodes from left to right. While traversing pushes the nodes to a user-defined stack. Then pops the top n values from the stack and adds them.
Java
// Java implementation to find the sum of last
// 'n' nodes of the Linked List
import java.util.*;
class GFG{
// A Linked list node
static class Node
{
int data;
Node next;
};
// Function to insert a node at the
// beginning of the linked list
static Node 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 to the new node
new_node.next = head_ref;
// Move the head to point to the
// new node
head_ref = new_node;
return head_ref;
}
// Utility function to find the sum of
// last 'n' nodes
static int sumOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
Stack<Integer> st = new Stack<Integer>();
int sum = 0;
// Traverses the list from left to right
while (head != null)
{
// Push the node's data onto the
// stack 'st'
st.push(head.data);
// Move to next node
head = head.next;
}
// Pop 'n' nodes from 'st' and
// add them
while (n-- >0)
{
sum += st.peek();
st.pop();
}
// Required sum
return sum;
}
// Driver code
public static void main(String[] args)
{
Node head = null;
// Create linked list 10.6.8.4.12
head = push(head, 12);
head = push(head, 4);
head = push(head, 8);
head = push(head, 6);
head = push(head, 10);
int n = 2;
System.out.print("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by 29AjayKumar
Output:
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(n), stack size
Method 3: (Reversing the linked list)
Following are the steps:
- Reverse the given linked list.
- Traverse the first n nodes of the reversed linked list.
- While traversing add them.
- Reverse the linked list back to its original order.
- Return the added sum.
Java
// Java implementation to find the sum of last
// 'n' nodes of the Linked List
import java.util.*;
class GFG{
// A Linked list node
static class Node
{
int data;
Node next;
};
static Node head;
// Function to insert a node at the
// beginning of the linked list
static 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 to the new node
new_node.next = head_ref;
// Move the head to point to the
// new node
head_ref = new_node;
head=head_ref;
}
static void reverseList(Node head_ref)
{
Node current, prev, next;
current = head_ref;
prev = null;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head_ref = prev;
head = head_ref;
}
// Utility function to find the sum of
// last 'n' nodes
static int sumOfLastN_NodesUtil(int n)
{
// if n == 0
if (n <= 0)
return 0;
// Reverse the linked list
reverseList(head);
int sum = 0;
Node current = head;
// Traverse the 1st 'n' nodes of the
// reversed linked list and add them
while (current != null && n-- >0)
{
// Accumulate node's data to 'sum'
sum += current.data;
// Move to next node
current = current.next;
}
// Reverse back the linked list
reverseList(head);
// Required sum
return sum;
}
// Driver code
public static void main(String[] args)
{
// Create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
int n = 2;
System.out.println("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(n));
}
}
// This code is contributed by PrinciRaj1992
Output:
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Method 4: (Using the length of linked list)
Following are the steps:
- Calculate the length of the given Linked List. Let it be len.
- First, traverse the (len - n) nodes from the beginning.
- Then traverse the remaining n nodes and while traversing add them.
- Return the added sum.
Java
// Java implementation to find the sum of last
// 'n' nodes of the Linked List
class GFG{
// A Linked list node
static class Node
{
int data;
Node next;
};
static Node head;
// Function to insert a node at the
// beginning of the linked list
static 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 to the new node
new_node.next = head_ref;
// Move the head to point to the
// new node
head_ref = new_node;
head = head_ref;
}
// Utility function to find the sum of
// last 'n' nodes
static int sumOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
int sum = 0, len = 0;
Node temp = head;
// Calculate the length of the
// linked list
while (temp != null)
{
len++;
temp = temp.next;
}
// Count of first (len - n) nodes
int c = len - n;
temp = head;
// Just traverse the 1st 'c' nodes
while (temp != null&&c-- >0)
{
// Move to next node
temp = temp.next;
}
// Now traverse the last 'n' nodes and
// add them
while (temp != null)
{
// Accumulate node's data to sum
sum += temp.data;
// Move to next node
temp = temp.next;
}
// Required sum
return sum;
}
// Driver code
public static void main(String[] args)
{
// Create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
int n = 2;
System.out.println("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by 29AjayKumar
Output:
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Method 5: (Use of two pointers requires single traversal)
Maintain two pointers – reference pointer and main pointer. Initialize both reference and main pointers to head. First, move reference pointer to n nodes from head and while traversing accumulate node's data to some variable, say sum. Now move both pointers simultaneously until the reference pointer reaches the end of the list and while traversing accumulate all node's data to sum pointed by the reference pointer and accumulate all node's data to some variable, say, temp, pointed by the main pointer. Now, (sum - temp) is the required sum of the last n nodes.
Java
// Java implementation to find the sum of last
// 'n' nodes of the Linked List
class GfG
{
// Defining structure
static class Node
{
int data;
Node next;
}
static Node head;
static void printList(Node start)
{
Node temp = start;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
System.out.println();
}
// Push function
static void push(Node start, int info)
{
// Allocating node
Node node = new Node();
// Info into node
node.data = info;
// Next of new node to head
node.next = start;
// head points to new node
head = node;
}
private static int sumOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
int sum = 0, temp = 0;
Node ref_ptr, main_ptr;
ref_ptr = main_ptr = head;
// Traverse 1st 'n' nodes through 'ref_ptr'
// and accumulate all node's data to 'sum'
while (ref_ptr != null && (n--) > 0)
{
sum += ref_ptr.data;
// Move to next node
ref_ptr = ref_ptr.next;
}
// Traverse to the end of the linked list
while (ref_ptr != null)
{
// Accumulate all node's data to 'temp'
// pointed by the 'main_ptr'
temp += main_ptr.data;
// Accumulate all node's data to 'sum'
// pointed by the 'ref_ptr'
sum += ref_ptr.data;
// Move both the pointers to their
// respective next nodes
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
// Required sum
return (sum - temp);
}
// Driver code
public static void main(String[] args)
{
head = null;
// Adding elements to Linked List
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
printList(head);
int n = 2;
System.out.println("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by shubham96301
Output:
Sum of last 2 nodes = 16
Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Please refer complete article on Find the sum of last n nodes of the given Linked List for more details!
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