找元组中差值最大最小个数

Posted 2021-03-22 C_hp

题目：有n个数，两两组成二元组，相差最小的有多少对呢？相差最大呢？

例如 ar[] = {1,3,4,9}，返回 1  　1

br[] = {1,2,3,10,12,21}，返回 2　　1

我的思路是将元素两两做差，将差值保存在一个数组内，把数组进行排序，即可找出最小差值多少个，最大差值多少个。（看起来我的时间空间复杂度低不了）

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
void func(int n, int ar[], int br[])
{
	int size = (n*n-n)/2;
	vector<int> vec;
	for(int i=0; i<n-1; ++i)
	{
		for(int j=i+1; j<n; ++j)
		{
			if(ar[i]>ar[j])
				vec.push_back(ar[i]-ar[j]);
			else
				vec.push_back(ar[j]-ar[i]);
		}
	}
	sort(vec.begin(), vec.end());
	for(auto i=vec.begin(); i!=vec.end(); ++i)
		cout<<*i<<" ";
	cout<<endl;
	cout<<"vec.size() = "<<vec.size()<<endl;
	if(vec[0] != vec[size-1])
	{
		int min = 0, max = 0;
		int i = 0;
		while(vec[i] == *vec.begin())
		{++min;++i;}
		i = size-1;
		while(vec[i] == vec[size-1])
		{++max; --i;}
		br[0] = min;
		br[1] = max;
		return;
	}
	br[0] = size;
	br[1] = size;
	return;
}
int main()
{
	int ar[] = {1,2,3,4,5,6,7,8,9,0,234, -132, 354325, 52354, 123}, br[2]={0};
	func(sizeof(ar)/sizeof(ar[0]), ar, br);
	cout<<br[0]<<" "<<br[1]<<endl;
	return 0;
}

 运行结果

 

 因该还有更好地思路来节省空间时间

于是我换了一种思路，在每次拿到差值的时候就比较它是不是最大或最小值，是就计数，不是就跳过得到下面的算法

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
void func(int n, int ar[], int br[])
{
	int min = -1, max = -1, maxc = 0, minc = 0, tmp = 0;
	for(int i=0; i<n-1; ++i)
	{
		if(!i)
			min = ((ar[0]-ar[1])>0 ? ar[0]-ar[1] : ar[1]-ar[0]);
		for(int j=i+1; j<n; ++j)
		{
			tmp = ar[i]-ar[j];
			if(tmp<0)
				tmp = -tmp;

			if(tmp<min){min=tmp;minc=1;continue;}
			else if(tmp==min){++minc;continue;}
			else if(tmp>max){max=tmp;maxc=1;continue;}
			else if(tmp==max){++maxc;continue;}
			else continue;	
		}
	}
	br[0]=minc;
	br[1]=maxc;
}

int main()
{
	int ar[] = {1,2,3,4,5,6,7,8,9,0,234, -132, 354325, 52354, 123}, br[2]={0};
	func(sizeof(ar)/sizeof(ar[0]), ar, br);
	cout<<br[0]<<" "<<br[1]<<endl;
	return 0;
}

 执行结果

 

看起来节省了空间