<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* reverse(ListNode* head){
        if(head==NULL || head->next==NULL)
            return head;
        
        ListNode* newHead = reverse(head->next);
        ListNode* front = head->next;
        front->next = head;
        head->next = NULL;
        return newHead;
        
    }

//Aporoach 2
    bool isPalindrome(ListNode* head) {
        if(head->next==NULL)
            return true;
        int cnt =0;
        ListNode* mover = head;
        while(mover){
            mover = mover->next;
            cnt++;
        }
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* temp =NULL;
        while(fast && fast->next){
            temp = slow;
            slow = slow->next;
            fast = fast->next->next;
        }if(cnt%2==0){
            fast = slow;
            temp->next = NULL;
        }
        else {
            fast = slow->next;
            temp->next= NULL;
        }
        slow = reverse(head);
        while(slow&& fast){
            if(slow->val!=fast->val)
                return false;
            slow= slow->next;
            fast = fast->next;
        }
        if((slow==NULL && fast==NULL)||(slow->next==NULL&&fast==NULL)||(fast->next==NULL && slow==NULL))
            return true;
        return false;
            
        
    }

//Approach 1:
//This solution requires O(n) time complexity and O(n) space.
    // bool isPalindrome(ListNode* head) {
        
    //     ListNode* head2 = reverse(head);

    //     while(head && head2){
    //         if(head->val != head2->val)
    //             return false;
    //         head = head->next;
    //         head2 = head2->next;
    //     }

    //     return true;
        
    // }
};
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