求单向链表倒数第K个元素

typedef struct NodeList {
	int value;
	struct NodeList *next;
}NodeList;

有如上节点类型的单向链表，求链表的倒数第K个节点。

（1）递归

NodeList* find_node(NodeList *node, int K)
{
	//记录倒数第几个节点
	static int index = 0;
	NodeList *tmp = NULL;;
	
	if (node != NULL)
	{
		tmp = find_node(node->next);
		if (K == index)
		{
			tmp = node;
		}
	}
	
	index++;
	return tmp;
}

（2）通过两个指针

NodeList* find_node(NodeList *node, int K)
{
	int len = 0;
	NodeList *node_last = NULL;
	NodeList *node_k = NULL;

	node_last = node;
	while (node_last != NULL)
	{
		len++;
		node_last = node_last->next;
		if (len == K)
		{
			node_k = node;
		}
		else if (len > K)
		{
			node_k = node_k->next;
		}
	}
	return node_k;
}