K个节点的组内逆序调整

题目

给定一个单链表的头节点head，和一个正数k实现k个节点的小组内部逆序，如果最后一组不够k个就不调整
例子：
调整前：1->2->3->4->5->6->7->8，k=3
调整后：3->2->1->6->5->4->7->8

tips：功能拆分：获取当前组的结尾节点、反转、拼接

• 定义链表

"""单链表"""  
class ListNode:  def __init__(self, value):  self.next = Noneself.val = value

• 获取当前组的结尾节点

def get_end(node, k):  while k > 1 and node:  node = node.next  k -= 1  return node

• 反转

def reverse(start, end):  end = end.next  prev = None  cur = start  while cur != end:  next = cur.next  cur.next = prev  prev = cur  cur = next  start.next = end

• 拼接

def reverse_all(head, k):  start = head  end = get_end(head, k)  if end is None:  return head  head = end  reverse(start, end)  last_end = start  while last_end.next:  start = last_end.next  end = get_end(start, k)  if end is None:  return head  reverse(start, end)  last_end.next = end  last_end = start  return head