<p>Given the <code>head</code> of a singly linked list, return <code>true</code><em> if it is a </em><span data-keyword="palindrome-sequence"><em>palindrome</em></span><em> or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal1linked-list.jpg" style="width: 422px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2,2,1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/pal2linked-list.jpg" style="width: 182px; height: 62px;" />
<pre>
<strong>Input:</strong> head = [1,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do it in <code>O(n)</code> time and <code>O(1)</code> space?
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
  def isPalindrome(self, head: Optional[ListNode]) -> bool:
    def reverseList(node):
        prev = None
        current = node
        while current:
            next_node = current.next
            current.next = prev
            prev = current
            current = next_node
        return prev

    # Function to find the middle of the linked list
    def findMiddle(head):
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

    if not head or not head.next:
        return True

    # Step 1: Find the middle of the linked list
    mid = findMiddle(head)

    # Step 2: Reverse the second half of the linked list
    mid = reverseList(mid)

    # Step 3: Compare the first and the second half nodes
    first_half = head
    second_half = mid
    result = True
    while second_half:
        if first_half.val != second_half.val:
            result = False
            break
        first_half = first_half.next
        second_half = second_half.next

    # Optional: Restore the list to its original form by reversing the second half again
    reverseList(mid)

    return result
        
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