Given a list, rotate the list to the right by k places, where k is non-negative.For example:Given 1->2->3->4->5->NULL and k = 2,return 4->5->1->2->3->NULL.

Anaylisis: Linked List题目的做法其实比较固定，就是Runner Technique(Two pointer) 方法加上 Dummy Node(设置一个head之前的虚假节点)两种方法，这道题就这样搞一搞就出来了。

需要注意的是：处理null 输入需要特别考虑，然后，需要找到新的linked list的头

第二遍做法：注意15行需要特别处理，wrap up了之后n=0的情况

1 public class Solution { 2     public ListNode rotateRight(ListNode head, int n) { 3         if (head == null) return null; 4         ListNode dummy = new ListNode(-1); 5         dummy.next = head; 6         ListNode cursor = dummy; 7         ListNode walker = dummy; 8         ListNode runner = dummy; 9         int count = 0;10         while (cursor.next != null) {11             cursor = cursor.next;12             count ++;13         }14         n %= count;15         if (n == 0) return head;16         while (n > 0) {17             runner = runner.next;18             n--;19         }20         while (runner.next != null) {21             runner = runner.next;22             walker = walker.next;23         }24         dummy.next = walker.next;25         walker.next = runner.next;26         runner.next = head;27         return dummy.next;28     }29 }