Dada una lista enlazada y un número n . Encuentre la suma de los últimos n Nodes de la lista enlazada.
Restricciones: 0 <= n <= número de Nodes en la lista enlazada.
Ejemplos:
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
Método 1: (Enfoque recursivo usando la pila de llamadas del sistema):
Atraviesa recursivamente la lista enlazada hasta el final. Ahora, durante el retorno de las llamadas a funciones, sume los últimos n Nodes. La suma se puede acumular en alguna variable pasada por referencia a la función oa alguna variable global.
Implementación:
C++
// 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;
}
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
Python3
# Python3 implementation to find the sum of
# last 'n' nodes of the Linked List
# Linked List node
class Node:
def __init__(self, data):
self.data = data
self.next = None
head = None
n = 0
sum = 0
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
global head
# allocate node
new_node = Node(0)
# 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
def sumOfLastN_Nodes(head):
global sum
global n
# if head = None
if (head == None):
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 = n - 1
# utility function to find the sum of last 'n' nodes
def sumOfLastN_NodesUtil(head, n):
global sum
# 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
head = None
# 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
print("Sum of last " , n ,
" nodes = ", sumOfLastN_NodesUtil(head, n))
# This code is contributed by Arnab Kundu
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
Javascript
<script>
// JavaScript implementation to find the sum of
// last 'n' nodes of the Linked List
/* A Linked list node */
class Node
{
constructor()
{
this.data;
this.next;
}
}
let head;
let n, sum;
// function to insert a node at the
// beginning of the linked list
function push(head_ref,new_data)
{
/* allocate node */
let 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
function sumOfLastN_Nodes(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
function sumOfLastN_NodesUtil(head,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
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;
document.write("Sum of last " + n +
" nodes = " +
sumOfLastN_NodesUtil(head, n));
// This code is contributed by unknown2108
</script>
Producción
Sum of last 2 nodes = 16
Complejidad de tiempo: O(n) , donde n es el número de Nodes en la lista enlazada.
Espacio auxiliar: O(n), si se está considerando la pila de llamadas del sistema.
Método 2 (enfoque iterativo utilizando la pila definida por el usuario):
Es un procedimiento iterativo del enfoque recursivo explicado en el Método 1 de esta publicación. Atraviesa los Nodes de izquierda a derecha. Mientras atraviesa empuja los Nodes a una pila definida por el usuario. Luego extrae los n valores superiores de la pila y los agrega.
Implementación:
C++
// 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;
}
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 program to test above
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
Python3
# Python3 implementation to find the sum of
# last 'n' nodes of the Linked List
# Linked List node
class Node:
def __init__(self, data):
self.data = data
self.next = None
head = None
n = 0
sum = 0
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
global head
# allocate node
new_node = Node(0)
# 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
def sumOfLastN_NodesUtil(head, n):
global sum
# if n == 0
if (n <= 0):
return 0
st = []
sum = 0
# traverses the list from left to right
while (head != None):
# push the node's data onto the stack 'st'
st.append(head.data)
# move to next node
head = head.next
# pop 'n' nodes from 'st' and
# add them
while (n):
n -= 1
sum += st[0]
st.pop(0)
# required sum
return sum
# Driver Code
head = None
# 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
print("Sum of last" , n ,
"nodes =", sumOfLastN_NodesUtil(head, n))
# This code is contributed by shubhamsingh10
C#
// C# implementation to find the sum of last
// 'n' nodes of the Linked List
using System;
using System.Collections.Generic;
class GFG
{
/* A Linked list node */
class Node
{
public int data;
public 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<int> st = new Stack<int>();
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;
Console.Write("Sum of last " + n+ " nodes = "
+ sumOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation to find the sum of last
// 'n' nodes of the Linked List
/* A Linked list node */
class Node
{
constructor()
{
let data,next;
}
}
// function to insert a node at the
// beginning of the linked list
function push(head_ref,new_data)
{
/* allocate node */
let 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
function sumOfLastN_NodesUtil(head,n)
{
// if n == 0
if (n <= 0)
return 0;
let st = [];
let 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[st.length-1];
st.pop();
}
// required sum
return sum;
}
// Driver program to test above
let 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);
let n = 2;
document.write("Sum of last " + n+ " nodes = "
+ sumOfLastN_NodesUtil(head, n));
// This code is contributed by patel2127
</script>
Producción
Sum of last 2 nodes = 16
Complejidad de tiempo: O(n), donde n es el número de Nodes en la lista enlazada.
