From e75195df41613616ed08e6cd8c4c4d71c906948a Mon Sep 17 00:00:00 2001
From: Wonder <hezhaohui0807@163.com>
Date: Sat, 4 Oct 2025 12:43:05 +0800
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+# 排序链表
+
+## 题目
+
+给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。
+
+ 
+
+示例 1：
+
+
+输入：head = [4,2,1,3]
+输出：[1,2,3,4]
+示例 2：
+
+
+输入：head = [-1,5,3,4,0]
+输出：[-1,0,3,4,5]
+示例 3：
+
+输入：head = []
+输出：[]
+ 
+
+提示：
+
+链表中节点的数目在范围 [0, 5 * 104] 内
+-105 <= Node.val <= 105
+ 
+
+进阶：你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？
+
+## 思路
+
+- 归并排序
+  - 拆分
+    - 快慢指针
+  - 合并
+
+
+## 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 sortList(ListNode head) {
+        // -- Divide -> Sort -> Merge --
+        // Special
+        if (head == null || head.next == null) {
+            return head;
+        }
+
+        // Init: slow, fast, dummyHead
+        ListNode dummyHead = new ListNode(-1);
+        dummyHead.next = head;
+        ListNode slow = dummyHead;
+        ListNode fast = dummyHead;
+        while (fast.next != null && fast.next.next != null) {
+            fast = fast.next.next;
+            slow = slow.next;
+        }
+        // Divide: [headA .... tailA(slow)] [headB(slow.next) ... tailB]
+        ListNode headA = head;
+        ListNode headB = slow.next;
+        slow.next = null;
+        // Sort
+        headA = sortList(headA);
+        headB = sortList(headB);
+        // Merge
+        return mergeList(headA, headB);
+    }
+
+    public ListNode mergeList(ListNode headA, ListNode headB) {
+        // Init: dummyHead, tail
+        ListNode dummyHead = new ListNode(-1);
+        ListNode tail = dummyHead;
+        // Traverse: Merge
+        while (headA != null && headB != null) {
+            int valA = headA.val;
+            int valB = headB.val;
+            if (valA < valB) {
+                tail.next = headA;
+                headA = headA.next;
+            } else {
+                tail.next = headB;
+                headB = headB.next;
+            }
+            tail = tail.next;
+        }
+        // Operation: After
+        tail.next = headA == null ? headB : headA;
+        return dummyHead.next;
+    }
+}
+```
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