Python 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 (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 : 

Python3

# Python3 program to check if linked
# list is palindrome using stack
class Node:
    def __init__(self, data):
        self.data = data
        self.ptr = None

# Function to check if the linked list
# is palindrome or not
def ispalindrome(head):

    # Temp pointer
    slow = head

    # Declare a stack
    stack = []

    ispalin = True

    # Push all elements of the list
    # to the stack
    while slow != None:
        stack.append(slow.data)

        # Move ahead
        slow = slow.ptr

    # Iterate in the list again and
    # check by popping from the stack
    while head != None:

        # Get the top most element
        i = stack.pop()

        # Check if data is not
        # same as popped element
        if head.data == i:
            ispalin = True
        else:
            ispalin = False
            break

        # Move ahead
        head = head.ptr

    return ispalin

# Driver Code

# Addition of linked list
one = Node(1)
two = Node(2)
three = Node(3)
four = Node(4)
five = Node(3)
six = Node(2)
seven = Node(1)

# Initialize the next pointer
# of every current pointer
one.ptr = two
two.ptr = three
three.ptr = four
four.ptr = five
five.ptr = six
six.ptr = seven
seven.ptr = None

# Call function to check
# palindrome or not
result = ispalindrome(one)

print("isPalindrome:", result)

# This code is contributed by Nishtha Goel

Output:

isPalindrome: true

Time complexity: O(n), where n represents the length of the given linked list.

Auxiliary Space: O(n), for using a stack, where n represents the length of the given linked list.

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 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’. 

Python3

# Python 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 palindrome 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_ptrr 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 the 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 of the new one
        new_node.next = self.head

        # Move the head to point to the
        # 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")
        else:
            print("Not Palindrome")
        print()

# This code is contributed by MuskanKalra1

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)

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