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:任何问题都要先检查validation，special case  

注意：【这道题我的速度是18ms，还有要优化的地方。未解决？】

class Solution { public:    ListNode *rotateRight(ListNode *head, int k) {        //check validation        if(head==NULL || k==0) return head;        // special case : only one node        if(head->next==NULL) return head;        //compute length, becase we need to figure out :k, lenght, who is bigger?        int length=0;        ListNode *p=head;        while(p){            length++;            p=p->next;        }        //compute the offset        int offset=0;        if(k%length==0) return head;        else offset=k%length;                int movesteps=length-offset;                //rotate start        ListNode *phead=head;        while(--movesteps>0){            phead=phead->next;        }        p=phead;        phead=phead->next;        p->next=NULL;        p=phead;                while(p->next!=NULL) p=p->next;        p->next=head;                return phead;     }};