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  3->2->1->5->null and n = 2, return node  whose value is 1.

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;
    }
}