Dada una lista de caracteres enlazados individualmente, escriba una función que devuelva verdadero si la lista dada es un palíndromo, de lo contrario, falso.
MÉTODO 1 (Usar una pila)
- Una solución simple es usar una pila de Nodes de lista. Esto implica principalmente tres pasos.
- Recorra la lista dada de principio a fin y empuje cada Node visitado para apilar.
- Recorra la lista de nuevo. Por cada Node visitado, extraiga un Node de la pila y compare los datos del Node extraído con el Node visitado actualmente.
- Si todos los Nodes coinciden, devuelve verdadero, de lo contrario, falso.
La imagen de abajo es una ejecución en seco del enfoque anterior:
A continuación se muestra la implementación del enfoque anterior:
C++
#include<bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node(int d){
data = d;
}
Node *ptr;
};
// Function to check if the linked list
// is palindrome or not
bool isPalin(Node* head){
// Temp pointer
Node* slow= head;
// Declare a stack
stack <int> s;
// Push all elements of the list
// to the stack
while(slow != NULL){
s.push(slow->data);
// Move ahead
slow = slow->ptr;
}
// Iterate in the list again and
// check by popping from the stack
while(head != NULL ){
// Get the top most element
int i=s.top();
// Pop the element
s.pop();
// Check if data is not
// same as popped element
if(head -> data != i){
return false;
}
// Move ahead
head=head->ptr;
}
return true;
}
// Driver Code
int main(){
// Addition of linked list
Node one = Node(1);
Node two = Node(2);
Node three = Node(3);
Node four = Node(2);
Node five = Node(1);
// Initialize the next pointer
// of every current pointer
five.ptr = NULL;
one.ptr = &two;
two.ptr = &three;
three.ptr = &four;
four.ptr = &five;
Node* temp = &one;
// Call function to check palindrome or not
int result = isPalin(&one);
if(result == 1)
cout<<"isPalindrome is true\n";
else
cout<<"isPalindrome is true\n";
return 0;
}
// This code has been contributed by Striver
Java
/* Java program to check if linked list is palindrome recursively */
import java.util.*;
class linkedList {
public static void main(String args[])
{
Node one = new Node(1);
Node two = new Node(2);
Node three = new Node(3);
Node four = new Node(4);
Node five = new Node(3);
Node six = new Node(2);
Node seven = new Node(1);
one.ptr = two;
two.ptr = three;
three.ptr = four;
four.ptr = five;
five.ptr = six;
six.ptr = seven;
boolean condition = isPalindrome(one);
System.out.println("isPalidrome :" + condition);
}
static boolean isPalindrome(Node head)
{
Node slow = head;
boolean ispalin = true;
Stack<Integer> stack = new Stack<Integer>();
while (slow != null) {
stack.push(slow.data);
slow = slow.ptr;
}
while (head != null) {
int i = stack.pop();
if (head.data == i) {
ispalin = true;
}
else {
ispalin = false;
break;
}
head = head.ptr;
}
return ispalin;
}
}
class Node {
int data;
Node ptr;
Node(int d)
{
ptr = null;
data = d;
}
}
Python3
# Python3 program to check if linked
# list is palindrome using stack
class Node:
def __init__(self,data):
self.data = data
self.ptr = None
# Function to check if the linked list
# is palindrome or not
def ispalindrome(head):
# Temp pointer
slow = head
# Declare a stack
stack = []
ispalin = True
# Push all elements of the list
# to the stack
while slow != None:
stack.append(slow.data)
# Move ahead
slow = slow.ptr
# Iterate in the list again and
# check by popping from the stack
while head != None:
# Get the top most element
i = stack.pop()
# Check if data is not
# same as popped element
if head.data == i:
ispalin = True
else:
ispalin = False
break
# Move ahead
head = head.ptr
return ispalin
# Driver Code
# Addition of linked list
one = Node(1)
two = Node(2)
three = Node(3)
four = Node(4)
five = Node(3)
six = Node(2)
seven = Node(1)
# Initialize the next pointer
# of every current pointer
one.ptr = two
two.ptr = three
three.ptr = four
four.ptr = five
five.ptr = six
six.ptr = seven
seven.ptr = None
# Call function to check palindrome or not
result = ispalindrome(one)
print("isPalindrome:", result)
# This code is contributed by Nishtha Goel
C#
// C# program to check if linked list
// is palindrome recursively
using System;
using System.Collections.Generic;
class linkedList{
// Driver code
public static void Main(String []args)
{
Node one = new Node(1);
