<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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* temp=head;
        ListNode* prev=NULL;
        while(head!=NULL && head->next!=NULL){
            temp=head->next;
            head->next=prev;
            prev=head;
            head=temp;

        }
        if(head!=NULL && head->next==NULL && prev!=NULL){
            head->next=prev;
        }

        return head;
    }
    bool isPalindrome(ListNode* head) {
    

    //optimal approach
    if (head == NULL || head->next == NULL) {
        return true;
    }

    ListNode* slow = head;
    ListNode* fast = head;
    stack<int> firstHalf;

    // Move fast pointer twice as fast as slow pointer
    // and push the first half values onto the stack
    while (fast != NULL && fast->next != NULL) {
        firstHalf.push(slow->val);
        slow = slow->next;
        fast = fast->next->next;
    }

    // If the list has an odd number of elements, skip the middle element
    if (fast != NULL) {
        slow = slow->next;
    }

    ListNode* head2=reverseList(slow);
    
    while(head && head2){
        if(head->val!=head2->val){
            return false;
        }
        head=head->next;
        head2=head2->next;
    }

    return true;

    }
};
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