c++代码：请编写算法判断链表是否为回文链表。以1 2和1 2 2 1和1 2 3 2 1为例验证代码成功

#include <iostream>
#include <stack>

using namespace std;

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

bool isPalindrome(ListNode* head) {
    if (head == NULL || head->next == NULL)
        return true;

    ListNode *slow = head, *fast = head;
    stack<int> s;

    while (fast != NULL && fast->next != NULL) {
        s.push(slow->val);
        slow = slow->next;
        fast = fast->next->next;
    }

    if (fast != NULL)
        slow = slow->next;

    while (slow != NULL) {
        if (s.top() != slow->val)
            return false;
        s.pop();
        slow = slow->next;
    }

    return true;
}

int main() {
    ListNode* head1 = new ListNode(1);
    head1->next = new ListNode(2);
    cout << isPalindrome(head1) << endl; // 输出 0

    ListNode* head2 = new ListNode(1);
    head2->next = new ListNode(2);
    head2->next->next = new ListNode(2);
    head2->next->next->next = new ListNode(1);
    cout << isPalindrome(head2) << endl; // 输出 1

    ListNode* head3 = new ListNode(1);
    head3->next = new ListNode(2);
    head3->next->next = new ListNode(3);
    head3->next->next->next = new ListNode(2);
    head3->next->next->next->next = new ListNode(1);
    cout << isPalindrome(head3) << endl; // 输出 1

    return 0;
}