本题有两种做法：迭代和递归
本题的本质是：将链表中后k个结点变为前k个，然后将头结点连接到尾节点

迭代

考察知识：

• 边界条件判断
• 链表倒k结点寻找
• Get思想：结环

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null) {
            return null;
        }
        ListNode p = head;
        int count = 1;
        while (p.next != null) {
            count++;
            p = p.next;
        }
        k = k % count;
  
        // 成环，寻找 k 结点
        p.next = head;
        for (int i = 0; i < count - k; i++) {
            p = p.next;
        }
        head = p.next;
        p.next = null;
        return head;
    }
}

递归

暂时没思考