输入一个单向链表，输出该链表中倒数第k个结点

思路：利用快慢指针能实现在时间复杂度为O（n）的情况下，找到第k个节点

1，快慢指针共同指向头结点

2，快指针先走k步

3，快慢指针一起走，直到快指针指向null时，慢指针所在位置就是倒数第k个节点

public static void main(String[] args) {
       ListNode eight = new ListNode(8,null);
       ListNode sevent = new ListNode(7,eight);
       ListNode six = new ListNode(6,sevent);
       ListNode firth = new ListNode(5,six);
       ListNode fourth = new ListNode(4,firth);
       ListNode three = new ListNode(3,fourth);
       ListNode two = new ListNode(2,three);
       ListNode one = new ListNode(1,two);

       int k = 3;
        ListNode low = one;
        ListNode fast = one;
       while (k > 0){
           fast = fast.next;
           k--;
       }
       while (fast != null){
           fast = fast.next;
           low = low.next;
       }
        System.out.println(fast);
        System.out.println(low.val);
    }

static class ListNode{
    public Integer val;
    public ListNode next;

    public ListNode(Integer val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}