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().
Python3
# Python3 program to remove duplicate
# nodes from a sorted linked list
# Node class
class Node:
# Constructor to initialize
# the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# Function to insert a new node
# at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Given a reference to the head of a
# list and a key, delete the first
# occurrence of key in linked list
def deleteNode(self, key):
# Store head node
temp = self.head
# If head node itself holds the
# key to be deleted
if (temp is not None):
if (temp.data == key):
self.head = temp.next
temp = None
return
# Search for the key to be deleted,
# keep track of the previous node as
# we need to change 'prev.next'
while(temp is not None):
if temp.data == key:
break
prev = temp
temp = temp.next
# If key was not present in
# linked list
if(temp == None):
return
# Unlink the node from linked list
prev.next = temp.next
temp = None
# Utility function to print the
# linked LinkedList
def printList(self):
temp = self.head
while(temp):
print(temp.data , end = ' ')
temp = temp.next
# This function removes duplicates
# from a sorted list
def removeDuplicates(self):
temp = self.head
if temp is None:
return
while temp.next is not None:
if temp.data == temp.next.data:
new = temp.next.next
temp.next = None
temp.next = new
else:
temp = temp.next
return self.head
# Driver Code
llist = LinkedList()
llist.push(20)
llist.push(13)
llist.push(13)
llist.push(11)
llist.push(11)
llist.push(11)
print ("Created Linked List: ")
llist.printList()
print()
print("Linked List after removing",
"duplicate elements:")
llist.removeDuplicates()
llist.printList()
# This code is contributed by Dushyant Pathak.
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(1)
Enfoque recursivo:
Python3
# Python3 Program to remove duplicates
# from a sorted linked list
import math
# Link list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# The function removes duplicates
# from a sorted list
def removeDuplicates(head):
# Pointer to store the pointer
# of a node to be deleted to_free
# Do nothing if the list is empty
if (head == None):
return
# Traverse the list till last node
if (head.next != None):
# 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
# free(to_free)
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
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 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
def printList(node):
while (node != None):
print(node.data, end = " ")
node = node.next
# Driver code
if __name__=='__main__':
# Start with the empty list
head = None
# 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)
print("Linked list before duplicate removal ",
end = "")
printList(head)
# Remove duplicates from linked list
removeDuplicates(head)
print("Linked list after duplicate removal ",
end = "")
printList(head)
# This code is contributed by Srathore
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:
Python3
# Python3 program to remove duplicates
# from a sorted linked list
import math
# Link list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# The function removes duplicates
# from the given linked list
def removeDuplicates(head):
# Do nothing if the list consist of
# only one element or empty
if (head == None and
head.next == None):
return
# Construct a pointer
# pointing towards head
current = head
# Initialise a while loop till the
# second last node of the linkedlist
while (current.next):
# If the data of current and next
# node is equal we will skip the
# node between them
if current.data == current.next.data:
current.next = current.next.next
# If the data of current and
# next node is different move
# the pointer to the next node
else:
current = current.next
return
# UTILITY FUNCTIONS
# 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 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
def printList(node):
while (node != None):
print(node.data, end = " ")
node = node.next
# Driver code
if __name__=='__main__':
head = None
head = push(head, 20)
head = push(head, 13)
head = push(head, 13)
head = push(head, 11)
head = push(head, 11)
head = push(head, 11)
print("List before removal of "
"duplicates ", end = "")
printList(head)
removeDuplicates(head)
print("List after removal of "
"elements ", end = "")
printList(head)
# This code is contributed by MukulTomar
Producción:
List before removal of duplicates 11 11 11 13 13 20 List after removal of elements 11 13 20
Complejidad de 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.
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