Node two = new Node(2);
Node three = new Node(3);
Node four = new Node(4);
Node five = new Node(3);
Node six = new Node(2);
Node seven = new Node(1);
one.ptr = two;
two.ptr = three;
three.ptr = four;
four.ptr = five;
five.ptr = six;
six.ptr = seven;
bool condition = isPalindrome(one);
Console.WriteLine("isPalidrome :" + condition);
}
static bool isPalindrome(Node head)
{
Node slow = head;
bool ispalin = true;
Stack<int> stack = new Stack<int>();
while (slow != null)
{
stack.Push(slow.data);
slow = slow.ptr;
}
while (head != null)
{
int i = stack.Pop();
if (head.data == i)
{
ispalin = true;
}
else
{
ispalin = false;
break;
}
head = head.ptr;
}
return ispalin;
}
}
class Node
{
public int data;
public Node ptr;
public Node(int d)
{
ptr = null;
data = d;
}
}
// This code is contributed by amal kumar choubey
Javascript
<script>
/* JavaScript program to check if
linked list is palindrome recursively */
class Node {
constructor(val) {
this.data = val;
this.ptr = null;
}
}
var one = new Node(1);
var two = new Node(2);
var three = new Node(3);
var four = new Node(4);
var five = new Node(3);
var six = new Node(2);
var seven = new Node(1);
one.ptr = two;
two.ptr = three;
three.ptr = four;
four.ptr = five;
five.ptr = six;
six.ptr = seven;
var condition = isPalindrome(one);
document.write("isPalidrome: " + condition);
function isPalindrome(head) {
var slow = head;
var ispalin = true;
var stack = [];
while (slow != null) {
stack.push(slow.data);
slow = slow.ptr;
}
while (head != null) {
var i = stack.pop();
if (head.data == i) {
ispalin = true;
} else {
ispalin = false;
break;
}
head = head.ptr;
}
return ispalin;
}
// This code is contributed by todaysgaurav
</script>
Producción
isPalindrome: true
Complejidad temporal: O(n)
Espacio auxiliar : O (n) ya que estamos usando una pila auxiliar
MÉTODO 2 (Invirtiendo la lista)
Este método requiere O(n) tiempo y O(1) espacio extra.
1) Obtenga el medio de la lista enlazada.
2) Invierta la segunda mitad de la lista enlazada.
3) Compruebe si la primera mitad y la segunda mitad son idénticas.
4) Construya la lista enlazada original invirtiendo la segunda mitad nuevamente y vinculándola nuevamente a la primera mitad
Para dividir la lista en dos mitades, se usa el método 2 de esta publicación.
Cuando varios Nodes son pares, la primera y la segunda mitad contienen exactamente la mitad de los Nodes. Lo desafiante de este método es manejar el caso cuando el número de Nodes es impar. No queremos que el Node medio forme parte de las listas, ya que vamos a compararlos por igualdad. Para casos extraños, usamos una variable separada ‘Node medio’.
C++
// C++ program to check if a linked list is palindrome
#include <bits/stdc++.h>
using namespace std;
// Link list node
struct Node
{
char data;
struct Node* next;
};
void reverse(struct Node**);
bool compareLists(struct Node*, struct Node*);
// Function to check if given linked list is
// palindrome or not
bool isPalindrome(struct Node* head)
{
struct Node *slow_ptr = head, *fast_ptr = head;
struct Node *second_half, *prev_of_slow_ptr = head;
// To handle odd size list
struct Node* midnode = NULL;
// initialize result
bool res = true;
if (head != NULL && head->next != NULL)
{
// Get the middle of the list. Move slow_ptr by 1
// and fast_ptr by 2, slow_ptr will have the middle
// node
while (fast_ptr != NULL && fast_ptr->next != NULL)
{
fast_ptr = fast_ptr->next->next;
// We need previous of the slow_ptr for
// linked lists with odd elements
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr->next;
}
// fast_ptr would become NULL when there
// are even elements in list. And not NULL
// for odd elements. We need to skip the
// middle node for odd case and store it
// somewhere so that we can restore the
// original list
if (fast_ptr != NULL)
{
midnode = slow_ptr;
slow_ptr = slow_ptr->next;
}
// Now reverse the second half and
// compare it with first half
second_half = slow_ptr;
// NULL terminate first half
prev_of_slow_ptr->next = NULL;
// Reverse the second half
reverse(&second_half);
// compare
res = compareLists(head, second_half);
// Construct the original list back
reverse(&second_half); // Reverse the second half again
// If there was a mid node (odd size case)
// which was not part of either first half
// or second half.
