25.Reverse Nodes in k-Group

lc 23 要求分组翻转链表

题干：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

解释：

依旧是非常直白的一个描述，之前我们使用pre，current，next 3个指针完成了链表的翻转，现在要求的是K个翻转

思考：

无论怎么翻转，中间的过程都是一样的使用3个指针从前往后，使用一个count来实现统计个数，最后要把当前段的第一个节点（翻转后的尾节点）指向递归调用这段程序的头节点。

答案：

class Solution(object):
    def reverseKGroup(self, head, k):
        #count how many nodes
        count =0
        tmp=head
        while tmp!=None:
            tmp =tmp.next
            count+=1
        if count<k:
            return head
        pre =None
        current =head
        count=0
        while current!=None:
            next =current.next
            current.next=pre
            pre=current
            current=next
            count+=1
            if count ==k:
                break
        if current ==None:
            return pre
        else:
            head.next =self.reverseKGroup(self,current,k)
            return pre

答案补充：

注意一定要计数，最后一段是要保持原状的。