<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 {
private:
    ListNode* findMid(ListNode* head){
        if(!head){
            return head;
        }
        ListNode* slow=head;
        ListNode* fast=head;
        while(fast!=NULL && fast->next!=NULL &&  fast->next->next!=NULL){
            fast=fast->next->next;
            slow=slow->next;
        }
        return slow;
    }


    ListNode* rev(ListNode* head){
        if(!head){
            return head;
        }
        ListNode* prev=NULL;
        ListNode* curr=head;
        ListNode* front=head->next;
        while(curr!=NULL && front!=NULL){
            curr->next=prev;
            prev=curr;
            curr=front;
            front=front->next;
        }
        curr->next=prev;
        return curr;
    }


public:
    bool isPalindrome(ListNode* head) {
        if(!head || !head->next){
            return true;
        }
        ListNode* start=head;
        ListNode* mid=findMid(head);
        ListNode* revHead=rev(mid->next);
       while(revHead!=NULL){
            if(revHead->val!=start->val){
                rev(revHead);
                return false;
            }
            start=start->next;
            revHead=revHead->next;
       } 
       rev(revHead);
       return true;
    }
};
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