Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2Output: 4->5->1->2->3->NULLExplanation:rotate 1 steps to the right: 5->1->2->3->4->NULLrotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4Output: 2->0->1->NULLExplanation:rotate 1 steps to the right: 2->0->1->NULLrotate 2 steps to the right: 1->2->0->NULLrotate 3 steps to the right: 0->1->2->NULLrotate 4 steps to the right: 2->0->1->NULL

难度：medium

题目：给定链表，向右旋转k个位置，k为非负整数。

思路：首先统计结点个数，并记录尾结点。然后用前后指针找出倒数第k + 1个结点。

Runtime: 7 ms, faster than 92.47% of Java online submissions for Rotate List.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode rotateRight(ListNode head, int k) {        if (null == head || k <= 0) {            return head;        }        int cnt = 0;        ListNode ptr = head,tail = ptr;        while (ptr != null) {            tail = ptr;            ptr = ptr.next;            cnt++;        }        k = k % cnt;        ListNode dummyHead = new ListNode(0), lastKPtr = dummyHead;        dummyHead.next = head;        ptr = dummyHead.next;        while (ptr != null) {            if (--k < 0) {                lastKPtr = lastKPtr.next;            }            ptr = ptr.next;        }        tail.next = dummyHead.next;        dummyHead.next = lastKPtr.next;        lastKPtr.next = null;                return dummyHead.next;    }}