一道变形的链表反转题，使用lua实现

给定一个单链表的头节点 head,实现一个调整单链表的函数，使得每K个节点之间为一组进行逆序，并且从链表的尾部开始组起，头部剩余节点数量不够一组的不需要逆序。（不能使用队列或者栈作为辅助）

例如：链表:1->2->3->4->5->6->7->8->null, K = 3。那么 6->7->8，3->4->5，1->2各位一组。调整后：1->2->5->4->3->8->7->6->null。其中 1，2不调整，因为不够一组。

local t = {
	next = nil,
	data = 1,
}

local tt = t;
local max = 8;
while(max > 0) do 
	local t2 = {
		next = nil,
		data = 10 -max 
	}
	tt.next = t2;
	tt = t2;
	max = max -1;
end 


function revertTable( tab,limit )
	-- 递归拆分子链

	local tempHead = nil;
	local tempHead2 = nil;
	local tempK = 0
	function revertSub(nextNode,fatherNode)

		if nextNode == nil then 
			return
		end 

		revertSub(nextNode.next,nextNode)
		tempK = tempK +1;
	 	-- print(nextNode.data,tempK);

		if tempK == limit then 
			nextNode.next = tempHead2
			tempHead2 = tempHead
			tempHead = nil ;
			tempK = 0
		elseif tempK == 1 then 
			tempHead = nextNode
			
			nextNode.next = fatherNode
		else
			nextNode.next = fatherNode
		end 

	end


	revertSub(tab,nil)
	if tempHead ~= nil then 
		print('==========')
		local max = 10
		local tempNext = tempHead.next
		local tempNext2 = nil 
		tempHead.next = tempHead2;
		while tempNext and max > 0 do 
			max = max - 1;
			print('2222======= ',tempHead.data)
			tempNext2 = tempNext.next
			
			tempNext.next  =  tempHead 
			tempHead = tempNext;
			tempNext = tempNext2
			
		end 
	else
		tempHead = tempHead2
	end 
	return tempHead

end


local head = revertTable(t,3);

local k = 20
while(head and k > 0)
do
	k = k -1;
	print(head.data);
	head = head.next;
end