摘要: // at least two nodesif(!head || !head->next) return head;// find the middle nodeListNode *slow = head;ListNode *fast = head;while(fast->next && fast-... 阅读全文
posted @ 2015-10-08 16:53 daijkstra 阅读(333) 评论(0) 推荐(0) 编辑