if (midnode != NULL)
{
prev_of_slow_ptr->next = midnode;
midnode->next = second_half;
}
else
prev_of_slow_ptr->next = second_half;
}
return res;
}
// Function to reverse the linked list
// Note that this function may change
// the head
void reverse(struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
struct Node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
// Function to check if two input
// lists have same data
bool compareLists(struct Node* head1,
struct Node* head2)
{
struct Node* temp1 = head1;
struct Node* temp2 = head2;
while (temp1 && temp2)
{
if (temp1->data == temp2->data)
{
temp1 = temp1->next;
temp2 = temp2->next;
}
else
return 0;
}
// Both are empty return 1
if (temp1 == NULL && temp2 == NULL)
return 1;
// Will reach here when one is NULL
// and other is not
return 0;
}
// Push a node to linked list. Note
// that this function changes the head
void push(struct Node** head_ref, char new_data)
{
// Allocate node
struct Node* new_node = (struct Node*)malloc(
sizeof(struct 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;
}
// A utility function to print a given linked list
void printList(struct Node* ptr)
{
while (ptr != NULL)
{
cout << ptr->data << "->";
ptr = ptr->next;
}
cout << "NULL" << "\n";
}
// Driver code
int main()
{
// Start with the empty list
struct Node* head = NULL;
char str[] = "abacaba";
int i;
for(i = 0; str[i] != '\0'; i++)
{
push(&head, str[i]);
printList(head);
isPalindrome(head) ? cout << "Is Palindrome"
<< "\n\n" : cout << "Not Palindrome"
<< "\n\n";
}
return 0;
}
// This code is contributed by Shivani
C
/* Program to check if a linked list is palindrome */
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node {
char data;
struct Node* next;
};
void reverse(struct Node**);
bool compareLists(struct Node*, struct Node*);
/* Function to check if given linked list is
palindrome or not */
bool isPalindrome(struct Node* head)
{
struct Node *slow_ptr = head, *fast_ptr = head;
struct Node *second_half, *prev_of_slow_ptr = head;
struct Node* midnode = NULL; // To handle odd size list
bool res = true; // initialize result
if (head != NULL && head->next != NULL) {
/* Get the middle of the list. Move slow_ptr by 1
and fast_ptr by 2, slow_ptr will have the middle
node */
while (fast_ptr != NULL && fast_ptr->next != NULL) {
fast_ptr = fast_ptr->next->next;
/*We need previous of the slow_ptr for
linked lists with odd elements */
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr->next;
}
/* fast_ptr would become NULL when there are even elements in list.
And not NULL for odd elements. We need to skip the middle node
for odd case and store it somewhere so that we can restore the
original list*/
if (fast_ptr != NULL) {
midnode = slow_ptr;
slow_ptr = slow_ptr->next;
}
// Now reverse the second half and compare it with first half
second_half = slow_ptr;
prev_of_slow_ptr->next = NULL; // NULL terminate first half
reverse(&second_half); // Reverse the second half
res = compareLists(head, second_half); // compare
/* Construct the original list back */
reverse(&second_half); // Reverse the second half again
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
if (midnode != NULL) {
prev_of_slow_ptr->next = midnode;
midnode->next = second_half;
}
else
prev_of_slow_ptr->next = second_half;
}
return res;
}
/* Function to reverse the linked list Note that this
function may change the head */
void reverse(struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
struct Node* next;
while (current != NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to check if two input lists have same data*/
bool compareLists(struct Node* head1, struct Node* head2)
{
struct Node* temp1 = head1;
struct Node* temp2 = head2;
while (temp1 && temp2) {
if (temp1->data == temp2->data) {
temp1 = temp1->next;
temp2 = temp2->next;
}
else
return 0;
}
/* Both are empty return 1*/
if (temp1 == NULL && temp2 == NULL)
return 1;
/* Will reach here when one is NULL
and other is not */
return 0;
}
/* Push a node to linked list. Note that this function
changes the head */
void push(struct Node** head_ref, char new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct 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 pochar to the new node */
(*head_ref) = new_node;
}
// A utility function to print a given linked list
void printList(struct Node* ptr)
{
while (ptr != NULL) {
printf("%c->", ptr->data);
ptr = ptr->next;
}
printf("NULL\n");
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
char str[] = "abacaba";
int i;
for (i = 0; str[i] != '\0'; i++) {
push(&head, str[i]);
printList(head);
isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");
