From b814779b7d7ac7d0ab17b7f50c7545484ecbe5e6 Mon Sep 17 00:00:00 2001
From: zhangsan <646228430@qq.com>
Date: Fri, 27 Jun 2025 21:24:55 +0800
Subject: [PATCH] =?UTF-8?q?6.27=20=E4=BA=8C=E5=88=B7=E9=93=BE=E8=A1=A8?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 src/main/java/linkedlist/CopyRandomList.java |  1 +
 src/main/java/linkedlist/ReverseKGroup.java  | 10 +++---
 src/main/java/linkedlist/SortList.java       | 32 +++++++++++++++++++-
 3 files changed, 37 insertions(+), 6 deletions(-)

diff --git a/src/main/java/linkedlist/CopyRandomList.java b/src/main/java/linkedlist/CopyRandomList.java
index 0a8ecca..5ddbad0 100644
--- a/src/main/java/linkedlist/CopyRandomList.java
+++ b/src/main/java/linkedlist/CopyRandomList.java
@@ -22,6 +22,7 @@ import java.util.HashMap;
 
  */
 //不会做 需要重做
+//二刷不会 ，很妙的思路
 public class CopyRandomList {
     public Node copyRandomList(Node head) {
         if (head == null)
diff --git a/src/main/java/linkedlist/ReverseKGroup.java b/src/main/java/linkedlist/ReverseKGroup.java
index 5f657f9..2d99a36 100644
--- a/src/main/java/linkedlist/ReverseKGroup.java
+++ b/src/main/java/linkedlist/ReverseKGroup.java
@@ -13,12 +13,12 @@ public class ReverseKGroup {
     // 反转 [head, tail] 这一段，返回新的头和新的尾
     private ListNode[] reverse(ListNode head, ListNode tail) {
         ListNode prev = tail.next;
-        ListNode p = head;
+        ListNode cur = head;
         while (prev != tail) {
-            ListNode nxt = p.next;
-            p.next = prev;
-            prev = p;
-            p = nxt;
+            ListNode nxt = cur.next;
+            cur.next = prev;
+            prev = cur;
+            cur = nxt;
         }
         // 反转后，tail 变成新头，head 变成新尾
         return new ListNode[]{ tail, head };
diff --git a/src/main/java/linkedlist/SortList.java b/src/main/java/linkedlist/SortList.java
index 16d6ffb..253f62f 100644
--- a/src/main/java/linkedlist/SortList.java
+++ b/src/main/java/linkedlist/SortList.java
@@ -8,11 +8,13 @@ import java.util.List;
  * 题目： 148. 排序链表  (sortList)
  * 描述：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 。
 
+ 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？
  * 链接：https://leetcode.cn/problems/sort-list/
 
  */
+//二刷思路会，但链表归并排序不会
 public class SortList {
-    public ListNode sortList(ListNode head) {
+    public ListNode sortList2(ListNode head) {
         ListNode cur=head;
         List<Integer>list=new ArrayList<>();
         while (cur!=null){
@@ -28,4 +30,32 @@ public class SortList {
         }
         return newnode.next;
     }
+
+    ListNode sortList(ListNode head) {
+        if (head == null || head.next == null) return head;
+        // ① 快慢指针找中点并断链
+        ListNode slow = head, fast = head.next;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        ListNode mid = slow.next;
+        slow.next = null;
+        // ② 递归排序左右两段
+        ListNode left  = sortList(head);
+        ListNode right = sortList(mid);
+        // ③ 合并
+        return merge(left, right);
+    }
+
+    ListNode merge(ListNode a, ListNode b) {
+        ListNode dummy = new ListNode(-1), p = dummy;
+        while (a != null && b != null) {
+            if (a.val <= b.val) { p.next = a; a = a.next; }
+            else               { p.next = b; b = b.next; }
+            p = p.next;
+        }
+        p.next = (a != null ? a : b);
+        return dummy.next;
+    }
 }