# 题目

实现一种算法，找出单向链表中倒数第 k 个节点。返回该节点的值。

示例 1：

	输入： 1->2->3->4->5 和 k = 2
	输出： 4

tips:

# 解法一：使用栈

利用栈先进后出的特点。

	/**
	* Definition for singly-linked list.
	* public class ListNode {
	* int val;
	* ListNode next;
	* ListNode(int x) { val = x; }
	* }
	*/
	class Solution {
	public int kthToLast(ListNode head, int k) {
	Stack<Integer> stack = new Stack();
	ListNode cur = head;
	while(cur != null){
	stack.add(cur.val);
	cur =cur.next;
	}
	int index = 1;
	while(index > 0){
	if(k == index++){
	return stack.pop();
	}
	stack.pop();
	}
	return 0;
	}
	}

	/**
	* Definition for singly-linked list.
	* public class ListNode {
	* int val;
	* ListNode next;
	* ListNode(int x) { val = x; }
	* }
	*/
	class Solution {
	public int kthToLast(ListNode head, int k) {
	ListNode fast = head;
	for (int i = 0; i < k; i++) {
	fast = fast.next;
	}
	ListNode slow = head;
	while (fast != null) {
	slow = slow.next;
	fast = fast.next;
	}

	return slow.val;
	}
	}