The Kth smallest (or largest) number in the unordered array (the Dutch flag problem method)

Find the K-th smallest (larger) number in an unordered array—using the method of the Dutch flag problem to solve
Time complexity (expected value): O(N)

#include <iostream>
#include <vector>
#include <ctime>
using namespace std;

void swap(vector<int>& arr, int n1, int n2) {
    
    
	int tmp = arr.at(n1);
	arr.at(n1) = arr.at(n2);
	arr.at(n2) = tmp;
}

vector<int> partition(vector<int>& arr, int left, int right) {
    
    
	// 随机选择一个值作为划分值
	int index = left + (rand() % (right - left + 1));
	// 将选择的随机值置换到末尾
	swap(arr, index, right);
	int tmpL = left - 1;
	int tmpR = right;
	index = left;
	while (index < tmpR) {
    
    
		if (arr.at(index) < arr.at(right)) {
    
    
			swap(arr, index++, ++tmpL);
		}
		else if (arr.at(index) > arr.at(right)) {
    
    
			swap(arr, index, --tmpR);
		}
		else {
    
    
			index++;
		}
	}
	swap(arr, index, right);
	vector<int> res{
    
     tmpL, tmpR + 1 };
	return res;
}

// 时间复杂度: O(N) 
// 最差情况每次只排好一个数O(N^2); 最好情况T(N)=T(N/2)+O(N)=O(N);     期望值O(N)
int getMinKth(vector<int>& arr, int k, int left, int right) {
    
    
	if (arr.empty() || arr.size() < k || k < left || k > right || right < left) {
    
    
		return -1;
	}
	// 进行划分: < 左，= 中，> 右
	vector<int> range = partition(arr, left, right);
	// 若K恰好在 等于区域 直接返回
	if (k > range.at(0) && k < range.at(1)) {
    
    
		return arr.at(k);
	}
	// 否则，如果k在 小于区域 小于区域继续递归，如果k在 大于区域 大于区域继续递归
	return k <= range.at(0) ? getMinKth(arr, k, left, range.at(0)) : getMinKth(arr, k, range.at(1), right);
}

int main() {
    
    
	srand(time(0));
	vector<int> arr{
    
     10, 4, 2, 9, 0, 12, 10, 22 };
	cout << getMinKth(arr, 6, 0, arr.size() - 1) << endl;

	system("pause");
	return 0;
}

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