Find the sum of last n nodes of the given Linked List

Given a linked list and a number n. Find the sum of 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 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.

C++

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// C++ implementation to find the sum of
// last 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct 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;
}
  
// function to recursively find the sum of last
// 'n' nodes of the given linked list
void sumOfLastN_Nodes(struct Node* head, int* n,
                                      int* sum)
{
    // if head = NULL
    if (!head)
        return;
  
    // recursively traverse the remaining nodes
    sumOfLastN_Nodes(head->next, n, sum);
  
    // 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
int sumOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    int sum = 0;
  
    // find the sum of last 'n' nodes
    sumOfLastN_Nodes(head, &n, &sum);
  
    // required sum
    return sum;
}
  
// Driver program to test above
int main()
{
    struct Node* 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);
  
    int n = 2;
    cout << "Sum of last " << n << " nodes = "
         << sumOfLastN_NodesUtil(head, n);
    return 0;
}

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Java

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// 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

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C#

// C# implementation to find the sum of
// last ‘n’ nodes of the Linked List
using System;

class GFG
{

/* A Linked list node */
public class Node
{
public int data;
public 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; Console.Write("Sum of last " + n + " nodes = " + sumOfLastN_NodesUtil(head, n)); } } // This code is contributed by Rajput-Ji [tabbyending] 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.

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// C++ implementation to find the sum of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct 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;
}
  
// utility function to find the sum of last 'n' nodes
int sumOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    stack<int> st;
    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--) {
        sum += st.top();
        st.pop();
    }
  
    // required sum
    return sum;
}
  
// Driver program to test above
int main()
{
    struct Node* 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);
  
    int n = 2;
    cout << "Sum of last " << n << " nodes = "
         << sumOfLastN_NodesUtil(head, n);
    return 0;
}

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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:

  1. Reverse the given linked list.
  2. Traverse the first n nodes of the reversed linked list.
  3. While traversing add them.
  4. Reverse the linked list back to its original order.
  5. Return the added sum.

C++

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// C++ implementation to find the sum of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct 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;
}
  
void reverseList(struct Node** head_ref)
{
    struct 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;
}
  
// utility function to find the sum of last 'n' nodes
int sumOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    // reverse the linked list
    reverseList(&head);
  
    int sum = 0;
    struct Node* current = head;
  
    // traverse the 1st 'n' nodes of the reversed
    // linked list and add them
    while (current != NULL && n--) {                   
  
        // 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 program to test above
int main()
{
    struct Node* 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);
  
    int n = 2;
    cout << "Sum of last " << n << " nodes = "
         << sumOfLastN_NodesUtil(head, n);
    return 0;
}

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Java

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// 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 */

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C#

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// C# implementation to find the sum of last
// 'n' nodes of the Linked List
using System; 
  
class GFG 
{
      
/* A Linked list node */
public class Node
{
    public int data;
    public 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;
    Console.WriteLine("Sum of last " + n + " nodes = "
        + sumOfLastN_NodesUtil(n));
}
}
  
// This code is contributed by Rajput-Ji

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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 length of linked list)
Following are the steps:

  1. Calculate the length of the given Linked List. Let it be len.
  2. First traverse the (len – n) nodes from the beginning.
  3. Then traverse the remaining n nodes and while traversing add them.
  4. Return the added sum.

C++

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// C++ implementation to find the sum of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct 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;
}
  
// utility function to find the sum of last 'n' nodes
int sumOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    int sum = 0, len = 0;
    struct 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--)                    
        // 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 program to test above
int main()
{
    struct Node* 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);
  
    int n = 2;
    cout << "Sum of last " << n << " nodes = "
         << sumOfLastN_NodesUtil(head, n);
    return 0;
}

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Java

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// 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

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C#

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// C# implementation to find the sum of last 
// 'n' nodes of the Linked List 
using System;
      
class GFG 
{
  
  
/* A Linked list node */
public class Node
    public int data; 
    public 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; 
    Console.WriteLine("Sum of last " + n + " nodes = "
        + sumOfLastN_NodesUtil(head, n)); 
}
}
  
// This code is contributed by Princi Singh

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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 reference pointer reaches to 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.

C++

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// C++ implementation to find the sum of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct 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;
}
  
// utility function to find the sum of last 'n' nodes
int sumOfLastN_NodesUtil(struct Node* head, int n)
{
    // if n == 0
    if (n <= 0)
        return 0;
  
    int sum = 0, temp = 0;
    struct 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--) {                   
        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 program to test above
int main()
{
    struct Node* 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);
  
    int n = 2;
    cout << "Sum of last " << n << " nodes = "
         << sumOfLastN_NodesUtil(head, n);
    return 0;
}

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Java

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// 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

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C#

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// C# implementation to find the sum of last 
// 'n' nodes of the Linked List 
using System;
  
class GfG 
  
    // Defining structure 
    public class Node 
    
        public int data; 
        public Node next; 
    
  
    static Node head; 
  
    static void printList(Node start) 
    
        Node temp = start; 
        while (temp != null
        
            Console.Write(temp.data + " "); 
            temp = temp.next; 
        
        Console.WriteLine(); 
    
  
    // 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; 
  
        Console.WriteLine("Sum of last " + n + 
                    " nodes = " + sumOfLastN_NodesUtil(head, n)); 
    
  
// This code contributed by Rajput-Ji

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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)

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