🐱(link): 92. 反转链表 II

Author: JalanJiang

docs/data-structure/linked_list/README.md

@@ -286,6 +286,80 @@ class Solution(object):
  return head
 ```
 
+## 92. 反转链表 II
+
+[原题链接](https://leetcode-cn.com/problems/reverse-linked-list-ii/)
+
+### 思路
+
+需要找到几个关键节点：
+
+1. 开始翻转的节点 `begin`
+2. 开始翻转节点的前一个节点 `begin_pre`
+3. 结束翻转的节点 `end`
+4. 结束翻转节点的下一个节点 `end_next`
+
+从 `begin` 到 `end` 位置执行反转操作。此段链表翻转后需要：
+
+1. 修改 `begin_pre` 节点的指针，指向 `end` 节点：`begin_pre.next = end`
+2. 修改 `begin` 节点的指针，指向 `end_next` 节点：`begin.next = end_next`
+
+注意：`m` 可能会等于 `n`。
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+
+class Solution:
+ def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
+ if head is None:
+ return head
+
+ if m == n:
+ return head
+
+ index = 0 
+ pre = None
+ cur = head
+ begin_pre = None
+ begin = None
+ end = None
+ end_next = None
+
+ while cur is not None:
+ index += 1
+ next_cur = cur.next
+
+ if index == m:
+ # 记录开始节点
+ begin = cur
+ if index == m - 1:
+ begin_pre = cur
+ if index == n:
+ # 记录结束位置
+ end = cur
+ if index == n + 1:
+ end_next = cur
+ 
+ if index > m and index <= n:
+ # 翻转
+ cur.next = pre
+ pre = cur
+ cur = next_cur
+
+ # 修改指针位置
+ begin.next = end_next
+ if begin_pre is None:
+ # 从头开始翻转了
+ return end
+ else:
+ begin_pre.next = end
+ return head
+```
+
 
 ## 138. 复制带随机指针的链表