Espacio auxiliar: O(n), tamaño de pila
Método 3 (Invertir la lista enlazada):
Los siguientes son los pasos:
- Invierta la lista enlazada dada.
- Atraviese los primeros n Nodes de la lista enlazada invertida.
- Mientras atraviesa, agréguelos.
- Invierta la lista enlazada a su orden original.
- Devuelve la suma añadida.
Implementación:
C++
// 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;
}
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 */
Python3
# Python implementation to find the sum of last
# 'n' Nodes of the Linked List
''' A Linked list Node '''
# A Linked list Node
class Node:
def __init__(self, x):
self.data = x
self.next = None
head = None
# Function to insert a Node at the
# beginning of the linked list
def push(head_ref, new_data):
# Allocate Node
new_Node = Node(new_data)
# 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
return head
def reverseList():
global head;
current, prev, next = None, None, None;
current = head;
prev = None;
while (current != None):
next = current.next;
current.next = prev;
prev = current;
current = next;
head = prev;
# utility function to find the sum of last 'n' Nodes
def sumOfLastN_NodesUtil(n):
# if n == 0
if (n <= 0):
return 0;
# reverse the linked list
reverseList();
sum = 0;
current = head;
# traverse the 1st 'n' Nodes of the reversed
# linked list and add them
while (current != None and n > 0):
# accumulate Node's data to 'sum'
sum += current.data;
# move to next Node
current = current.next;
n -= 1;
# reverse back the linked list
reverseList();
# required sum
return sum;
# Driver code
if __name__ == '__main__':
# create linked list 10.6.8.4.12
# 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)
n = 2;
print("Sum of last " , n , " Nodes = " , sumOfLastN_NodesUtil(n));
# This code contributed by Princi Singh
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;
// 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
Javascript
<script>
// javascript implementation to find the sum of last
// 'n' nodes of the Linked List
/* A Linked list node */
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
var head;
// function to insert a node at the
// beginning of the linked list
function push(head_ref , new_data)
{
/* allocate node */
var 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 reverseList(head_ref)
{
var 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
function sumOfLastN_NodesUtil(n)
{
// if n == 0
if (n <= 0)
return 0;
// reverse the linked list
reverseList(head);
var sum = 0;
var 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
// create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
var n = 2;
document.write("Sum of last " + n + " nodes = "
+ sumOfLastN_NodesUtil(n));
// This code contributed by umadevi9616
</script>
Producción
Sum of last 2 nodes = 16
Complejidad de tiempo: O(n) , donde n es el número de Nodes en la lista enlazada.
Espacio Auxiliar: O(1)
Método 4 (usando la longitud de la lista enlazada):
Los siguientes son los pasos:
- Calcule la longitud de la lista enlazada dada. Que sea len .
- Primero, recorra los Nodes (len – n) desde el principio.
- Luego, recorra los n Nodes restantes y, mientras los recorre, agréguelos.
- Devuelve la suma añadida.
