<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?
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if (head == NULL || head->next == NULL) return true;
        
        // Step 1: Find the middle of the linked list using two pointers
        ListNode *slow = head, *fast = head;
        
        while (fast->next != NULL && fast->next->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        
        // Step 2: Reverse the second half of the list
        ListNode* secondHalf = reverseList(slow->next);
        ListNode* firstHalf = head;
        
        // Step 3: Compare the first and the second half
        ListNode* secondHalfCopy = secondHalf; // Save the head of reversed part to restore later
        while (secondHalf != NULL) {
            if (firstHalf->val != secondHalf->val) {
                return false; // Not a palindrome
            }
            firstHalf = firstHalf->next;
            secondHalf = secondHalf->next;
        }
        
        // Step 4 (optional): Restore the second half of the list (optional step)
        slow->next = reverseList(secondHalfCopy);
        
        return true;
    }
    
    // Helper function to reverse a linked list
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = NULL;
        while (head != NULL) {
            ListNode* nextNode = head->next;
            head->next = prev;
            prev = head;
            head = nextNode;
        }
        return prev;
    }
};

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