<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:


//     bool isPalindrome(ListNode* head) {
//         // vector<int> ls ;
//         // ListNode* temp = head;
//         // while(temp!= NULL){
//         //     ls.push_back(temp->val);
//         //     temp = temp->next;
//         // }
//         // int left = 0;
//         // int right = ls.size()-1;
//         // while(left< right && ls[left] == ls[right] ){
//         //     left++;
//         //     right --;
//         // }
//         // return left>= right;

        
//     // }
// };
class Solution {
public:
    ListNode* reverse(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* curr = head;
        while (curr != nullptr) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
    
    bool isPalindrome(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        if(head== NULL || head->next == NULL){
            return true;
        }
        while (fast->next!= nullptr && fast->next->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* rev = reverse(slow->next);
        while (rev != nullptr) {
            if (head->val != rev->val) {
                reverse(rev);
                return false;
            }
            head = head->next;
            rev = rev->next;
        }
        reverse(rev);
        return true;
    }
};
94755