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.
先算出链表的长度，然后将链表的尾部与头部连起来形成环，
然后再开始的len-k%len处断开，即可
ListNode *rotateRight(ListNode *head, int k) { if(head == NULL || k == 0 ) return head; ListNode *p = head; int len = 1; while(p->next){len++;p=p->next;}; p->next = head; k = len-k; int step = 0; while(step < k){ p=p->next; step++; } head = p->next; p->next = NULL; return head; }
Leetcode Rotate List,布布扣,bubuko.com
原文：http://www.cnblogs.com/xiongqiangcs/p/3804943.html