🐱(link): 25. K 个一组翻转链表

Author: JalanJiang

docs/data-structure/linked_list/README.md

@@ -199,6 +199,73 @@ class Solution(object):
  return first.next
 ```
 
+## 25. K 个一组翻转链表
+
+[原题链接](https://leetcode-cn.com/problems/reverse-nodes-in-k-group/)
+
+### 思路
+
+模拟过程。
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+
+class Solution:
+ def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
+ cur_idx = 0
+ ans = ListNode(0)
+
+ # 计算链表长度
+ length = 0
+ tmp = head
+ while tmp is not None:
+ length += 1
+ tmp = tmp.next
+
+ # 上一个节点
+ pre = None
+ pre_range_tail, range_head = ans, ans
+ while head is not None and cur_idx + k <= length:
+ for i in range(k):
+ if i == 0:
+ # 记录头部指针
+ range_head = head
+ if i == k - 1:
+ # 记录尾部节点
+ range_tail = head
+ # 记录已遍历节点数量
+ cur_idx += 1
+ # 获取下一个节点
+ next_node = head.next
+ # 反转链表，指向上一个节点
+ head.next = pre
+ # 当前节点成为下一轮的「上一个节点」
+ pre = head
+ # 继续遍历下一个节点
+ head = next_node
+ # 前后两个链表相连
+ pre_range_tail.next = range_tail
+ # print(ans)
+ pre_range_tail = range_head
+ # print(ans)
+ # 一轮结束，改变指针连接
+ # range_head.next = head
+ # 一轮结束后，改变 pre 指向
+ pre = None
+
+ # 如果有剩余节点，继续连接
+ range_head.next = head
+
+ return ans.next
+```
+
+- 事件复杂度：O(n)
+- 空间复杂度：O(1)
+
 ## 61. 旋转链表
 
 [原题链接](https://leetcode-cn.com/problems/rotate-list/solution/chuan-zhen-yin-xian-by-liweiwei1419/)