PAT 1101 Quick Sort[一般上]
1101 Quick Sort(25 分)
There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. Given N distinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?
For example, given N=5 and the numbers 1, 3, 2, 4, and 5. We have:
- 1 could be the pivot since there is no element to its left and all the elements to its right are larger than it;
- 3 must not be the pivot since although all the elements to its left are smaller, the number 2 to its right is less than it as well;
- 2 must not be the pivot since although all the elements to its right are larger, the number 3 to its left is larger than it as well;
- and for the similar reason, 4 and 5 could also be the pivot.
Hence in total there are 3 pivot candidates.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤105). Then the next line contains N distinct positive integers no larger than 109. The numbers in a line are separated by spaces.
Output Specification:
For each test case, output in the first line the number of pivot candidates. Then in the next line print these candidates in increasing order. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:
5
1 3 2 4 5
Sample Output:
3
1 4 5
题目大意:找出pivot,即其左边都小于它,右边都大于它。并且输出。数据量还挺大的。
//我的想法就是遍历,那么一定会超时的啊。所以想看看大佬的写法
代码来自:https://www.liuchuo.net/archives/1917
#include <iostream>
#include <algorithm>
#include <vector>
#include<stdio.h>
int a[], b[];
using namespace std;
int main() {
int n;
scanf("%d", &n);
for (int i = ; i < n; i++) {
scanf("%d", &a[i]);
b[i] = a[i];
}
sort(a, a + n);//进行排序
int v[], max = , cnt = ;
for (int i = ; i < n; i++) {
if(a[i] == b[i] && b[i] > max)
v[cnt++] = b[i];
if (b[i] > max)
max = b[i];
}
printf("%d\n", cnt);
for(int i = ; i < cnt; i++) {
if (i != ) printf(" ");
printf("%d", v[i]);
}
printf("\n");
return ;
}
思想:很容易证明如果一个数是pivot,那么排序之后其位置肯定是不变的,(不排之前左边都比它小,右边比它大;排完之后,也是这么个情况)
1.另外当special case主元个数为0的时候,是特殊情况。
2.如果最终不输出printf("\n");会有一个格式错误,以后需要注意。
另一种解法,因为pivot拥有的性质是比左边最大的数大,比右边最小的数小。
所以一次正向遍历,求出那些比左边最大的数大的数标记为1,反向遍历一次将那些比右边最小数小的&&被标记为1的存进pivot数组,然后倒序输出(从小到大)。
代码来自:https://blog.csdn.net/kid_lwc/article/details/54780837
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std; const int INF = ;
const int N = ;
int a[N];
int la[N];
int flag[N];
int ans[N];
int main()
{
int n;
cin>>n;
int ma=;
for(int i=;i<n;i++){
cin>>a[i];
ma=max(ma,a[i]);
if(ma==a[i]) flag[i]=;//表示比左边最大的大。
}
int mm=INF;
int cnt=;
for(int i=n-;i>=;i--){
mm=min(mm,a[i]);
if(mm==a[i] && flag[i]==)//相等表示最小的是它。
{
ans[cnt++]=a[i];//由大到小放进去
} }
cout<<cnt<<endl;
for(int i=cnt-;i>=;i--){
cout<<ans[i];
if(i) cout<<" ";
}
cout<<endl;
return ;
}
PAT 1101 Quick Sort[一般上]的更多相关文章
- PAT 1101 Quick Sort
There is a classical process named partition in the famous quick sort algorithm. In this process we ...
- PAT甲1101 Quick Sort
1101 Quick Sort (25 分) There is a classical process named partition in the famous quick sort algorit ...
- PAT甲级——1101 Quick Sort (快速排序)
本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90613846 1101 Quick Sort (25 分) ...
- PAT 甲级 1101 Quick Sort
https://pintia.cn/problem-sets/994805342720868352/problems/994805366343188480 There is a classical p ...
- 1101. Quick Sort (25)
There is a classical process named partition in the famous quick sort algorithm. In this process we ...
- 1101 Quick Sort
There is a classical process named partition in the famous quick sort algorithm. In this process we ...
- 1101 Quick Sort(25 分
There is a classical process named partition in the famous quick sort algorithm. In this process we ...
- PAT (Advanced Level) 1101. Quick Sort (25)
树状数组+离散化 #include<cstdio> #include<cstring> #include<cmath> #include<map> #i ...
- PAT甲题题解-1101. Quick Sort (25)-大水题
快速排序有一个特点,就是在排序过程中,我们会从序列找一个pivot,它前面的都小于它,它后面的都大于它.题目给你n个数的序列,让你找出适合这个序列的pivot有多少个并且输出来. 大水题,正循环和倒着 ...
随机推荐
- Git学习之Git 暂存区
============================= 修改文件后是否可以直接提交 ============================ (1) 向文件中追加一行内容 $ echo &quo ...
- RH318之域控服务器
Windows2012域控服务器 一.安装域服务及DNS 1.配置静态IP 2.点击左下角 3.进入--->服务器管理器 4.点击角色和功能 勾选Active Directory域服务与DNS服 ...
- Elasticsearch学习之深入搜索一 --- 提高查询的精准度
1. 为帖子增加标题字段 POST /forum/article/_bulk { "} } { "doc" : {"title" : "th ...
- sencha touch tpl 实现按钮功能
js如下: Ext.define('app.view.message.Info', { alternateClassName: 'messageInfo', extend: 'Ext.Containe ...
- jquery validator
jQuery.validate是一款非常不错的表单验证工具,简单易上手,而且能达到很好的体验效果,虽然说在项目中早已用过,但看到这篇文章写得还是不错的,转载下与大家共同分享. 一.用前必备 官方网站: ...
- html5media 视频
官网: https://html5media.info/ 二.引入script <script src="//api.html5media.info/1.1.8/html5media. ...
- 【BZOJ1294】[SCOI2009]围豆豆Bean 射线法+状压DP+SPFA
[BZOJ1294][SCOI2009]围豆豆Bean Description Input 第一行两个整数N和M,为矩阵的边长. 第二行一个整数D,为豆子的总个数. 第三行包含D个整数V1到VD,分别 ...
- 【CF896E】Welcome home, Chtholly 暴力+分块+链表
[CF896E]Welcome home, Chtholly 题意:一个长度为n的序列ai,让你支持两种操作: 1.l r x:将[l,r]中ai>x的ai都减去x.2.l r x:询问[l,r ...
- 配置java环境变量后没有生效的解决办法
参考文章:https://blog.csdn.net/tooky_poom/article/details/60768458 系统安装了jdk1.7,环境变量正常,但是安装jdk1.8后,修改环境变量 ...
- RabbitMQ服务端配置详解(转自:http://www.cnblogs.com/zhen-rh/p/6884297.html)
RabbitMQ支持三种配置方式: 1) 读取环境变量中配置, 这包括shell中环境变量和rabbitmq-env.conf/rabbitmq-env-conf.bat文件中配置的环境变量 可配置如 ...