## [234\. Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/)

Given the `head` of a singly linked list, return `true`_ if it is a _

_palindrome_

_ or _`false`_ otherwise_.

**Example 1:**


**Input:** head = \[1,2,2,1\]
**Output:** true

**Example 2:**


**Input:** head = \[1,2\]
**Output:** false

**Constraints:**

- The number of nodes in the list is in the range `[1, 105]`.
- `0 <= Node.val <= 9`

判斷 linked list 是否有迴文

解法一：遞迴

/**
 * 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) {
        ListNode* cur = head;
        return helper(head, cur);
    }
    bool helper(ListNode* head, ListNode*& cur) {
        if (!head) return true;
        bool res = helper(head->next, cur) && (cur->val == head->val);
        cur = cur->next;
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::

/**
 * 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 || !head->next) return true;
        ListNode* slow = head;
        ListNode* fast = head;
        stack<int> stk;
        stk.push(head->val);
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
            stk.push(slow->val);
        }
        if (!fast->next) stk.pop();

        while (slow && slow->next) {
            slow = slow->next;
            int tmp = stk.top(); stk.pop();
            if (tmp != slow->val) return false;            
        }
        return true;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::

不用 stack，空間複雜度 $O(1)$ 的解法

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(1)$
:::