}
return 0;
}
Java
/* Java program to check if linked list is palindrome */
class LinkedList {
Node head; // head of list
Node slow_ptr, fast_ptr, second_half;
/* Linked list Node*/
class Node {
char data;
Node next;
Node(char d)
{
data = d;
next = null;
}
}
/* Function to check if given linked list is
palindrome or not */
boolean isPalindrome(Node head)
{
slow_ptr = head;
fast_ptr = head;
Node prev_of_slow_ptr = head;
Node midnode = null; // To handle odd size list
boolean res = true; // initialize result
if (head != null && head.next != null) {
/* Get the middle of the list. Move slow_ptr by 1
and fast_ptr by 2, slow_ptr will have the middle
node */
while (fast_ptr != null && fast_ptr.next != null) {
fast_ptr = fast_ptr.next.next;
/*We need previous of the slow_ptr for
linked lists with odd elements */
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr.next;
}
/* fast_ptr would become NULL when there are even elements
in the list and not NULL for odd elements. We need to skip
the middle node for odd case and store it somewhere so that
we can restore the original list */
if (fast_ptr != null) {
midnode = slow_ptr;
slow_ptr = slow_ptr.next;
}
// Now reverse the second half and compare it with first half
second_half = slow_ptr;
prev_of_slow_ptr.next = null; // NULL terminate first half
reverse(); // Reverse the second half
res = compareLists(head, second_half); // compare
/* Construct the original list back */
reverse(); // Reverse the second half again
if (midnode != null) {
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
prev_of_slow_ptr.next = midnode;
midnode.next = second_half;
}
else
prev_of_slow_ptr.next = second_half;
}
return res;
}
/* Function to reverse the linked list Note that this
function may change the head */
void reverse()
{
Node prev = null;
Node current = second_half;
Node next;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
second_half = prev;
}
/* Function to check if two input lists have same data*/
boolean compareLists(Node head1, Node head2)
{
Node temp1 = head1;
Node temp2 = head2;
while (temp1 != null && temp2 != null) {
if (temp1.data == temp2.data) {
temp1 = temp1.next;
temp2 = temp2.next;
}
else
return false;
}
/* Both are empty return 1*/
if (temp1 == null && temp2 == null)
return true;
/* Will reach here when one is NULL
and other is not */
return false;
}
/* Push a node to linked list. Note that this function
changes the head */
public void push(char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);
/* link the old list off the new one */
new_node.next = head;
/* Move the head to point to new Node */
head = new_node;
}
// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null) {
System.out.print(ptr.data + "->");
ptr = ptr.next;
}
System.out.println("NULL");
}
/* Driver program to test the above functions */
public static void main(String[] args)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
char str[] = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
String string = new String(str);
for (int i = 0; i < 7; i++) {
llist.push(str[i]);
llist.printList(llist.head);
if (llist.isPalindrome(llist.head) != false) {
System.out.println("Is Palindrome");
System.out.println("");
}
else {
System.out.println("Not Palindrome");
System.out.println("");
}
}
}
}
Python3
# Python3 program to check if
# linked list is palindrome
# 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 check if given
# linked list is palindrome or not
def isPalindrome(self, head):
slow_ptr = head
fast_ptr = head
prev_of_slow_ptr = head
# To handle odd size list
midnode = None
# Initialize result
res = True
if (head != None and head.next != None):
# Get the middle of the list.
# Move slow_ptr by 1 and
# fast_ptr by 2, slow_ptr
# will have the middle node
while (fast_ptr != None and
fast_ptr.next != None):
# We need previous of the slow_ptr
# for linked lists with odd
# elements
fast_ptr = fast_ptr.next.next
prev_of_slow_ptr = slow_ptr
slow_ptr = slow_ptr.next
# fast_ptr would become NULL when
# there are even elements in the
# list and not NULL for odd elements.
# We need to skip the middle node for
# odd case and store it somewhere so
# that we can restore the original list
if (fast_ptr != None):
midnode = slow_ptr
slow_ptr = slow_ptr.next
# Now reverse the second half
# and compare it with first half
second_half = slow_ptr
# NULL terminate first half
prev_of_slow_ptr.next = None
# Reverse the second half
second_half = self.reverse(second_half)
# Compare
res = self.compareLists(head, second_half)
# Construct the original list back
# Reverse the second half again
second_half = self.reverse(second_half)
if (midnode != None):
