Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

代码如下：

public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        ListNode start = new ListNode(0);        ListNode l1 = start,l2 = start;        l1.next = head;        l2.next = head;//l2作为辅助工具                int i=0;        while(i++<=n){//l2先走n步            l2=l2.next;        }                while(l2 != null){//此时，l1和l2同步走，那么l1走的就是l2剩下的步数，l2走到尽头时，l1刚好走到倒数第n个数            l2 = l2.next;            l1 = l1.next;        }                l1.next = l1.next.next;//移除倒数第n个数                return start.next;    }}