From 63f9b0faac9c85fa698cfe3efbc09716628a923f Mon Sep 17 00:00:00 2001
From: fanxb <fleyx20@outlook.com>
Date: Wed, 9 Feb 2022 18:01:22 +0800
Subject: [PATCH] add

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+---
+id: "20220209"
+date: "2022/02/09 10:38:05"
+title: "leetcode.Q25.k个一组翻转链表"
+tags: ["java", "leetcode", "链表"]
+categories:
+  - "算法"
+  - "leetcode刷题"
+---
+
+## 解析思路
+
+leetcode 困难，题目描述[点击这里](https://leetcode-cn.com/problems/reverse-nodes-in-k-group/)。
+
+本题属于链表翻转的进阶题，思路如下：
+
+<!-- more -->
+
+1. 添加一个虚拟节点头，方便编码编写
+2. 遍历链表，当遍历到k个时，对这k个进行处理，处理过程如下：
+   1. 对这k个节点进行翻转
+   2. 翻转后处理头尾节点的关系，注意需要记录头节点、头节点之前的一个节点、尾节点和尾节点的下一节点共四个节点，细节看代码
+3. 一轮处理完毕后继续遍历，当遍历到k个时继续第2步，直到结束遍历
+
+## 代码
+
+```java
+public class Q25 {
+    public ListNode reverseKGroup(ListNode head, int k) {
+        ListNode res = new ListNode(), index = head;
+        res.next = head;
+        int count = 0;
+        //left:开始翻转节点的父节点
+        //right:结束翻转的节点
+        ListNode beforeL = res, l = beforeL.next, r, afterR;
+        while (index != null) {
+            if (++count == k) {
+                //进行翻转
+                r = index;
+                afterR = index.next;
+                reverse(l, count);
+                index = l;
+                //处理头尾节点关系
+                l.next = afterR;
+                beforeL.next = r;
+                //进行下一轮循环
+                beforeL = index;
+                l = index.next;
+                count = 0;
+            }
+            index = index.next;
+        }
+        return res.next;
+    }
+
+    /**
+     * 翻转start后的n个节点
+     *
+     * @param start
+     * @param n
+     */
+    private void reverse(ListNode start, int n) {
+        //反转节点
+        ListNode prev = null;
+        for (int i = 0; i < n; i++) {
+            ListNode next = start.next;
+            start.next = prev;
+            prev = start;
+            start = next;
+        }
+    }
+
+    public static void main(String[] args) {
+        ListNode node = new ListNode(1);
+        node.next = new ListNode(2);
+        node.next.next = new ListNode(3);
+        node.next.next.next = new ListNode(4);
+        node.next.next.next.next = new ListNode(5);
+        new Q25().reverseKGroup(node, 3);
+    }
+}
+
+```
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