Palindrome Linked List in Python

Suppose we have a linked list. We have to check whether the list elements are forming a palindrome or not. So if the list element is like [1,2,3,2,1], then this is a palindrome.

To solve this, we will follow these steps −

• fast := head, slow := head, rev := None and flag := 1

• if the head is empty, then return true

• while fast and next of fast is available

  • if next of the next of fast is available, then set flag := 0 and break the loop

  • fast := next of the next of fast

  • temp := slow, slow := next of slow

  • next of temp := rev, and rev := temp

• fast := next of slow, and next of slow := rev

• if flag is set, then slow := next of slow

• while fast and slow are not None,

  • if the value of fast is not the same as the value of slow, then return false

  • fast := next of fast, and slow := next of slow

• return true

Example (Python)

Let us see the following implementation to get a better understanding −

 Live Demo

class ListNode:
   def __init__(self, data, next = None):
      self.data = data
      self.next = next
def make_list(elements):
   head = ListNode(elements[0])
   for element in elements[1:]:
      ptr = head
      while ptr.next:
         ptr = ptr.next
      ptr.next = ListNode(element)
   return head
class Solution(object):
   def isPalindrome(self, head):
      fast,slow = head,head
      rev = None
      flag = 1
      if not head:
         return True
      while fast and fast.next:
         if not fast.next.next:
            flag = 0
            break
         fast = fast.next.next
         temp = slow
         slow = slow.next
         temp.next = rev
         rev = temp
      #print(fast.val)
      fast = slow.next
      slow.next = rev
      if flag:
         slow = slow.next
      while fast and slow:
         if fast.data != slow.data:
            return False
         fast = fast.next
         slow = slow.next
      return True
head = make_list([1,2,3,2,1])
ob1 = Solution()
print(ob1.isPalindrome(head))

Input

[1,2,3,2,1]

Output

True