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 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.
1) Traverse the given list from head to tail and push every visited node to stack.
2) Traverse the list again. For every visited node, pop a node from stack and compare data of popped node with currently visited node.
3) If all nodes matched, then return true, else false.

Time complexity of above method is O(n), but it requires O(n) extra space. Following methods solve this with constant extra space.

METHOD 2 (By reversing the list)

This method takes O(n) time and O(1) extra space.
1) Get the middle of the linked list.
2) Reverse the second half of the linked list.
3) Check if the first half and second half are identical.
4) Construct the original linked list by reversing the second half again and attaching it back to the first half

To divide the list in two halves, method 2 of this post is used.
When number of nodes are even, the first and second half contain exactly half nodes. The challenging thing in this method is to handle the case when number of nodes are odd. We don’t want the middle node as part of any of the lists as we are going to compare them for equality. For odd case, we use a separate variable ‘midnode’.

C

```/* Program to check if a linked list is palindrome */
#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>

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 */
{
struct Node *second_half, *prev_of_slow_ptr = head;
struct Node *midnode = NULL;  // To handle odd size list
bool res = true; // initialize result

{
/* Get the middle of the list. Move slow_ptr by 1
and fast_ptrr 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 */
{
struct Node* prev   = NULL;
struct Node* next;
while (current != NULL)
{
next  = current->next;
current->next = prev;
prev = current;
current = next;
}
}

/* Function to check if two input lists have same data*/
{

while (temp1 && temp2)
{
if (temp1->data == temp2->data)
{
temp1 = temp1->next;
temp2 = temp2->next;
}
else return 0;
}

/* Both are empty reurn 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
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 */

/* move the head to pochar to the 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");
}

/* Drier program to test above function*/
int main()
{
char str[] = "abacaba";
int i;

for (i = 0; str[i] != '\0'; i++)
{
printf("Not Palindrome\n\n");
}

return 0;
}
```

Java

```/* Java program to check if linked list is palindrome */

{
Node slow_ptr, fast_ptr,second_half;

class Node
{
char data;
Node next;

Node(char d)
{
data = d;
next = null;
}
}

/* Function to check if given linked list is
palindrome or not */
{
Node midnode = null;  // To handle odd size list
boolean res = true; // initialize result

{
/* Get the middle of the list. Move slow_ptr by 1
and fast_ptrr 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*/
{

while (temp1 != null && temp2 != null)
{
if (temp1.data == temp2.data)
{
temp1 = temp1.next;
temp2 = temp2.next;
} else
return false;
}

/* Both are empty reurn 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
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 */

/* Move the head to point to 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) {

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]);
{
System.out.println("Is Palindrome");
System.out.println("");
}
else
{
System.out.println("Not Palindrome");
System.out.println("");
}
}
}
}

```

Output:
```a->NULL
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```

Time 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 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 container. Recursively traverse till the end of list. When we return from last NULL, we will be at 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 head. We advance the head pointer in previous call, to refer to next node in the list.

However, the trick in identifying double pointer. Passing 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 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 <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

struct node
{
char data;
struct node* next;
};

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()
{
}

/* Push a node to linked list. Note that this function
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 */

/* move the head to pochar to the 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");
}

/* Drier program to test above function*/
int main()
{
char str[] = "abacaba";
int i;

for (i = 0; str[i] != '\0'; i++)
{
printf("Not Palindrome\n\n");
}

return 0;
}
```

Java

```/* Java program to check if linked list is palindrome recursively */

{
Node left;

class Node
{
char data;
Node next;

Node(char d)
{
data = d;
next = null;
}
}

boolean isPalindromeUtil(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 */
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()
{
return result;
}

/* Push a node to linked list. Note that this function
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 */

/* Move the head to point to 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)
{

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]);
{
System.out.println("Is Palindrome");
System.out.println("");
}
else
{
System.out.println("Not Palindrome");
System.out.println("");
}
}
}
}

// This code has been contributed by Mayank Jaiswal(mayank_24)
```

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 Palindrome```

Time Complexity: O(n)
Auxiliary Space: O(n) if Function Call Stack size is considered, otherwise O(1).

Asked in: Amazon, KLA Tencor, Kritikal Solutions, Microsoft, Snapdeal, Yodlee Infotech

Please comment if you find any bug in the programs/algorithms or a better way to do the same.

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