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找到第K小的元素(分治)

程序员文章站 2022-04-21 23:26:30
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先做一个简单好理解的,找到第一小和第二小的数

#include<iostream>
using namespace std;

void select(int *ar,int n)
{
	if(ar == NULL && n < 2)
		return ;
	int min1 = ar[0] < ar[1] ? ar[0]:ar[1];
	int min2 = ar[0] < ar[1] ? ar[1]:ar[0];
	for(int i = 2;i < n; i++)
	{
		if(ar[i] < min1)
		{
			min2 = min1;
			min1 = ar[i];
		}
		else if(ar[i] < min2)
		{
			min2 = ar[i];	
		}
	}
	cout << "min1:"<< min1<<" min2:"<< min2<<endl;
}
int main()
{
	int ar[]={12,23,5,65,78,25,32,15,96,99};
	int len =sizeof(ar)/sizeof(ar[0]);
        select(ar,len);
	return 0;
}

运行结果:

找到第K小的元素(分治)

找到第K小的元素:

​​​​#include<iostream>
#include<assert.h>
using namespace std;

template<class Type>
int Partition(Type *ar,int left,int right)
{
	int i = left,j = right;
	Type tmp = ar[i]; 
	while(i < j)
	{
		while(i<j && ar[j] > tmp) --j;
		if(i<j) ar[i] = ar[j];
		while(i<j && ar[i] <= tmp) ++i;
		if(i<j) ar[j] = ar[i];
	}
	ar[i] = tmp;
	return i;
}

template<class Type>
const Type & Select_K(Type *ar,int left,int right,int k)
{
	if(left == right && k == 1) return ar[left];
	int index = Partition(ar,left,right);
	int pos = index - left + 1;
	if(k <= pos) return Select_K(ar,left,index,k);
	else return Select_K(ar,index+1,right,k-pos);
}

template<class Type>
const Type & Select_K_Min(Type *ar,int n, int k)
{
	assert(ar != NULL && n > 0 && k>=1 && k <= n);

	return Select_K(ar,0,n-1,k);
}

int main()
{
	int ar[] = {12,23,5,65,78,25,32,15,96,99};
	int n = sizeof(ar)/sizeof(ar[0]);
	for(int k = 1; k <= n; k++)
	{
		cout<<k<<" => "<<Select_K_Min(ar,n,k)<<endl;
	}
	return 0;
}

运行结果:

 

找到第K小的元素(分治)

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