# If there was a mid node (odd size
# case) which was not part of either
# first half or second half.
prev_of_slow_ptr.next = midnode
midnode.next = second_half
else:
prev_of_slow_ptr.next = second_half
return res
# Function to reverse the linked list
# Note that this function may change
# the head
def reverse(self, second_half):
prev = None
current = second_half
next = None
while current != None:
next = current.next
current.next = prev
prev = current
current = next
second_half = prev
return second_half
# Function to check if two input
# lists have same data
def compareLists(self, head1, head2):
temp1 = head1
temp2 = head2
while (temp1 and temp2):
if (temp1.data == temp2.data):
temp1 = temp1.next
temp2 = temp2.next
else:
return 0
# Both are empty return 1
if (temp1 == None and temp2 == None):
return 1
# Will reach here when one is NULL
# and other is not
return 0
# Function to insert a new node
# at the beginning
def push(self, new_data):
# Allocate the Node &
# Put in the data
new_node = Node(new_data)
# Link the old list off the new one
new_node.next = self.head
# Move the head to point to new Node
self.head = new_node
# A utility function to print
# a given linked list
def printList(self):
temp = self.head
while(temp):
print(temp.data, end = "->")
temp = temp.next
print("NULL")
# Driver code
if __name__ == '__main__':
l = LinkedList()
s = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]
for i in range(7):
l.push(s[i])
l.printList()
if (l.isPalindrome(l.head) != False):
print("Is Palindrome\n")
else:
print("Not Palindrome\n")
print()
# This code is contributed by MuskanKalra1
C#
/* C# program to check if linked list is palindrome */
using System;
class LinkedList {
Node head; // head of list
Node slow_ptr, fast_ptr, second_half;
/* Linked list Node*/
public class Node {
public char data;
public Node next;
public Node(char d)
{
data = d;
next = null;
}
}
/* Function to check if given linked list is
palindrome or not */
Boolean isPalindrome(Node head)
{
slow_ptr = head;
fast_ptr = head;
Node prev_of_slow_ptr = head;
Node midnode = null; // To handle odd size list
Boolean res = true; // initialize result
if (head != null && head.next != null) {
/* Get the middle of the list. Move slow_ptr by 1
and fast_ptr by 2, slow_ptr will have the middle
node */
while (fast_ptr != null && fast_ptr.next != null) {
fast_ptr = fast_ptr.next.next;
/*We need previous of the slow_ptr for
linked lists with odd elements */
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr.next;
}
/* fast_ptr would become NULL when there are even elements
in the list and not NULL for odd elements. We need to skip
the middle node for odd case and store it somewhere so that
we can restore the original list */
if (fast_ptr != null) {
midnode = slow_ptr;
slow_ptr = slow_ptr.next;
}
// Now reverse the second half and compare it with first half
second_half = slow_ptr;
prev_of_slow_ptr.next = null; // NULL terminate first half
reverse(); // Reverse the second half
res = compareLists(head, second_half); // compare
/* Construct the original list back */
reverse(); // Reverse the second half again
if (midnode != null) {
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
prev_of_slow_ptr.next = midnode;
midnode.next = second_half;
}
else
prev_of_slow_ptr.next = second_half;
}
return res;
}
/* Function to reverse the linked list Note that this
function may change the head */
void reverse()
{
Node prev = null;
Node current = second_half;
Node next;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
second_half = prev;
}
/* Function to check if two input lists have same data*/
Boolean compareLists(Node head1, Node head2)
{
Node temp1 = head1;
Node temp2 = head2;
while (temp1 != null && temp2 != null) {
if (temp1.data == temp2.data) {
temp1 = temp1.next;
temp2 = temp2.next;
}
else
return false;
}
/* Both are empty return 1*/
if (temp1 == null && temp2 == null)
return true;
/* Will reach here when one is NULL
and other is not */
return false;
}
/* Push a node to linked list. Note that this function
changes the head */
public void push(char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);
/* link the old list off the new one */
new_node.next = head;
/* Move the head to point to new Node */
head = new_node;
}
// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null) {
Console.Write(ptr.data + "->");
ptr = ptr.next;
}
Console.WriteLine("NULL");
}
/* Driver program to test the above functions */
public static void Main(String[] args)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
for (int i = 0; i < 7; i++) {
llist.push(str[i]);
llist.printList(llist.head);
if (llist.isPalindrome(llist.head) != false) {
Console.WriteLine("Is Palindrome");
Console.WriteLine("");
}
else {
Console.WriteLine("Not Palindrome");
Console.WriteLine("");
}
}
}
}
// This code is contributed by Arnab Kundu
Javascript
<script>
/* javascript program to check if linked list is palindrome */
var head; // head of list
var slow_ptr, fast_ptr, second_half;
/* Linked list Node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
/*
* Function to check if given linked list is palindrome or not
*/
function isPalindrome(head) {
slow_ptr = head;
fast_ptr = head;
var prev_of_slow_ptr = head;
var midnode = null; // To handle odd size list
var res = true; // initialize result
if (head != null && head.next != null) {
/*
* Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr
* will have the middle node
*/
while (fast_ptr != null && fast_ptr.next != null) {
fast_ptr = fast_ptr.next.next;
/*
* We need previous of the slow_ptr for linked lists with odd elements
*/
prev_of_slow_ptr = slow_ptr;
slow_ptr = slow_ptr.next;
}
/*
* fast_ptr would become NULL when there are even elements in the list and not
* NULL for odd elements. We need to skip the middle node for odd case and store
* it somewhere so that we can restore the original list
*/
if (fast_ptr != null) {
midnode = slow_ptr;
slow_ptr = slow_ptr.next;
}
// Now reverse the second half and compare it with first half
second_half = slow_ptr;
prev_of_slow_ptr.next = null; // NULL terminate first half
reverse(); // Reverse the second half
res = compareLists(head, second_half); // compare
/* Construct the original list back */
reverse(); // Reverse the second half again
if (midnode != null) {
// If there was a mid node (odd size case) which
// was not part of either first half or second half.
prev_of_slow_ptr.next = midnode;
midnode.next = second_half;
} else
prev_of_slow_ptr.next = second_half;
}
return res;
}
/*
* Function to reverse the linked list Note that this function may change the
* head
*/
function reverse() {
var prev = null;
var current = second_half;
var next;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
second_half = prev;
}
/* Function to check if two input lists have same data */
function compareLists(head1, head2) {
var temp1 = head1;
var temp2 = head2;
while (temp1 != null && temp2 != null) {
if (temp1.data == temp2.data) {
temp1 = temp1.next;
temp2 = temp2.next;
} else
return false;
}
/* Both are empty return 1 */
if (temp1 == null && temp2 == null)
return true;
/*
* Will reach here when one is NULL and other is not
*/
return false;
}
/*
* Push a node to linked list. Note that this function changes the head
*/
function push( new_data) {
/*
* Allocate the Node & Put in the data
*/
var new_node = new Node(new_data);
/* link the old list off the new one */
new_node.next = head;
/* Move the head to point to new Node */
head = new_node;
}
// A utility function to print a given linked list
function printList(ptr) {
while (ptr != null) {
document.write(ptr.data + "->");
ptr = ptr.next;
}
document.write("NULL<br/>");
}
/* Driver program to test the above functions */
/* Start with the empty list */
var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ];
var string = str.toString();
for (i = 0; i < 7; i++) {
push(str[i]);
printList(head);
if (isPalindrome(head) != false) {
document.write("Is Palindrome");
document.write("<br/>");
} else {
document.write("Not Palindrome");
document.write("<br/>");
}
}
// This code contributed by gauravrajput1
</script>
Producción:
a->NULL Is Palindrome b->a->NULL Not Palindrome a->b->a->NULL Is Palindrome c->a->b->a->NULL Not Palindrome a->c->a->b->a->NULL Not Palindrome b->a->c->a->b->a->NULL Not Palindrome a->b->a->c->a->b->a->NULL Is Palindrome
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
MÉTODO 3 (Uso de la recursividad)
Use dos punteros a la izquierda y a la derecha. Muévase hacia la derecha y hacia la izquierda usando la recursividad y verifique el seguimiento en cada llamada recursiva.
1) La sublista es un palíndromo.
2) Los valores a la izquierda y a la derecha actuales coinciden.
Si las dos condiciones anteriores son verdaderas, devuelva verdadero.
La idea es usar la pila de llamadas de función como un contenedor. Atraviesa recursivamente hasta el final de la lista. Cuando regresemos del último NULL, estaremos en el último Node. El último Node que se comparará con el primer Node de la lista.
Para acceder al primer Node de la lista, necesitamos que el encabezado de la lista esté disponible en la última llamada de recursividad. Por lo tanto, pasamos de cabeza también a la función recursiva. Si ambos coinciden, necesitamos comparar (2, n-2) Nodes. Nuevamente, cuando la recursión vuelve al (n-2) Node, necesitamos una referencia al segundo Node desde la cabeza. Avanzamos el puntero de cabecera en la llamada anterior, para referirnos al siguiente Node de la lista.
Sin embargo, el truco está en identificar un doble puntero. Pasar un solo puntero es tan bueno como pasar por valor, y pasaremos el mismo puntero una y otra vez. Necesitamos pasar la dirección del puntero principal para reflejar los cambios en las llamadas recursivas principales.
Gracias a Sharad Chandra por sugerir este enfoque.
