Lintcode: Nth to Last Node in List

Find the nth to last element of a singly linked list. 

The minimum number of nodes in list is n.

Example
Given a List  ->->->->null and n = , return node  whose value is .

Runner Technique: 两个指针都从Dummy node出发，结束条件是runner.next!=null

 public class Solution {
     /**
      * @param head: The first node of linked list.
      * @param n: An integer.
      * @return: Nth to last node of a singly linked list.
      */
     ListNode nthToLast(ListNode head, int n) {
         // write your code here
         ListNode dummy = new ListNode(0);
         dummy.next = head;
         ListNode walker = dummy;
         ListNode runner = dummy;
         while (runner.next != null && n>0) {
             runner = runner.next;
             n--;
         }
         while (runner.next != null) {
             runner = runner.next;
             walker = walker.next;
         }
         return walker.next;
     }
 }