> 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.
解题思路：
找出分隔点，注意对输入k值的处理，k可能超过链表长度。另外别忘了链表结尾要手动置为NULL。
int length(ListNode* head)
{
int len = 0;
while(head)
++len, head = head->next;
return len;
}
ListNode *rotateRight(ListNode *head, int k)
{
if(head == NULL || head->next == NULL || k == 0) return head;
int len = length(head);
k = k % len;
if(k == 0) return head;
ListNode * newHead = NULL;
ListNode * headbak = head;
int i = 1;
while(i < len-k)
head = head->next, i++;
newHead = head->next;
head->next = NULL; //!!! important
ListNode *tail = newHead;
while(tail->next)
tail = tail->next;
tail->next = headbak;
return newHead;
}