利用快排思想找数组中第K小的数

#include<iostream>
#include<vector>
using namespace std;
int QP(vector<int>arr, int begin, int end,int k)
{
	if (begin == end)return arr[begin];
	int first = begin;
	int last = end;
	int key = arr[first];
	while (first < last)
	{
		while (first < last&&key <= arr[last])
			last--;
		arr[first] = arr[last];
		while (first < last&&key >= arr[first])
			first++;
		arr[last] = arr[first];
	}
	arr[first] = key;
	if (first+1 == k)return arr[first];
	else if (first+1 < k)return QP(arr, first + 1, end, k);
	else return QP(arr, begin, first - 1, k);
}
int findk(vector<int>arr,int k)
{
	int len = arr.size();
	return QP(arr, 0, len - 1, k);
}
int main()
{
	vector<int>a = { 6,5,7,8,1,3,4,9,2,0 };
	int No = findk(a, 9);
	cout << No;
}

我在想数组从小到大前K个数是不是可以先把第K个数找出来，然后再遍历一遍数组把所有比该数字小的数字全部输出出来？