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

    ListNode* revMidNode(ListNode* mid){
        ListNode*prev = NULL;
        ListNode*temp = mid;
        ListNode*front = mid->next;
        while(front != NULL){
            temp->next = prev;
            prev = temp;
            temp = front;
            front = front->next;
        }
        temp->next = prev;
        prev = temp;
        temp = front;
        delete temp;
        delete front;
        return prev;
        
    }

    bool isPalindrome(ListNode* head) {
      ListNode* temp = head;
      ListNode* mid = midNode(head); 
      ListNode*revHead = revMidNode(mid);
      int cnt = 0; 
      while(temp != mid  && revHead != NULL){
        if(temp->val != revHead->val){
            cnt++;
            break;
        }
        temp = temp->next;
        revHead = revHead->next;
      }
      revMidNode(temp);
      if(cnt == 1) return false;
      return true;
    }  

};
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