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🐱(link): 25. K 个一组翻转链表

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`‎docs/data-structure/linked_list/README.md‎`

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`@@ -199,6 +199,73 @@ class Solution(object):`
`199`	`199`	`return first.next`
`200`	`200`	```
`201`	`201`
	`202`	`+## 25. K 个一组翻转链表`
	`203`	`+`
	`204`	`+[原题链接](https://leetcode-cn.com/problems/reverse-nodes-in-k-group/)`
	`205`	`+`
	`206`	`+### 思路`
	`207`	`+`
	`208`	`+模拟过程。`
	`209`	`+`
	`210`	+```python
	`211`	`+# Definition for singly-linked list.`
	`212`	`+# class ListNode:`
	`213`	`+# def __init__(self, x):`
	`214`	`+# self.val = x`
	`215`	`+# self.next = None`
	`216`	`+`
	`217`	`+class Solution:`
	`218`	`+ def reverseKGroup(self, head: ListNode, k: int) -> ListNode:`
	`219`	`+ cur_idx = 0`
	`220`	`+ ans = ListNode(0)`
	`221`	`+`
	`222`	`+ # 计算链表长度`
	`223`	`+ length = 0`
	`224`	`+ tmp = head`
	`225`	`+ while tmp is not None:`
	`226`	`+ length += 1`
	`227`	`+ tmp = tmp.next`
	`228`	`+`
	`229`	`+ # 上一个节点`
	`230`	`+ pre = None`
	`231`	`+ pre_range_tail, range_head = ans, ans`
	`232`	`+ while head is not None and cur_idx + k <= length:`
	`233`	`+ for i in range(k):`
	`234`	`+ if i == 0:`
	`235`	`+ # 记录头部指针`
	`236`	`+ range_head = head`
	`237`	`+ if i == k - 1:`
	`238`	`+ # 记录尾部节点`
	`239`	`+ range_tail = head`
	`240`	`+ # 记录已遍历节点数量`
	`241`	`+ cur_idx += 1`
	`242`	`+ # 获取下一个节点`
	`243`	`+ next_node = head.next`
	`244`	`+ # 反转链表,指向上一个节点`
	`245`	`+ head.next = pre`
	`246`	`+ # 当前节点成为下一轮的「上一个节点」`
	`247`	`+ pre = head`
	`248`	`+ # 继续遍历下一个节点`
	`249`	`+ head = next_node`
	`250`	`+ # 前后两个链表相连`
	`251`	`+ pre_range_tail.next = range_tail`
	`252`	`+ # print(ans)`
	`253`	`+ pre_range_tail = range_head`
	`254`	`+ # print(ans)`
	`255`	`+ # 一轮结束,改变指针连接`
	`256`	`+ # range_head.next = head`
	`257`	`+ # 一轮结束后,改变 pre 指向`
	`258`	`+ pre = None`
	`259`	`+`
	`260`	`+ # 如果有剩余节点,继续连接`
	`261`	`+ range_head.next = head`
	`262`	`+`
	`263`	`+ return ans.next`
	`264`	+```
	`265`	`+`
	`266`	`+- 事件复杂度:O(n)`
	`267`	`+- 空间复杂度:O(1)`
	`268`	`+`
`202`	`269`	`## 61. 旋转链表`
`203`	`270`
`204`	`271`	`[原题链接](https://leetcode-cn.com/problems/rotate-list/solution/chuan-zhen-yin-xian-by-liweiwei1419/)`

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