寻找链表倒数第k个元素，只遍历一遍（编程之美）

class LNode
{
    public LNode next;
    public int data;
}
/*找出倒数第k个元素，只遍历一遍*/
class Kk
{
    private static LNode head = new LNode();;
    private static LNode node;
    private static LNode tail;
    private static LNode fast;
    private static LNode slow;
    private static int index;
    private static int k;
    public static void main(String[] args){
        int[] nums = {1,2,3,4,5,6,7,8,9,10};
        head.data = nums[0];
        tail = head;
        createLine(nums);
        printLine();

        //假设要找的倒数K为4,也就是7
        //让fast先走K步，随后slow跟上，同步后移，当fast到达最后，slow的位置就是倒数K
        k=4;
        fast = head;
        for (int i=0;i<4 ;i++ )
        {
            fast = fast.next;
            System.out.println("fast开始："+fast.data);
        }

        //当fast到达第K个元素，slow就从第一个元素开始
        slow = head.next;

        while (fast!=null&&fast.next!=null)
        {
            fast = fast.next;
            System.out.println("fast="+fast.data);
            slow = slow.next;
            System.out.println("slow="+slow.data);
        }

        System.out.println("链表的倒数第"+k+"个元素是:"+slow.data);

    }

    private static void createLine(int[] nums){
        while (index<10)
        {
            node = new LNode();
            tail.next = node;
            node.data = nums[index];
            node.next = null;
            tail = node;
            index ++;

        }
    }

    private static void printLine(){
        node = head;
        while(node!=null&&node.next!=null){
            node = node.next;
            System.out.println(node.data);
        }
    }
}