# C Program To Check If A Singly Linked List Is Palindrome

• Last Updated : 24 Dec, 2021

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 ``#include ``#include ` `// 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 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)``    ``{``        ``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 ``#include ``#include ` `// 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)``    ``{``        ``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!

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