题目：
Given a list, rotate the list to the right by k places, where k is non-negative.
Example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
题意及分析：从右到左第n个点进行翻转链表。主要是要找到翻转点和翻转点前面一个点，然后将最后一个点的next设置为head，翻转点前一个点的next设置为null，返回翻转点即可。
代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode rotateRight(ListNode head, int n) {
if(head == null || n==0) return head;
int length = 1; //记录链表有多长
ListNode p = head;
while(p.next != null){
p = p.next;
length++;
}
//此时p指向链表最后一个点
n=n%length;
if(n==0) return head;
ListNode rotedNode = null;
ListNode preRotate = head;
int k=0;
while(k<length-n-1){
preRotate = preRotate.next;
k++;
}
rotedNode = preRotate.next;
preRotate.next = null;
p.next = head;
return rotedNode;
}
}