C++
// Recursive program to check if a given linked list is palindrome
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
struct node {
char data;
struct node* next;
};
// Initial parameters to this function are &head and head
bool isPalindromeUtil(struct node** left, struct node* right)
{
/* stop recursion when right becomes NULL */
if (right == NULL)
return true;
/* If sub-list is not palindrome then no need to
check for current left and right, return false */
bool isp = isPalindromeUtil(left, right->next);
if (isp == false)
return false;
/* Check values at current left and right */
bool isp1 = (right->data == (*left)->data);
/* Move left to next node */
*left = (*left)->next;
return isp1;
}
// A wrapper over isPalindromeUtil()
bool isPalindrome(struct node* head)
{
isPalindromeUtil(&head, head);
}
/* Push a node to linked list. Note that this function
changes the head */
void push(struct node** head_ref, char new_data)
{
/* allocate node */
struct node* new_node = (struct node*)malloc(sizeof(struct 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;
}
// A utility function to print a given linked list
void printList(struct node* ptr)
{
while (ptr != NULL) {
cout << ptr->data << "->";
ptr = ptr->next;
}
cout << "NULL\n" ;
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
char str[] = "abacaba";
int i;
for (i = 0; str[i] != '\0'; i++) {
push(&head, str[i]);
printList(head);
isPalindrome(head) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n";
}
return 0;
}
//this code is contributed by shivanisinghss2110
C
// Recursive program to check if a given linked list is palindrome
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct node {
char data;
struct node* next;
};
// Initial parameters to this function are &head and head
bool isPalindromeUtil(struct node** left, struct node* right)
{
/* stop recursion when right becomes NULL */
if (right == NULL)
return true;
/* If sub-list is not palindrome then no need to
check for current left and right, return false */
bool isp = isPalindromeUtil(left, right->next);
if (isp == false)
return false;
/* Check values at current left and right */
bool isp1 = (right->data == (*left)->data);
/* Move left to next node */
*left = (*left)->next;
return isp1;
}
// A wrapper over isPalindromeUtil()
bool isPalindrome(struct node* head)
{
isPalindromeUtil(&head, head);
}
/* Push a node to linked list. Note that this function
changes the head */
void push(struct node** head_ref, char new_data)
{
/* allocate node */
struct node* new_node = (struct node*)malloc(sizeof(struct 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 pochar to the new node */
(*head_ref) = new_node;
}
// A utility function to print a given linked list
void printList(struct node* ptr)
{
while (ptr != NULL) {
printf("%c->", ptr->data);
ptr = ptr->next;
}
printf("NULL\n");
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
char str[] = "abacaba";
int i;
for (i = 0; str[i] != '\0'; i++) {
push(&head, str[i]);
printList(head);
isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");
}
return 0;
}
Java
// Java program for the above approach
public class LinkedList{
// Head of the list
Node head;
Node left;
public class Node
{
public char data;
public Node next;
// Linked list node
public Node(char d)
{
data = d;
next = null;
}
}
// Initial parameters to this function are
// &head and head
boolean isPalindromeUtil(Node right)
{
left = head;
// Stop recursion when right becomes null
if (right == null)
return true;
// If sub-list is not palindrome then no need to
// check for the current left and right, return
// false
boolean isp = isPalindromeUtil(right.next);
if (isp == false)
return false;
// Check values at current left and right
boolean isp1 = (right.data == left.data);
left = left.next;
// Move left to next node;
return isp1;
}
// A wrapper over isPalindrome(Node head)
boolean isPalindrome(Node head)
{
boolean result = isPalindromeUtil(head);
return result;
}
// Push a node to linked list. Note that
// this function changes the head
public void push(char new_data)
{
// Allocate the node and put in the data
Node new_node = new Node(new_data);
// Link the old list off the the new one
new_node.next = head;
// Move the head to point to new node
head = new_node;
}
// A utility function to print a
// given linked list
void printList(Node ptr)
{
while (ptr != null)
{
System.out.print(ptr.data + "->");
ptr = ptr.next;
}
System.out.println("Null");
}
// Driver Code
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
for(int i = 0; i < 7; i++)
{
llist.push(str[i]);
llist.printList(llist.head);
if (llist.isPalindrome(llist.head))
{
System.out.println("Is Palindrome");
System.out.println("");
}
else
{
System.out.println("Not Palindrome");
System.out.println("");
}
}
}
}
// This code is contributed by abhinavjain194
Python3
# Python program for the above approach
# Head of the list
head = None
left = None
class Node:
def __init__(self, val):
self.data = val
self.next = None
# Initial parameters to this function are
# &head and head
def isPalindromeUtil(right):
global head, left
left = head
# Stop recursion when right becomes null
if (right == None):
return True
# If sub-list is not palindrome then no need to
# check for the current left and right, return
# false
isp = isPalindromeUtil(right.next)
if (isp == False):
return False
# Check values at current left and right
isp1 = (right.data == left.data)
left = left.next
# Move left to next node;
return isp1
# A wrapper over isPalindrome(Node head)
def isPalindrome(head):
result = isPalindromeUtil(head)
return result
# Push a node to linked list. Note that
# this function changes the head
def push(new_data):
global head
# Allocate the node and put in the data
new_node = Node(new_data)
# Link the old list off the the new one
new_node.next = head
# Move the head to point to new node
head = new_node
# A utility function to print a
# given linked list
def printList(ptr):
while (ptr != None):
print(ptr.data, end="->")
ptr = ptr.next
print("Null ")
# Driver Code
str = ['a', 'b', 'a', 'c', 'a', 'b', 'a']
for i in range(0, 7):
push(str[i])
printList(head)
if (isPalindrome(head) and i != 0):
print("Is Palindrome\n")
else:
print("Not Palindrome\n")
