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.

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 stack and compare data of popped node with 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 : 

Output 

 isPalindrome: true

The time complexity of the above method is O(n).

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’. 

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 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.  

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

class LinkedList {
    Node head; // head of list
    Node left;

    /* Linked list Node*/
    class Node {
        char data;
        Node next;

        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) {
            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("");
            }
        }
    }
}

// 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).
 

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