<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) {
        if(!head->next) return true;
        int length = 0;
        ListNode* node = head;
        while(node!=nullptr){
            node = node->next;
            length++;
        }
        ListNode* prev = nullptr;
        ListNode* next = nullptr;
        for(int i = 0; i < length / 2; i++){
            next = head->next;
            head->next = prev;
            prev = head;
            head = next;
        }

        ListNode* left_head = prev;
        ListNode* right_head;
        if(length%2==0){
            right_head = head;
        }else{
            right_head = head->next;
        }

        while(left_head != nullptr && right_head != nullptr){
            if(left_head->val!=right_head->val){
                return false;
            }
            left_head = left_head->next;
            right_head = right_head->next;
        }

        return left_head == nullptr && right_head == nullptr;




    }
};
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