给定一个链表，向右旋转链表 k 次，k 大于等于 0。

例一：
输入： 1->2->3->4->5->NULL, k = 2
输出： 4->5->1->2->3->NULL
解释： 向右旋转一次：5->1->2->3->4->NULL，向右旋转两次：4->5->1->2->3->NULL

例二：
输入： 0->1->2->NULL, k = 4
输出： 2->0->1->NULL
解释： 向右旋转一次：2->0->1->NULL，向右旋转两次：1->2->0->NULL，向右旋转三次：0->1->2->NULL，向右旋转四次：2->0->1->NULL

/*
 * 61. Rotate List
 * https://leetcode.com/problems/rotate-list/
 * https://www.whosneo.com/61-rotate-list/
 */

class RotateList {
    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        ListNode node = head;
        for (int i = 2; i < 10; i++) {
            node.next = new ListNode(i);
            node = node.next;
        }

        RotateList solution = new RotateList();

        solution.print(head);
        System.out.println("Rotate 1");
        head = solution.rotateRight(head, 1);
        solution.print(head);
        System.out.println("Rotate 1");
        head = solution.rotateRight(head, 1);
        solution.print(head);
        System.out.println("Rotate 3");
        head = solution.rotateRight(head, 3);
        solution.print(head);
    }

    private void print(ListNode node) {
        for (; node != null; node = node.next) {
            System.out.print(node.val);
            System.out.print("->");
        }
        System.out.println("null");
    }

    private ListNode rotateRight(ListNode head, int k) {
        if (k == 0 || head == null || head.next == null)
            return head;
        ListNode node = head;
        int length = 1;
        while (node.next != null) {
            node = node.next;
            length++;
        }
        k = k % length;
        k = length - k; //值的范围 [1, length]，表示要进行移动的次数
        //循环链表
        node.next = head;
        while (k > 0) {
            head = head.next;
            node = node.next;
            k--;
        }
        node.next = null;
        return head;
    }
}