From d0fd932593b94f33154f64c053b9773a97ddd8cd Mon Sep 17 00:00:00 2001
From: Wonder <hezhaohui0807@163.com>
Date: Fri, 3 Oct 2025 11:41:02 +0800
Subject: [PATCH] =?UTF-8?q?=20=F0=9F=94=84Update:=20=E5=9B=9E=E6=96=87?=
 =?UTF-8?q?=E9=93=BE=E8=A1=A8?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 链表/3. 回文链表.md | 76 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 76 insertions(+)
 create mode 100644 链表/3. 回文链表.md

diff --git a/链表/3. 回文链表.md b/链表/3. 回文链表.md
new file mode 100644
index 0000000..a915bba
--- /dev/null
+++ b/链表/3. 回文链表.md	
@@ -0,0 +1,76 @@
+# 回文链表
+
+## 题目
+
+给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。
+
+ 
+
+示例 1：
+
+
+输入：head = [1,2,2,1]
+输出：true
+示例 2：
+
+
+输入：head = [1,2]
+输出：false
+ 
+
+提示：
+
+链表中节点数目在范围[1, 105] 内
+0 <= Node.val <= 9
+ 
+
+进阶：你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？
+
+## 思路
+
+- 初始化
+  - `ArrayList`
+- 双指针
+  - 左右对称判断
+
+## 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) {
+        // Special
+        if (head == null) {
+            return true;
+        }
+        
+        // Init: ArrayList
+        List<Integer> list = new ArrayList<>();
+
+        // Traverse: Add Vals
+        while (head != null) {
+            list.add(head.val);
+            head = head.next;
+        }
+
+        // Traverse: Judge
+        int size = list.size();
+        for (int i = 0; i < size / 2; i++) {
+            if (list.get(i) != list.get(size - i - 1)) {
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```