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.

旋转一个链表，首先我们计算出链表的长度len，如果k % len == 0, 我们可以直接返回head，如果不为0，我们就让head移动 len - 1 - (k % len)次，将链表分为两部分，这样剩下的右边部分就是长度为k的一个链表，然后将后面长度为k的链表连接到左边部分的链表就完成了。代码如下：

/**
 * 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 k) {
        if(head == null) return null;
        int sum = 1;
        ListNode tem = head;
        while(tem.next != null) {
            sum ++;
            tem = tem.next;
        }
        k %= sum;
        if(k == 0) return head;
        int step = sum - k - 1;
        ListNode helper = head;
        while(step > 0) {
            helper = helper.next;
            step --;
        }
        ListNode node = helper.next;
        helper.next = null;
        tem.next = head;
        return node;
    }
}