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# 234. Palindrome Linked List

## problem description

Given a singly linked list, determine if it is a palindrome.

**Example 1:**

```
Input: 1->2
Output: false
```

**Example 2:**

```
Input: 1->2->2->1
Output: true
```

**Follow up:** Could you do it in O(n) time and O(1) space?

## algorithm thought

这里主要是找到链表的中点，将链表分成两半，然后匹配是否是回文串。找到中点用快慢指针。使用一个stack来辅助，就只需要一次遍历可的结果

## code

```cpp
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        stack<int> st;
        ListNode* fast=head,*slow=head;
        while(fast!=NULL&&fast->next!=NULL){
            st.push(slow->val);
            slow=slow->next;
            fast=fast->next->next;
        }
        if(fast)
            slow=slow->next;
        while(slow){
            if(slow->val!=st.top())
                return false;
            slow=slow->next;
            st.pop();
        }
        return true;
    }
};
```

## algorithm analysis

一次遍历，时间复杂度是O(n)，使用栈辅助，空间复杂度也是O(n)