Implementación:
C++
// 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;
}
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
Python3
# Python3 implementation to find the sum
# of last 'n' Nodes of the Linked List
# A Linked list Node
class Node:
def __init__(self, x):
self.data = x
self.next = None
head = None
# Function to insert a Node at the
# beginning of the linked list
def push(head_ref, new_data):
# Allocate Node
new_Node = Node(new_data)
# 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
return head
# Utility function to find the sum of
# last 'n' Nodes
def sumOfLastN_NodesUtil(head, n):
# If n == 0
if (n <= 0):
return 0
sum = 0
len = 0
temp = head
# Calculate the length of the linked list
while (temp != None):
len += 1
temp = temp.next
# Count of first (len - n) Nodes
c = len - n
temp = head
# Just traverse the 1st 'c' Nodes
while (temp != None and c > 0):
# Move to next Node
temp = temp.next
c -= 1
# Now traverse the last 'n' Nodes
# and add them
while (temp != None):
# Accumulate Node's data to sum
sum += temp.data
# Move to next Node
temp = temp.next
# Required sum
return sum
# Driver code
if __name__ == '__main__':
# 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)
n = 2
print("Sum of last ", n, " Nodes = ",
sumOfLastN_NodesUtil(head, n))
# This code is contributed by Princi Singh
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;
// 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
Javascript
<script>
// Javascript implementation to
// find the sum of last
// 'n' nodes of the Linked List
/* A Linked list node */
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
var head;
// function to insert a node at the
// beginning of the linked list
function push( head_ref , new_data)
{
/* allocate 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
function sumOfLastN_NodesUtil( head , n)
{
// if n == 0
if (n <= 0)
return 0;
var sum = 0, len = 0;
temp = head;
// calculate the length of the linked list
while (temp != null) {
len++;
temp = temp.next;
}
// count of first (len - n) nodes
var 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
// create linked list 10.6.8.4.12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
var n = 2;
document.write("Sum of last " + n
+ " nodes = " + sumOfLastN_NodesUtil(head, n));
// This code contributed by umadevi9616
</script>
Producción
Sum of last 2 nodes = 16
Complejidad de tiempo: O(n), donde n es el número de Nodes en la lista enlazada.
Espacio Auxiliar: O(1)
Método 5 (El uso de dos punteros requiere un solo recorrido):
Mantener dos punteros: puntero de referencia y puntero principal. Inicialice tanto los punteros de referencia como los principales en head. Primero, mueva el puntero de referencia a n Nodes desde la cabecera y, mientras atraviesa, acumule los datos del Node en alguna variable, digamos sum .
Ahora mueva ambos punteros simultáneamente hasta que el puntero de referencia llegue al final de la lista y, mientras se desplaza, acumule todos los datos del Node para sumar los señalados por el puntero de referencia y acumule todos los datos del Node en alguna variable, digamos, temp , señalada por el puntero principal. Ahora, (sum – temp) es la suma requerida de los últimos n Nodes.
Implementación:
C++
// 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;
}
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
Python3
# Python3 implementation to find the sum of last
# 'n' nodes of the Linked List
# include
class Node:
def __init__(self, x):
self.data = x
self.next = None
# function to insert a node at the
# beginning of the linked list
def push(head_ref,new_data):
# /* allocate node */
new_node = Node(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
def sumOfLastN_NodesUtil(head, n):
# if n == 0
if (n <= 0):
return 0
sum = 0
temp = 0
ref_ptr = None
main_ptr = None
ref_ptr = main_ptr = head
# traverse 1st 'n' nodes through 'ref_ptr' and
# accumulate all node's data to 'sum'
while (ref_ptr != None and n):
sum += ref_ptr.data
# move to next node
ref_ptr = ref_ptr.next
n -= 1
# traverse to the end of the linked list
while (ref_ptr != None):
# 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
if __name__ == '__main__':
head = None
# 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)
n = 2
print("Sum of last ",n," nodes = ",sumOfLastN_NodesUtil(head, n))
# This code is contributed by mohit kumar 29
C#
// 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
Javascript
<script>
// Javascript implementation to find the sum of last
// 'n' nodes of the Linked List
// Defining structure
class Node
{
constructor()
{
let node,next;
}
}
let head;
function printList(start)
{
let temp = start;
while (temp != null)
{
document.write(temp.data + " ");
temp = temp.next;
}
document.write("<br>");
}
// Push function
function push(start,info)
{
// Allocating node
let 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;
}
function sumOfLastN_NodesUtil(head,n)
{
// if n == 0
if (n <= 0)
return 0;
let sum = 0, temp = 0;
let 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
head = null;
// Adding elements to Linked List
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
let n = 2;
document.write("Sum of last " + n +
" nodes = " + sumOfLastN_NodesUtil(head, n));
// This code is contributed by avanitrachhadiya2155
</script>
Producción
Sum of last 2 nodes = 16
Complejidad de tiempo: O(n), donde n es el número de Nodes en la lista enlazada.
Espacio Auxiliar: O(1)
Este artículo es una contribución de Ayush Jauhari . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA