Dada una lista enlazada individualmente y un número entero K , la tarea es invertir todos los Nodes K de la lista enlazada dada.
Ejemplos:
Entrada: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULO, K = 3
Salida: 3 2 1 6 5 4 8 7Entrada: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULO, K = 5
Salida: 5 4 3 2 1 8 7 6
Enfoque: Se han discutido dos enfoques diferentes para resolver este problema en el Conjunto 1 y el Conjunto 2 de este artículo. En este artículo, se discutirá un enfoque basado en deque .
- Crea un deque.
- Almacene la dirección de los primeros k Nodes en el deque.
- Extraiga el primero y el último valor de la deque e intercambie los valores de datos en esas direcciones.
- Repita el paso 3 hasta que el deque no esté vacío.
- Repita el paso 2 para los siguientes k Nodes y hasta que no llegue al final de la lista enlazada.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Link list node
struct node {
int data;
struct node* next;
};
// Function to insert a node at
// the head of the linked list
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 off the new node */
new_node->next = (*head_ref);
/* Move the head to point to the new node */
(*head_ref) = new_node;
}
// Function to print the linked list
void printList(node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}
/* Function to reverse the linked list in groups of
size k and return the pointer to the new head node. */
struct node* reverse(struct node* head, int k)
{
if (head == NULL)
return head;
// Create deque to store the address
// of the nodes of the linked list
deque<node*> q;
// Store head pointer in current to
// traverse the linked list
node* current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != NULL) {
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k) {
if (current == NULL)
break;
q.push_back(current);
current = current->next;
i++;
}
/* pop first and the last value from
the deque and swap the data values at
those addresses
Do this till there exist an address in
the deque or deque is not empty*/
while (!q.empty()) {
node* front = q.front();
node* last = q.back();
swap(front->data, last->data);
// pop from the front if
// the deque is not empty
if (!q.empty())
q.pop_front();
// pop from the back if
// the deque is not empty
if (!q.empty())
q.pop_back();
}
}
return head;
}
// Driver code
int main()
{
// Start with the empty list
node* head = NULL;
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
push(&head, 10);
push(&head, 9);
push(&head, 8);
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = reverse(head, k);
printList(head);
return 0;
}
Java
// Java implementation of the above approach
import java.util.*;
class LinkedList{
static Node head;
// Creating node class
static class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
// Inserts a new Node at front of the list.
public void push(int new_data)
{
// Allocate the Node &
// Put in the data
Node new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Prints content of linked list
void printList(Node node)
{
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
}
// Function to reverse the linked list
// in groups of size k and return the
// pointer to the new head node.
Node reverse(Node head, int k)
{
if (head == null)
return head;
// Create deque to store the address
// of the nodes of the linked list
Deque<Node> q = new ArrayDeque<Node>();
// Store head pointer in current to
// traverse the linked list
Node current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != null)
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null)
break;
q.addLast(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (!q.isEmpty())
{
Node front = q.peekFirst();
Node last = q.peekLast();
// Swapping front and
// last nodes
int temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (!q.isEmpty())
q.removeFirst();
// pop from the back if
// the deque is not empty
if (!q.isEmpty())
q.removeLast();
}
}
return head;
}
// Driver code
public static void main(String[] args)
{
LinkedList list = new LinkedList();
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
list.push(10);
list.push(9);
list.push(8);
list.push(7);
list.push(6);
list.push(5);
list.push(4);
list.push(3);
list.push(2);
list.push(1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = list.reverse(head, k);
list.printList(head);
}
}
// This code is contributed by Bindu Madhav
Python3
