1074. Reversing Linked List (25)【链表翻转】——PAT (Advanced Level) Practise
2016-01-02 20:39
531 查看
题目信息
1074. Reversing Linked List (25)时间限制400 ms
内存限制65536 kB
代码长度限制16000 B
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 10^5) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
解题思路
提取出链表后分组翻转AC代码
[code]#include <cstdio> #include <vector> #include <map> #include <algorithm> using namespace std; struct { char front[8], next[8]; int value; }node[100005]; int main() { int begin, n, k; map<int, pair<int, int> > mp; scanf("%d%d%d", &begin, &n, &k); for (int i = 0; i < n; ++i){ scanf("%s%d%s", node[i].front, &node[i].value, node[i].next); mp[atoi(node[i].front)] = make_pair(atoi(node[i].next), i); } vector<int> rs; while (begin != -1){ rs.push_back(mp[begin].second); begin = mp[begin].first; } for (int i = 0; i + k <= rs.size(); i += k){ reverse(rs.begin() + i, rs.begin() + i + k); } printf("%s %d ", node[rs[0]].front, node[rs[0]].value); for (int i = 1; i < rs.size(); ++i){ printf("%s\n%s %d ", node[rs[i]].front, node[rs[i]].front, node[rs[i]].value); } printf("-1\n"); return 0; }
个人游戏推广:
《10云方》与方块来次消除大战!
相关文章推荐
- 四.OC基础--1.文档安装和方法重载,2.self和super&static,3.继承和派生,4.实例变量修饰符 ,5.私有变量&私有方法,6.description方法
- UML类图的关系
- 创建数据源的问题
- service通信
- (三) UART 串口通讯
- 安装archlinux
- 主系统linux + 次系统windows7
- 软件测试试题
- jquery操作select
- Pyhton 2.5 高级特性
- Doubango之Sip协议栈分析----阿冬专栏!!!
- display:none与visible:hidden的区别
- 栈与队列
- 瀑布模型、螺旋模型、敏捷开发
- Android 中的 Intent
- UIWindow的一些基本概念和注意点
- Vijos p1518 河流 转二叉树左儿子又兄弟
- Java中forEach, 用来遍历数组
- 编程的智慧(王垠)(http://www.cocoachina.com/programmer/20151125/14410.html)
- (转)一切伟大的创造都来源于行业间的相互借鉴模仿