【编程练习】快速select算法的实现

 代码来自:

http://blog.csdn.net/v_JULY_v

算法思想:

// Quick_select.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <time.h>

using namespace std;

const int num_array = 13;
const int num_med_array = num_array/5 + 1;

int array[num_array];
int midian_array[num_med_array];

/*
//插入排序算法伪代码
INSERTION-SORT(A)                                                 cost times
1 for j ← 2 to length[A]                                           c1 n
2 do key ← A[j]                                                    c2 n - 1
3 Insert A[j] into the sorted sequence A[1 ‥ j - 1]. 0...n - 1
4 i ← j - 1                                                        c4 n - 1
5 while i > 0 and A[i] > key                                        c5
6 do A[i + 1] ← A[i]                                               c6
7 i ← i - 1                                                        c7
8 A[i + 1] ← key                                                   c8 n - 1
*/


void insert_sort(int array[], int left, int loop_times)
{//这块的插入排序感觉有点问题,第一个数字没有排啊
	for (int j = left; j < left+loop_times; j++)
	{
		int key = array[j];
		int i = j - 1;

		while (i > left && array[i] > key)
		{
			array[i+1] = array[i];
			i--;
		}

		array[i+1] = key;
	}
}

void insertion_sort(int array[],int first,int last)
{
	int i,j;
	int temp;
	for(i = first + 1 ;i<=last;i++)
	{
		temp = array[i];
		j=i-1;
		//与已排序的数逐一比较,大于temp时,该数移后
		while((j>=0)&&(array[j]>temp))
		{
			array[j+1]=array[j];
			j--;
		}
		//存在大于temp的数
		if(j!=i-1)
		{array[j+1]=temp;}
	}

}

int find_median(int array[], int left, int right)
{
	if (left == right)
		return array[left];int index;
	for (index = left; index < right - 5; index += 5)
	{
		//insert_sort(array, index, 4);
		insertion_sort(array,index,4);
		int num = index - left;
		midian_array[num / 5] = array[index + 2];
	}
	// 处理剩余元素
	int remain_num = right - index + 1;
	if (remain_num > 0)
	{
		//insert_sort(array, index, remain_num - 1);
		insertion_sort(array,index,remain_num - 1);
		int num = index - left;
		midian_array[num / 5] = array[index + remain_num / 2];
	}
	int elem_aux_array = (right - left) / 5 - 1;
	if ((right - left) % 5 != 0)
		elem_aux_array++;
	// 如果剩余一个元素返回,否则继续递归
	if (elem_aux_array == 0)
		return midian_array[0];
	else
		return find_median(midian_array, 0, elem_aux_array);
}

// 寻找中位数的所在位置
int find_index(int array[], int left, int right, int median)
{
	for (int i = left; i <= right; i++)
	{
		if (array[i] == median)
			return i;
	}
	return -1;
}


int q_select(int array[], int left, int right, int k)
{
	// 寻找中位数的中位数
	int median = find_median(array, left, right);
	// 将中位数的中位数与最右元素交换
	int index = find_index(array, left, right, median);
	swap(array[index], array[right]);
	int pivot = array[right];
	// 申请两个移动指针并初始化
	int i = left;
	int j = right - 1;
	// 根据枢纽元素的值对数组进行一次划分
	while (true)
	{
		while(array[i] < pivot)
			i++;
		while(array[j] > pivot)
			j--;
		if (i < j)
			swap(array[i], array[j]);
		else
			break;
	}
	swap(array[i], array[right]);
	/* 对三种情况进行处理:(m = i - left + 1)
	1、如果m=k,即返回的主元即为我们要找的第k 小的元素,那么直接返回主元a[i]即可;
	2、如果m>k,那么接下来要到低区间A[0....m-1]中寻找,丢掉高区间;
	3、如果m<k,那么接下来要到高区间A[m+1...n-1]中寻找,丢掉低区间。
	*/
	int m = i - left + 1;
	if (m == k)
		return array[i];
	else if(m > k)
		//上条语句相当于if( (i-left+1) >k),即if( (i-left) > k-1 ),于此就与2.2 节里的
		//代码实现一、二相对应起来了。
		return q_select(array, left, i - 1, k);
	else
		return q_select(array, i + 1, right, k - m);
}


	
int _tmain(int argc, _TCHAR* argv[])
{
	//srand(unsigned(time(NULL)));
	//for (int j = 0; j < num_array; j++)
	int a[4] = {13,26,9,100};
	insert_sort(a,0,3);

	//insertion_sort(a,0,3);

	cout<<a[0]<<a[1]<<a[2]<<a[3]<<endl;


	//array[j] = rand();
	int array[num_array]={0,45,78,55,47,4,1,2,7,8,96,36,45};
	// 寻找第k 最小数
	int k = 13;
	int i = q_select(array, 0, num_array - 1, k);
	cout << i << endl;

	
	getchar();
	return 0;
}


 

原文地址:https://www.cnblogs.com/wuyida/p/6301408.html