<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:
ListNode* middle(ListNode* head){
if(head->next!=nullptr){
      ListNode* slow=head->next;
      ListNode* fast=slow->next; 
      while(fast!=nullptr && fast->next != nullptr){
        slow=slow->next;
        fast=fast->next;
        if(fast!=nullptr) fast=fast->next;
      }
      return slow;
 }
 else return head;
}
ListNode* reverseList(ListNode* head){
if(head==nullptr||head->next == nullptr)
return head;
ListNode* n=reverseList(head->next);
head->next->next=head;
head->next=nullptr;
return n;
}
bool isPalindrome(ListNode* head){
  ListNode* m=middle(head);
  ListNode* temp=reverseList(m);
  while(head!=m){
    if(head->val!=temp->val)
    return false;
    else{
        head=head->next; 
        temp=temp->next;
    }
  }
  return true;
  }
};
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