25. Reverse Nodes in k-Group ListNode 链表各个节点顺序倒过来，再复原顺序

# 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:
        if not head or not head.next:
            return head
        if k==1:
            return head
        
        dummy=ListNode(0)
        dummy.next=head
        record=dummy
        
        left=head
        right=head.next
        count=1
        
        while True:
            while right and count<k: #reverse links within a group
                tmp=right.next
                right.next=left
                prev=left
                left=right
                right=tmp
                count+=1
        
            if count==k:#rotate the group#####difficult
                count=1
                tmp=dummy.next  #这个group的第一个
                dummy.next=left  #这个group里面倒序之后的第一个，group原来的最后一个
                tmp.next=right # 原来group的第一个连接下一个group的第一个
                dummy=tmp
                left=right #准备下一个group的循环
                if right:
                    right=right.next
            else:#最后剩下不足k个
                
                if count>=2: 
                    left.next=None
                    if count>2:#需要将最后一个节点个数不足k的group顺序复原
                        while True:
                            tmp=prev.next
                            prev.next=left
                            if tmp.next==prev:
                                return record.next
                            left=prev
                            prev=tmp
                return record.next