输出最小的k个数

//题目：输入最小的n个数，找出其中最小的k个数
//思路：基于快排的思想，使比第k个数字小的数字小的都位于数组左边，比第k个数字大的所有数字都位于数组右边
//缺点：需要修改输入的数组
//时间复杂度：O(n)

#include<stdio.h>
#include<iostream>
#include <time.h>
using namespace std;

int RandomInRange(int start, int end)
{
	srand((int)time(0));
	int temp = rand();
	int number = temp % (end - start + 1) + start;
	return number;

}
void Swap(int * a, int * b)
{
	int temp = *a;
	*a = *b;
	*b = temp;
}

int Partition(int data[], int length, int start, int end)
{
	if (data == nullptr || length <= 0 || start < 0 || end >= length)
		throw new std::exception("error");

	int index = RandomInRange(start, end);

	Swap(&data[index], &data[end]); //交换index和end，end中存储分解值，index作为遍历下标

	int border = start - 1; //分界边界

	for (index = start;index < end; ++index)
	{
		if (data[index] < data[end])
		{
			border += 1;
			if (border != index) //只是为了不对同一个位置的数交换，减少操作，对算法无影响
				Swap(&data[index], &data[border]);
		}
	}
	++border;
	Swap(&data[border], &data[end]); //data[border]存储分解值，左边都< data[border], 右边都>=border
	return border;

}

void GetLeastNumbers1(int * input, int length, int * output, int k)
{
	int start = 0;
	int end = length - 1;
	int index = Partition(input, length, start, end);
	while (index != (k-1))
	{
		if (index < (k-1))
		{
			start = index + 1;
			index = Partition(input, length, start, end);

		}
		else
		{
			end = index - 1;
			index = Partition(input, length, start, end);
		}
			
	}
	for (int i = 0; i < k;++i)
	{
		output[i] = input[i];
	}
}

void test()
{
	int length = 10;
	int input[10] = { 2,3,5,1,7,9,1,5,2,8 };
	int k = 5;
	int output[5];
	GetLeastNumbers1(input, length, output, k);
	for (int i = 0; i < k;++i)
	{
		printf("%d ", output[i]);
	}
	printf("\n");
}

int main()
{
	test();
	system("pause");
	return 0;
}