83 lines
2.4 KiB
Markdown
83 lines
2.4 KiB
Markdown
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---
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id: "20220209"
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date: "2022/02/09 10:38:05"
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title: "leetcode.Q25.k个一组翻转链表"
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tags: ["java", "leetcode", "链表"]
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categories:
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- "算法"
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- "leetcode刷题"
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---
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## 解析思路
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leetcode 困难，题目描述[点击这里](https://leetcode-cn.com/problems/reverse-nodes-in-k-group/)。
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本题属于链表翻转的进阶题，思路如下：
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<!-- more -->
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1. 添加一个虚拟节点头，方便编码编写
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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：
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1. 对这k个节点进行翻转
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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
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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
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## 代码
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```java
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public class Q25 {
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public ListNode reverseKGroup(ListNode head, int k) {
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ListNode res = new ListNode(), index = head;
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res.next = head;
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int count = 0;
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//left:开始翻转节点的父节点
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//right:结束翻转的节点
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ListNode beforeL = res, l = beforeL.next, r, afterR;
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while (index != null) {
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if (++count == k) {
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//进行翻转
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r = index;
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afterR = index.next;
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reverse(l, count);
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index = l;
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//处理头尾节点关系
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l.next = afterR;
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beforeL.next = r;
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//进行下一轮循环
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beforeL = index;
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l = index.next;
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count = 0;
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}
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index = index.next;
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}
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return res.next;
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}
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/**
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* 翻转start后的n个节点
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*
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* @param start
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* @param n
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*/
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private void reverse(ListNode start, int n) {
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//反转节点
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ListNode prev = null;
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for (int i = 0; i < n; i++) {
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ListNode next = start.next;
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start.next = prev;
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prev = start;
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start = next;
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}
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}
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public static void main(String[] args) {
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ListNode node = new ListNode(1);
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node.next = new ListNode(2);
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node.next.next = new ListNode(3);
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node.next.next.next = new ListNode(4);
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node.next.next.next.next = new ListNode(5);
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new Q25().reverseKGroup(node, 3);
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}
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}
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```
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