From 268ea2fa72fd4e7a8efaa9a32c05a7932b6743a6 Mon Sep 17 00:00:00 2001
From: Wonder <hezhaohui0807@163.com>
Date: Fri, 3 Oct 2025 21:12:17 +0800
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 链表/9. 两两交换链表中的节点.md | 75 +++++++++++++++++++++++++++++++++
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+# 两两交换链表中的节点
+
+## 题目
+
+给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。
+
+ 
+
+示例 1：
+
+
+输入：head = [1,2,3,4]
+输出：[2,1,4,3]
+示例 2：
+
+输入：head = []
+输出：[]
+示例 3：
+
+输入：head = [1]
+输出：[1]
+ 
+
+提示：
+
+链表中节点的数目在范围 [0, 100] 内
+0 <= Node.val <= 100
+
+## 思路
+
+- 模拟
+- 结束条件
+  - cur.next == null || cur.next.next == null
+
+
+## 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 ListNode swapPairs(ListNode head) {
+        // Special
+        if (head == null || head.next == null) {
+            return head;
+        }
+        // Init: dummyHead, cur
+        ListNode dummyHead = new ListNode(-1);
+        dummyHead.next = head;
+        ListNode cur = dummyHead;
+        // Traverse: Swap
+        while (cur.next != null && cur.next.next != null) {
+            // Swap: A(cur.next) B(cur.next.next)
+            ListNode tmp1 = cur.next;
+            ListNode tmp2 = cur.next.next;
+            ListNode tmp3 = cur.next.next.next;
+            cur.next = tmp2;
+            cur.next.next = tmp1;
+            cur.next.next.next = tmp3;
+
+            cur = cur.next.next;
+        }
+
+        return dummyHead.next;
+    }
+}
+```
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