边工作边刷题：70天一遍leetcode: day 37-2

Reverse Nodes in k-Group

思路：这题最主要是犁清对每组操作的时更新链表指针的顺序。对每个group要记录pre，cur，tail。一种错误的记忆是仿照reverse把cur.next=pre。实际中只有group内reverse，对group外（i.e., pre和tail.next), 想象成做rotation: pre要链接原来的tail，而原来的head去链接tail.next。
k-Group内循环移动几步？这里思维的要点是counter表示的是移动的步数而不是结点数。当前group head移动k步，就会移到下一组的head

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        def reverseOne(head, tail):
            cur = head
            pre = None
            while cur!=tail:
                next=cur.next
                cur.next=pre
                pre=cur
                cur=next
                
            return pre
        
        dummy = ListNode(0)
        dummy.next = head
        cur = head
        tail = head
        pre = dummy
        while tail:
            n=k
            while tail and n>0: 
                tail=tail.next
                n-=1
            
            if n!=0:
                break
            
            ch = reverseOne(cur, tail)
            pre.next = ch
            cur.next = tail
            pre = cur
            cur = tail
            
        return dummy.next