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.

思路：

首先求出list的长度length，然后将最后一个结点指向初始的头结点，

然后从该初始结点出发，指针向后移动length-k次，

找到新链表的尾结点，然后把它的next置为NULL。

ListNode* rotateRight(ListNode* head, int k)    {        ListNode* result;        if (k == 0 || head == NULL) return head;        int length = 1;        ListNode* temp = head;        while (head != NULL && head->next != NULL)        {            head = head->next;            length++;        }        k %= length;        head->next = temp;//head 目前为最后一个结点的指针 指向初始时候的头部结点        for (int i = 0; i < length - k-1;i++)        {            temp = temp->next;        }         result = temp->next;        temp->next = NULL;        return result;    }