Function to check if a singly linked list is palindrome
Given a singly linked list of characters, write a function that returns true if the given list is a palindrome, else false.
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
METHOD 1 [Use a Stack]
- A simple solution is to use a stack of list nodes. This mainly involves three steps.
- Traverse the given list from head to tail and push every visited node to stack.
- Traverse the list again. For every visited node, pop a node from the stack and compare data of popped node with the currently visited node.
- If all nodes matched, then return true, else false.
Below image is a dry run of the above approach:
Below is the implementation of the above approach :
C++
|
#include
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 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]
coutnext;
// 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 data next;
}
cout 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 pallindrome 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
|
/* 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
"];
}
/* 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["
"];
} else {
document.write["Not Palindrome"];
document.write["
"];
}
}
// This code contributed by gauravrajput1
|
Output:
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 PalindromeTime Complexity: O[n]
Auxiliary Space: O[1]
METHOD 3 [Using Recursion]
Use two pointers left and right. Move right and left using recursion and check for following in each recursive call.
1] Sub-list is a palindrome.
2] Value at current left and right are matching.
If both above conditions are true then return true.
The idea is to use function call stack as a container. Recursively traverse till the end of list. When we return from last NULL, we will be at the last node. The last node to be compared with first node of list.
In order to access first node of list, we need list head to be available in the last call of recursion. Hence, we pass head also to the recursive function. If they both match we need to compare [2, n-2] nodes. Again when recursion falls back to [n-2]nd node, we need reference to 2nd node from the head. We advance the head pointer in the previous call, to refer to the next node in the list.
However, the trick is identifying a double-pointer. Passing a single pointer is as good as pass-by-value, and we will pass the same pointer again and again. We need to pass the address of the head pointer for reflecting the changes in parent recursive calls.
Thanks to Sharad Chandra for suggesting this approach.
C++
|
// Recursive program to check if a given linked list is palindrome
#include
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 data next;
}
cout 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
|
// 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["
"];
}
// 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["
"];
document.write["
"];
} else {
document.write["Not Palindrome"];
document.write["
"];
document.write["
"];
}
}
// This code contributed by aashish2995
|
Output:
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 PalindromeTime Complexity: O[n]
Auxiliary Space: O[n] if Function Call Stack size is considered, otherwise O[1].
Article Tags :
Linked List
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palindrome
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Practice Tags :
Accolite
Amazon
Microsoft
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Adobe
Yodlee Infotech
KLA Tencor
Kritikal Solutions
Linked List
palindrome
Read Full Article
Program to determine whether a singly linked list is the palindrome
Explanation
In this program, we need to check whether given singly linked list is a palindrome or not. A palindromic list is the one which is equivalent to the reverse of itself.
The list given in the above figure is a palindrome since it is equivalent to its reverse list, i.e., 1, 2, 3, 2, 1. To check whether a list is a palindrome, we traverse the list and check if any element from the starting half doesn't match with any element from the ending half, then we set the variable flag to false and break the loop.
In the last, if the flag is false, then the list is palindrome otherwise not. The algorithm to check whether a list is a palindrome or not is given below.
Algorithm
- Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
- Create another class Palindrome which has three attributes: head, tail, and size.
- addNode[] will add a new node to the list:
- Create a new node.
- It first checks, whether the head is equal to null which means the list is empty.
- If the list is empty, both head and tail will point to a newly added node.
- If the list is not empty, the new node will be added to end of the list such that tail's next will point to a newly added node. This new node will become the new tail of the list.
- reverseList[] will reverse the order of the node present in the list:
- Node current will represent a node from which a list needs to be reversed.
- Node prevNode represent the previous node to current and nextNode represent the node next to current.
- The list will be reversed by swapping the prevNode with nextNode for each node.
- isPalindrome[] will check whether given list is palindrome or not:
- Declare a node current which will initially point to head node.
- The variable flag will store a boolean value true.
- Calculate the mid-point of the list by dividing the size of the list by 2.
- Traverse through the list till current points to the middle node.
- Reverse the list after the middle node until the last node using reverseList[]. This list will be the second half of the list.
- Now, compare nodes of first half and second half of the list.
- If any of the nodes don't match then, set a flag to false and break the loop.
- If the flag is true after the loop which denotes that list is a palindrome.
- If the flag is false, then the list is not a palindrome.
- display[] will display the nodes present in the list:
- Define a node current which will initially point to the head of the list.
- Traverse through the list till current points to null.
- Display each node by making current to point to node next to it in each iteration.
Solution
Python
Output:
Nodes of the singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
C
Output:
Nodes of the singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
JAVA
Output:
Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
C#
Output:
Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
PHP
Output:
Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
Using Stack:
Algorithm
- Traverse the linked list and push the value in Stack.
- Now, again traverse the linked list and while doing so pop the values from the stack.If stack become empty it means given Linked list is pallindrom otherwise not.
Given below is a Singly Linked List:
Traversing the list:
Step 1:
Currently,pointer is at the beginning of the list which points to Node value 2
Step 2:
Now,After the Node 2 is visited pointer moves forward and the visited node will be push into the stack.
Step 3:
Node 3 is visited,push Node 3 into Stack and moves the pointer forward.
Step 4:
Node 3 is visited,push Node 3 into Stack and moves the pointer forward.
Step 5:
Node 2 is visited,push Node 2 into Stack and moves the pointer forward until it's NULL.
Step 6:
Now,our approach will be to traverse the list once again from the beginning and this time will be ke checking is the Stack.top[] is equal to the current node or not.Here,top value of stack is 2 and current node value of list is 2 they both are equal then pop the top value from the stack and move the pointer forward,repeat these steps until list found to be NULL.
Whenever you found that top value of stack is not equal to the current value of node then return false.