# Python3 implementation of the approach # Link list node class Node: def __init__(self): self.data = 0 self.next = None # Function to insert a node at # the head of the linked list def push(head_ref, new_data): # Allocate node new_node = Node() # Put in the data new_node.data = new_data # Link the old list off the new node new_node.next = (head_ref) # Move the head to point to the new node (head_ref) = new_node return head_ref # Function to print the linked list def printList( head): while (head != None) : print( head.data, end = " ") head = head.next # Function to reverse the linked list in groups of # size k and return the pointer to the new head node. def reverse( head, k): if (head == None): return head # Create deque to store the address # of the nodes of the linked list q = [] # Store head pointer in current to # traverse the linked list current = head i = 0 # Iterate through the entire linked # list by moving the current while (current != None) : i = 1 # Store addresses of the k # nodes in the deque while (i <= k) : if (current == None): break q.append(current) current = current.next i = i + 1 # pop first and the last value from # the deque and swap the data values at # those addresses # Do this till there exist an address in # the deque or deque is not empty while (len(q) > 0): front = q[-1] last = q[0] temp = front.data front.data = last.data last.data = temp # pop from the front if # the deque is not empty if (len(q) > 0): q.pop() # pop from the back if # the deque is not empty if (len(q)): q.pop(0) return head # Driver code # Start with the empty list head = None # Created Linked list is # 1.2.3.4.5.6.7.8.9.10 head = push(head, 10) head = push(head, 9) head = push(head, 8) head = push(head, 7) head = push(head, 6) head = push(head, 5) head = push(head, 4) head = push(head, 3) head = push(head, 2) head = push(head, 1) k = 2 # Get the new head after reversing the # linked list in groups of size k head = reverse(head, k) printList(head) # This code is contributed by Arnab Kundu
C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class List{
static Node head;
// Creating node class
class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
// Inserts a new Node at front of the list.
public void Push(int new_data)
{
// Allocate the Node &
// Put in the data
Node new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Prints content of linked list
void printList(Node node)
{
while (node != null)
{
Console.Write(node.data + " ");
node = node.next;
}
}
// Function to reverse the linked list
// in groups of size k and return the
// pointer to the new head node.
Node reverse(Node head, int k)
{
if (head == null)
return head;
// Create deque to store the address
// of the nodes of the linked list
List<Node> q = new List<Node>();
// Store head pointer in current to
// traverse the linked list
Node current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != null)
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null)
break;
q.Add(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (q.Count != 0)
{
Node front = q[0];
Node last = q[q.Count - 1];
// Swapping front and
// last nodes
int temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (q.Count != 0)
q.RemoveAt(0);
// pop from the back if
// the deque is not empty
if (q.Count != 0)
q.RemoveAt(q.Count - 1);
}
}
return head;
}
// Driver code
public static void Main(String[] args)
{
List list = new List();
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
list.Push(10);
list.Push(9);
list.Push(8);
list.Push(7);
list.Push(6);
list.Push(5);
list.Push(4);
list.Push(3);
list.Push(2);
list.Push(1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = list.reverse(head, k);
list.printList(head);
}
}
// This code is contributed by todaysgaurav
Javascript
<script>
// Javascript implementation of the above approach
var head = null;
// Creating node class
class Node
{
constructor(d)
{
this.data = d;
this.next = null;
}
}
// Inserts a new Node at front of the list.
function Push(new_data)
{
// Allocate the Node &
// Put in the data
var new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Prints content of linked list
function printList(node)
{
while (node != null)
{
document.write(node.data + " ");
node = node.next;
}
}
// Function to reverse the linked list
// in groups of size k and return the
// pointer to the new head node.
function reverse(head, k)
{
if (head == null)
return head;
// Create deque to store the address
// of the nodes of the linked list
var q = [];
// Store head pointer in current to
// traverse the linked list
var current = head;
var i;
// Iterate through the entire linked
// list by moving the current
while (current != null)
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null)
break;
q.push(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (q.length != 0)
{
var front = q[0];
var last = q[q.length - 1];
// Swapping front and
// last nodes
var temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (q.Count != 0)
q.shift();
// pop from the back if
// the deque is not empty
if (q.Count != 0)
q.pop();
}
}
return head;
}
// Driver code
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
Push(10);
Push(9);
Push(8);
Push(7);
Push(6);
Push(5);
Push(4);
Push(3);
Push(2);
Push(1);
var k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = reverse(head, k);
printList(head);
// This code is contributed by itsok
</script>
Producción:
2 1 4 3 6 5 8 7 10 9
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por Rohit_Dwivedi y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA