旋转链表

输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

/**
 * 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 || head.next == null) {
            return head;
        }
        // 链表长度
        int length = getLength(head);
        // 取余
        k = k % length;
        // 算出头节点右移多少之后得到新的头节点
        k = length - k;
        // 尾节点，后面需要连到旧头节点
        ListNode tail = getTail(head);
        ListNode temp = head;
        while (k > 1) {
            temp = temp.next;
            k--;
        }
        // 此时，temp.next是新的头节点
        // 旧尾节点连到旧头节点
        tail.next = head;
        // 新头节点
        head = temp.next;
        // temp是新头节点的上一个节点，断开与新头结点的链接
        temp.next = null;
        return head;
    }

    private ListNode getTail(ListNode head) {
        while (head != null) {
            if (head.next == null) {
                break;
            }
            head = head.next;
        }
        return head;
    }

    private int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            length++;
            head = head.next;
        }
        return length;
    }
}