2021-03-26 11:58:47 +00:00
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> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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今天我们来说一下反转链表 2，其实这个和 1 的思路差不多，今天先说一个比较好理解的方法，完全按照反转链表 1 的方法来解决，大家看这个题目之前要先看一下[【动画模拟】leetcode 206 反转链表](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E9%93%BE%E8%A1%A8%E7%AF%87/leetcode206%E5%8F%8D%E8%BD%AC%E9%93%BE%E8%A1%A8.md)
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下面我们先来看一下题目。
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#### [92. 反转链表 II](https://leetcode-cn.com/problems/reverse-linked-list-ii/)
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难度中等836
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给你单链表的头指针 `head` 和两个整数 `left` 和 `right` ，其中 `left <= right` 。请你反转从位置 `left` 到位置 `right` 的链表节点，返回 **反转后的链表** 。
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**示例 1：**
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> 输入：head = [1,2,3,4,5], left = 2, right = 4
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> 输出：[1,4,3,2,5]
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**示例 2：**
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> 输入：head = [5], left = 1, right = 1
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> 输出：[5]
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含义就是让我们反转部分链表。也就是上面蓝色的部分，那我们怎么借助 1 的方法来解决呢？
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我们主要通过两部分来解决，先截取需要翻转的部分，然后再头尾交换即可。下面我们通过一个动画来看看具体步骤。
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2021-03-27 08:38:40 +00:00
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2021-03-26 11:58:47 +00:00
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是不是很容易理解，下面我们来看代码吧。
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2021-04-28 10:28:00 +00:00
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**题目代码**
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Java Code:
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2021-03-26 11:58:47 +00:00
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```java
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class Solution {
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public ListNode reverseBetween(ListNode head, int left, int right) {
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//虚拟头节点
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ListNode temp = new ListNode(-1);
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temp.next = head;
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ListNode pro = temp;
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//来到 left 节点前的一个节点
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int i = 0;
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for (; i < left-1; ++i) {
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pro = pro.next;
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}
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// 保存 left 节点前的第一个节点
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ListNode leftNode = pro;
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for (; i < right; ++i) {
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pro = pro.next;
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}
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// 保存 right 节点后的节点
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ListNode rightNode = pro.next;
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//切断链表
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pro.next = null;
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ListNode newhead = leftNode.next;
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leftNode.next = null;
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leftNode.next = rever(newhead);
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//重新接头
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newhead.next = rightNode;
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return temp.next;
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}
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//和反转链表1代码一致
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public ListNode rever (ListNode head) {
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ListNode low = null;
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ListNode pro = head;
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while (pro != null) {
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ListNode temp = pro;
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pro = pro.next;
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temp.next = low;
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low = temp;
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}
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return low;
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}
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}
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```
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2021-04-28 10:28:00 +00:00
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C++ Code:
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```cpp
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class Solution {
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public:
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ListNode* reverseBetween(ListNode* head, int left, int right) {
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ListNode * temp = new ListNode(-1);
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temp->next = head;
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ListNode * pro = temp;
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//来到 left 节点前的一个节点
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int i = 0;
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for (; i < left-1; ++i) {
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pro = pro->next;
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}
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// 保存 left 节点前的第一个节点
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ListNode * leftNode = pro;
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for (; i < right; ++i) {
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pro = pro->next;
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}
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// 保存 right 节点后的节点
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ListNode * rightNode = pro->next;
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//切断链表
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pro->next = nullptr;
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ListNode * newhead = leftNode->next;
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leftNode->next = nullptr;
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leftNode->next = rever(newhead);
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//重新接头
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newhead->next = rightNode;
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return temp->next;
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}
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ListNode * rever (ListNode * head) {
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ListNode * low = nullptr;
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ListNode * pro = head;
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while (pro != nullptr) {
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ListNode * temp = pro;
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pro = pro->next;
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temp->next = low;
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low = temp;
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}
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return low;
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}
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};
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```
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