# This code is contributed by saurabh_jaiswal.
C#
/* C# program to check if linked list
is palindrome recursively */
using System;
public class LinkedList
{
Node head; // head of list
Node left;
/* Linked list Node*/
public class Node
{
public char data;
public Node next;
public Node(char d)
{
data = d;
next = null;
}
}
// Initial parameters to this function are &head and head
Boolean isPalindromeUtil(Node right)
{
left = head;
/* stop recursion when right becomes NULL */
if (right == null)
return true;
/* If sub-list is not palindrome then no need to
check for current left and right, return false */
Boolean isp = isPalindromeUtil(right.next);
if (isp == false)
return false;
/* Check values at current left and right */
Boolean isp1 = (right.data == (left).data);
/* Move left to next node */
left = left.next;
return isp1;
}
// A wrapper over isPalindromeUtil()
Boolean isPalindrome(Node head)
{
Boolean result = isPalindromeUtil(head);
return result;
}
/* Push a node to linked list. Note that this function
changes the head */
public void push(char new_data)
{
/* Allocate the Node &
Put in the data */
Node new_node = new Node(new_data);
/* link the old list off the new one */
new_node.next = head;
/* Move the head to point to new Node */
head = new_node;
}
// A utility function to print a given linked list
void printList(Node ptr)
{
while (ptr != null)
{
Console.Write(ptr.data + "->");
ptr = ptr.next;
}
Console.WriteLine("NULL");
}
/* Driver code */
public static void Main(String[] args)
{
/* Start with the empty list */
LinkedList llist = new LinkedList();
char []str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
//String string = new String(str);
for (int i = 0; i < 7; i++) {
llist.push(str[i]);
llist.printList(llist.head);
if (llist.isPalindrome(llist.head) != false)
{
Console.WriteLine("Is Palindrome");
Console.WriteLine("");
}
else
{
Console.WriteLine("Not Palindrome");
Console.WriteLine("");
}
}
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// javascript program for the above approach
// Head of the list
var head;
var left;
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Initial parameters to this function are
// &head and head
function isPalindromeUtil( right) {
left = head;
// Stop recursion when right becomes null
if (right == null)
return true;
// If sub-list is not palindrome then no need to
// check for the current left and right, return
// false
var isp = isPalindromeUtil(right.next);
if (isp == false)
return false;
// Check values at current left and right
var isp1 = (right.data == left.data);
left = left.next;
// Move left to next node;
return isp1;
}
// A wrapper over isPalindrome(Node head)
function isPalindrome( head) {
var result = isPalindromeUtil(head);
return result;
}
// Push a node to linked list. Note that
// this function changes the head
function push( new_data) {
// Allocate the node and put in the data
var new_node = new Node(new_data);
// Link the old list off the the new one
new_node.next = head;
// Move the head to point to new node
head = new_node;
}
// A utility function to print a
// given linked list
function printList( ptr) {
while (ptr != null) {
document.write(ptr.data + "->");
ptr = ptr.next;
}
document.write("Null ");
document.write("<br>");
}
// Driver Code
var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ];
for (var i = 0; i < 7; i++) {
push(str[i]);
printList(head);
if (isPalindrome(head)) {
document.write("Is Palindrome");
document.write("<br/>");
document.write("<br>");
} else {
document.write("Not Palindrome");
document.write("<br/>");
document.write("<br/>");
}
}
// This code contributed by aashish1995
</script>
Producción:
a->NULL Not Palindrome b->a->NULL Not Palindrome a->b->a->NULL Is Palindrome c->a->b->a->NULL Not Palindrome a->c->a->b->a->NULL Not Palindrome b->a->c->a->b->a->NULL Not Palindrome a->b->a->c->a->b->a->NULL Is Palindrome
Complejidad de tiempo: O(n)
Espacio auxiliar: O(n) si se considera el tamaño de la pila de llamadas de función; de lo contrario, O(1).
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA