单向链表反转，k个一组

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有链表1 2 3 4 5 6 7 8 ，3个一组反转结果：3 2 1 6 5 4 8 7

Node* KNodeReverse(Node* root,int k)
{
	Node *cur,*pre,*next;//必须有三个临时指针，因为cur需要知道连接的对象，同时不丢失它的下个对象
	cur = root;
	pre = nullptr;

	Node* revRoot = new Node(0);//新建一个节点作为起点，返回前再delete
	Node* first;
	Node* connect = revRoot;
	while(cur)
	{
		int n = k;
		first = cur;
		while(n-- && cur)
		{
			next = cur->pNext;
			cur->pNext = pre;
			pre = cur;
			cur = next;
		}
		connect->pNext = pre;
		connect = first;	
	}

	Node *tempt = revRoot;
	revRoot = revRoot->pNext;
	delete tempt;
	return revRoot;
}