pat甲级1101 Quick Sort(左右数组）（简单模拟）

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本文深入探讨了快速排序算法中的经典分区过程，分析了如何确定一个元素是否可以作为枢轴，以及如何计算满足条件的枢轴候选数量。通过具体实例说明了算法的实现步骤，并提供了一段C++代码示例，帮助读者理解并掌握这一关键概念。

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

/**
本题题意就是判断给定一组数字，判断这这组数字中，满足左边都比指定数小，右边的数都比指定的数大，
判断这组数字中有多少个这样的数字
本题思路：
   和P1039类似，设定left , right数组分别将每个数左边的最大值，右边的最小值存入数组中
   for(int i = 1; i < n; i++){
       leftmax[i] = max(leftmax[i - 1], a[i - 1]);
       rightmin[n - i - 1] = min(rightmin[n - i], a[n - i]);
   }
   最后进行遍历，将满足条件的数放入 v向量容器中.
**/

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxsize = 1e5 + 5;
const int inf = 0x7fffffff;
int n, a[maxsize], leftmax[maxsize], rightmin[maxsize];
vector<int> v;
int main(){
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%d", &a[i]);
	}
	rightmin[n - 1] = inf;
	for(int i = 1; i < n; i++){
		leftmax[i] = max(leftmax[i - 1], a[i - 1]);
		rightmin[n - i - 1] = min(rightmin[n - i], a[n - i]);
	}
	for(int i = 0; i < n; i++){
		if(a[i] < rightmin[i] && a[i] > leftmax[i])
			v.push_back(a[i]);
	}
	printf("%d\n", v.size());
	if(v.size() == 0) //不判断0的特殊情况会出现格式错误 
		printf("\n");
	for(int i = 0; i < v.size(); i++){
		printf("%d", v[i]);
		if(i != v.size() - 1)
			printf(" ");
	}
	return 0;
}