PAT1101:Quick Sort

1101. Quick Sort (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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

思路

实现快排中取划分点的这一部分代码（而且只是一种最简单的找划分点的方法）。

题目要求让你统计一组数中有哪些数满足划分点的要求。

1.从左至右遍历找不满足本身比左侧所有数大的点。

2.从右到左遍历找不满足本身比右侧所有数小的点(如果之前在从左到右的遍历中这个点已经不满足条件了就不用再计数了)。

3.打印结果就行

代码

#include<iostream>
#include<vector>
using namespace std;
vector<int> nums();
vector<bool> check(,true);
const int INITMAX = ;
const int INITMIN = -;
using namespace std;
int main()
{
   int N;
   while(cin >> N)
   {
      int countnum = N,lminnum = INITMIN,rmaxnum = INITMAX;
      for(int i = ;i < N;i++)
      {
          cin >> nums[i];
          if(nums[i] > lminnum)
          {
              lminnum = nums[i];
              continue;
          }
          check[i] = false;
          countnum--;
      }
      for(int i = N - ;i >=;i--)
      {
          if(nums[i] < rmaxnum)
          {
              rmaxnum = nums[i];
              continue;
          }
          if(check[i])  //避免重复计数
          {
              check[i] = false;
              countnum--;
          }

      }
      bool isFirst = true;
      cout << countnum << endl;
      for(int i = ;i < N;i++)
      {
          if(check[i])
          {
              if(isFirst)
              {
                cout << nums[i];
                isFirst = false;
              }
              else
                cout << " " << nums[i];
          }
      }
      cout << endl;
   }
}