Escriba una función que tome una lista ordenada en orden no decreciente y elimine cualquier Node duplicado de la lista. La lista solo debe recorrerse una vez.
Por ejemplo, si la lista vinculada es 11->11->11->21->43->43->60, removeDuplicates() debería convertir la lista a 11->21->43->60.
Algoritmo:
recorrer la lista desde el Node principal (o inicial). Mientras atraviesa, compare cada Node con su siguiente Node. Si los datos del siguiente Node son los mismos que los del Node actual, elimine el siguiente Node. Antes de eliminar un Node, debemos almacenar el siguiente puntero del Node
Implementación:
las funciones que no sean removeDuplicates() son solo para crear una lista vinculada y probar removeDuplicates().
C#
// C# program to remove duplicates
// from a sorted linked list
using System;
public class LinkedList
{
// Head of list
Node head;
// Linked list Node
class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
void removeDuplicates()
{
// Another reference to head
Node current = head;
/* Pointer to store the next
pointer of a node to be deleted*/
Node next_next;
// Do nothing if the list is empty
if (head == null)
return;
// Traverse list till the last node
while (current.next != null)
{
// Compare current node with the
// next node
if (current.data == current.next.data)
{
next_next = current.next.next;
current.next = null;
current.next = next_next;
}
// Advance if no deletion
else
current = current.next;
}
}
// Utility functions
// Inserts a new Node at front
// of the list.
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
// 3. Make next of new Node as head
new_node.next = head;
// 4. Move the head to point to new Node
head = new_node;
}
// Function to print linked list
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Driver code
public static void Main(String []args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(13);
llist.push(13);
llist.push(11);
llist.push(11);
llist.push(11);
Console.WriteLine(
"List before removal of duplicates");
llist.printList();
llist.removeDuplicates();
Console.WriteLine(
"List after removal of elements");
llist.printList();
}
}
// This code is contributed by 29AjayKumar
Producción:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Complejidad de tiempo: O(n) donde n es el número de Nodes en la lista enlazada dada.
Enfoque recursivo:
C#
// C# Program to remove duplicates
// from a sorted linked list
using System;
class GFG
{
// Link list node
public class Node
{
public int data;
public Node next;
};
// The function removes duplicates
// from a sorted list
static Node removeDuplicates(Node head)
{
/* Pointer to store the pointer
of a node to be deleted*/
Node to_free;
// Do nothing if the list is empty
if (head == null)
return null;
// Traverse the list till last node
if (head.next != null)
{
// Compare head node with next node
if (head.data == head.next.data)
{
/* The sequence of steps is important.
to_free pointer stores the next of
head pointer which is to be deleted.*/
to_free = head.next;
head.next = head.next.next;
removeDuplicates(head);
}
// This is tricky: only advance if no deletion
else
{
removeDuplicates(head.next);
}
}
return head;
}
// UTILITY FUNCTIONS
/* 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 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 nodes in
a given linked list */
static void printList(Node node)
{
while (node != null)
{
Console.Write(" " + node.data);
node = node.next;
}
}
// Driver code
public static void Main(String []args)
{
// Start with the empty list
Node head = null;
/* Let us create a sorted linked list
to test the functions. Created
linked list will be
11.11.11.13.13.20 */
head = push(head, 20);
head = push(head, 13);
head = push(head, 13);
head = push(head, 11);
head = push(head, 11);
head = push(head, 11);
Console.Write("Linked list before" +
" duplicate removal ");
printList(head);
// Remove duplicates from linked list
head = removeDuplicates(head);
Console.Write("Linked list after" +
" duplicate removal ");
printList(head);
}
}
// This code is contributed by PrinciRaj1992
Producción:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Complejidad de tiempo: O(n), donde n es el número de Nodes en la lista enlazada dada.
Espacio auxiliar: O(n), debido a la pila recursiva donde n es el número de Nodes en la lista enlazada dada.
Otro enfoque: cree un puntero que apunte hacia la primera aparición de cada elemento y otro puntero temporal que iterará a cada elemento y cuando el valor del puntero anterior no sea igual al puntero temporal, estableceremos el puntero del anterior puntero a la primera aparición de otro Node.
A continuación se muestra la implementación del enfoque anterior:
C#
// C# program to remove duplicates
// from a sorted linked list
using System;
class LinkedList
{
// Head of list
public Node head;
// Linked list Node
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
// Function to remove duplicates
// from the given linked list
void removeDuplicates()
{
// Two references to head temp
// will iterate to the whole
// Linked List prev will point
// towards the first occurrence
// of every element
Node temp = head, prev = head;
// Traverse list till the last node
while (temp != null)
{
// Compare values of both pointers
if(temp.data != prev.data)
{
/* if the value of prev is not
equal to the value of temp
that means there are no more
occurrences of the prev data.
So we can set the next of
prev to the temp node.*/
prev.next = temp;
prev = temp;
}
/*Set the temp to the next node*/
temp = temp.next;
}
/* This is the edge case if there are more
than one occurrences of the last element */
if(prev != temp)
{
prev.next = null;
}
}
// Utility functions
// Inserts a new Node at front
// of the list.
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
// 3. Make next of new Node as head
new_node.next = head;
// 4. Move the head to point to
// new Node
head = new_node;
}
// Function to print linked list
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Driver code
public static void Main(string []args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(13);
llist.push(13);
llist.push(11);
llist.push(11);
llist.push(11);
Console.Write("List before ");
Console.WriteLine("removal of duplicates");
llist.printList();
llist.removeDuplicates();
Console.WriteLine(
"List after removal of elements");
llist.printList();
}
}
// This code is contributed by rutvik_56
Producción:
List before removal of duplicates 11 11 11 13 13 20 List after removal of elements 11 13 20
Complejidad del tiempo:O(n) donde n es el número de Nodes en la lista enlazada dada.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.
Otro enfoque: uso de mapas
La idea es empujar todos los valores en un mapa e imprimir sus claves.
A continuación se muestra la implementación del enfoque anterior:
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class Node
{
public int data;
public Node next;
public Node()
{
data = 0;
next = null;
}
}
public class GFG
{
/* 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 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 nodes
in a given linked list */
static void printList(Node node)
{
while (node != null)
{
Console.Write(node.data + " ");
node = node.next;
}
}
// Function to remove duplicates
static void removeDuplicates(Node head)
{
Dictionary<int, bool> track =
new Dictionary<int, bool>();
Node temp = head;
while(temp != null)
{
if(!track.ContainsKey(temp.data))
{
Console.Write(temp.data + " ");
track.Add(temp.data , true);
}
temp = temp.next;
}
}
// Driver Code
static public void Main ()
{
Node head = null;
/* Created linked list will be
11->11->11->13->13->20 */
head = push(head, 20);
head = push(head, 13);
head = push(head, 13);
head = push(head, 11);
head = push(head, 11);
head = push(head, 11);
Console.Write(
"Linked list before duplicate removal ");
printList(head);
Console.Write(
"Linked list after duplicate removal ");
removeDuplicates(head);
}
}
// This code is contributed by rag2127
Producción:
Linked list before duplicate removal 11 11 11 13 13 20
Complejidad de tiempo: O (Número de Nodes)
Complejidad espacial: O (Número de Nodes)
Consulte el artículo completo sobre Eliminar duplicados de una lista ordenada ordenada para obtener más detalles.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA