C Program 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.

METHOD 1 (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 into two halves, method 2 of this post is used. 

When a 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 the number of nodes is odd. We don’t want the middle node as part of the lists as we are going to compare them for equality. For odd cases, we use a separate variable ‘midnode’.

C

// 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;

    // 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_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;

        // 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 the second half again
        reverse(&second_half);

        // 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 of 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)
    {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL");
}

// Driver code
int main()
{
    // Start with the empty list
    struct Node* head = NULL;
    char str[] = "abacaba";
    int i;

    for (i = 0; str[i] != ''; i++)
    {
        push(&head, str[i]);
        printList(head);
        isPalindrome(head) ? printf("Is Palindrome") : printf("Not Palindrome");
    }
    return 0;
}

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 Palindrome

Time Complexity: O(n) 
Auxiliary Space: O(1)  

METHOD 2 (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 the list. When we return from the last NULL, we will be at the last node. The last node is to be compared with the first node of the list.

In order to access the first node of the list, we need the 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 a 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 <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 of 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)
    {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("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) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");
    }

    return 0;
}

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).

Please refer complete article on Function to check if a singly linked list is palindrome for more details!