Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

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 (≤10​3​​). 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 in [−10​5​​,10​5​​], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

 #include <stdio.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=;
struct node{
int data;
int pos=;
int address;
int next;
int rank;
}nodes[maxn];
vector<node> v;
int n,k,root;
bool cmp(node n1,node n2){
if(n1.rank<n2.rank) return true;
else if(n1.rank>n2.rank) return false;
else{
return n1.pos<n2.pos;
}
}
int main(){
scanf("%d %d %d",&root,&n,&k);
for(int i=;i<n;i++){
int first,data,next;
scanf("%d %d %d",&first,&data,&next);
node tmp;
tmp.address = first;
tmp.data = data;
tmp.next = next;
if(data<) tmp.rank=;
else if(data<=k) tmp.rank=;
else tmp.rank=;
nodes[first]=tmp;
//v.push_back(tmp);
}
int j=;
while(root!=-){
nodes[root].pos=j;
v.push_back(nodes[root]);
root=nodes[root].next;
j++;
}
sort(v.begin(),v.end(),cmp);
for(int i=;i<v.size();i++){
if(i!=v.size()-){
printf("%05d %d %05d\n",v[i].address,v[i].data,v[i+].address);
}
else {
printf("%05d %d -1\n",v[i].address,v[i].data);
}
}
}

注意点:看到只想着排序了,看大佬的方法真简单,直接三个vector,再合起来打印输出。

一开始最后第二个测试点,最后一个超时,超时是因为一开始存数据直接存在vector里,然后要得到正确的顺序每次都要遍历一遍整个vector,时间复杂度很高。而且最开始的pos其实设置的也是有问题的,直接根据输入顺序得到了,神奇的是结果居然前几个都是对的。

最后存储链表还是要用静态数组,不能用vector,找next太耗时

PAT A1133 Splitting A Linked List (25 分)——链表的更多相关文章

  1. PAT 甲级 1074 Reversing Linked List (25 分)(链表部分逆置,结合使用双端队列和栈,其实使用vector更简单呐)

    1074 Reversing Linked List (25 分)   Given a constant K and a singly linked list L, you are supposed ...

  2. PAT A1133 Splitting A Linked List (25) [链表]

    题目 Given a singly linked list, you are supposed to rearrange its elements so that all the negative v ...

  3. PAT甲级1074 Reversing Linked List (25分)

    [程序思路] 先根据地址按顺序读入节点,入栈,当栈里的元素个数等于k时全部出栈,并按出栈顺序保存,最后若栈不为空,则全部出栈并按出栈的稀饭顺序保存,最后输出各节点 注意:输入的节点中有可能存在无用节点 ...

  4. PTA 02-线性结构3 Reversing Linked List (25分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/664 5-2 Reversing Linked List   (25分) Given a ...

  5. PAT乙级:1090危险品装箱(25分)

    PAT乙级:1090危险品装箱(25分) 题干 集装箱运输货物时,我们必须特别小心,不能把不相容的货物装在一只箱子里.比如氧化剂绝对不能跟易燃液体同箱,否则很容易造成爆炸. 本题给定一张不相容物品的清 ...

  6. PAT乙级:1070 结绳 (25分)

    PAT乙级:1070 结绳 (25分) 题干 给定一段一段的绳子,你需要把它们串成一条绳.每次串连的时候,是把两段绳子对折,再如下图所示套接在一起.这样得到的绳子又被当成是另一段绳子,可以再次对折去跟 ...

  7. PAT-2019年冬季考试-甲级 7-2 Block Reversing (25分) (链表转置)

    7-2 Block Reversing (25分)   Given a singly linked list L. Let us consider every K nodes as a block ( ...

  8. PAT 1133 Splitting A Linked List[链表][简单]

    1133 Splitting A Linked List(25 分) Given a singly linked list, you are supposed to rearrange its ele ...

  9. PAT 甲级 1040 Longest Symmetric String (25 分)(字符串最长对称字串,遍历)

    1040 Longest Symmetric String (25 分)   Given a string, you are supposed to output the length of the ...

随机推荐

  1. HDU6215

    Brute Force Sorting Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Othe ...

  2. tortoisegit安装

    1.下载tortoisegit:https://tortoisegit.org/download/ 2.下载git 64位 3. 双击开始安装,选择默认,点击下一步 4.接着是选择安装目录,可以保持默 ...

  3. MySql Host is blocked because of many connection errors; unblock with 'mysqladmi

    原因: 同一个ip在短时间内产生太多(超过mysql数据库max_connection_errors的最大值)中断的数据库连接而导致的阻塞: 解决方法: 1.提高允许的max_connection_e ...

  4. (网页)JS中的小技巧,但十分的实用!

    转自CSDN: 1.document.write(”"); 输出语句2.JS中的注释为//3.传统的HTML文档顺序是:document->html->(head,body)4. ...

  5. angular升级指南

    废话少说 上链接 angular 升级指南

  6. 我的博客:C# PHP J2ee Java Android js WP Asp.net mvc Python

    <p><A target="_blank" href="http://blog.163.com/hr_company_product/" &g ...

  7. Spring Boot 中配置文件application.properties使用

    一.配置文档配置项的调用(application.properties可放在resources,或者resources下的config文件夹里) package com.my.study.contro ...

  8. 将爬取的数据保存到mysql中

    为了把数据保存到mysql费了很多周折,早上再来折腾,终于折腾好了 安装数据库 1.pip install pymysql(根据版本来装) 2.创建数据 打开终端 键入mysql -u root -p ...

  9. sql server 2008R2无人值守批处理脚本自动化安装

    ▲版权声明:本文为博主原创文章,未经博主允许不得转载. Microsoft SQL Server 2008 R2是一款软件,提供完整的企业级技术与工具,帮助您以最低的总拥有成本获得最有价值的信息.您可 ...

  10. 【PAT】B1072 开学寄语(20 分)

    代码注释应该很清晰 先存下违禁品,放到数组中,未使用map #include<cstdio> #include<string.h> int wupin[10],N,M; boo ...