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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 −
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
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