Team Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1294    Accepted Submission(s): 442 Problem Description Queues and Priority Queues are data structures which are known to most computer…

题目大意: 不同的人在不同的队伍里,插入链表的时候假设这个链表里有他的队友,就把它放到最后一个队友的最后.假设没有队友,就把它放到整个链表的最后面. 出链表的时候把第一个人拿出来. 思路分析: 要模拟这个链表就要记录这整个链表中的队伍存在的情况. 所以要再开一个链表的头指针和尾指针,在整个大的链表中再模拟小区域的链表. 然后就是deque部分,也就是注意head的推断以及更新. #include <cstdio> #include <iostream> #include <a…

Team Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1889    Accepted Submission(s): 639 Problem Description Queues and Priority Queues are data structures which are known to most computer…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1387 Team Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1518    Accepted Submission(s): 511 Problem Description Queues and Priority Queues…

Team Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2051    Accepted Submission(s): 713 Problem Description Queues and Priority Queues are data structures which are known to most computer…

Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for exa…

利用C++ 单向链表实现数据结构队列,其实和上一篇基本内容相同,仅仅是插入的时候在链表的尾部插入,取元素都是一样的,都从头部取. #pragma once #include "stdio.h" //利用链表来实现队列,先进先出 class queue { public: queue(void); queue(int value); ~queue(void); private: int m_value; queue* m_pnext; public: void push(int valu…

单向链表: class error(Exception): def __init__(self,msg): super(error,self).__init__() self.msg=msg def __str__(self): return self.msg class Node: def __init__(self,ele): self.ele=ele self.next=None class mylist: def __init__(self,ele=None): self.index=0…

0 问题描述 原题点击这里. 将单向链表第m个位置到第n个位置倒序连接.例如, 原链表:1->2->3->4->5, m=2, n =4 新链表:1->4->3->2->1 (注:最终的新链表记为head,过程中临时使用的一个链表头记为h) 1 基本思路 首先考虑整个链表的情况.见到单向链表的第一反应自然是轮询链表head中每个节点,轮询过程中按需要建立一个新链表h,每次访问一个节点,就将这个节点放在前一个访问的节点之后,这样便实现了倒序. 然后再考虑部分倒…

第 13 题(链表):题目:输入一个单向链表,输出该链表中倒数第 k 个结点.链表的倒数第 0 个结点为链表的尾指针.链表结点定义如下: struct ListNode {int m_nKey;ListNode* m_pNext;}; 我的思路:先翻转链表,再从翻转后的链表的头向尾数k-1个,返回,再次翻转链表. 代码如下:注意这个思路非常差.差的原因是:如果只是用最原始的方法,先遍历一遍计数,再遍历一遍找倒数第k个,需要遍历两遍.但我的思路,翻转两次链表就要遍历两遍.还要在走k-1步找倒数第k…