【leetcode】234. 回文链表（遍历head装入list，双指针检查list判断回文）

>题目:[https://leetcode.cn/problems/palindrome-linked-list/](https://leetcode.cn/problems/palindrome-linked-list/)

>解题：[https://leetcode.cn/problems/palindrome-linked-list/solution/zhong-yong-xie-fa-by-yi-qi-he-fen-da-ov7k/](https://leetcode.cn/problems/palindrome-linked-list/solution/zhong-yong-xie-fa-by-yi-qi-he-fen-da-ov7k/)
### 解题思路
中庸写法：遍历head装入list，双指针检查list判断回文

### 执行结果

执行用时：7 ms, 在所有 Java 提交中击败了48.80%的用户

内存消耗：53.7 MB, 在所有 Java 提交中击败了74.98%的用户

### 代码

```java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> list = new ArrayList<Integer>();
        do{
            list.add(head.val);
            head = head.next;
        }while (head!=null);
        
        for (int i = 0,j=list.size()-1; i <=j ; i++,j--) {
            if(!Objects.equals(list.get(i), list.get(j))){
                return false;
            }
        }
        return true;
    }
}
```