top_k问题

这个方法适用于数组比较小的情况，可以用快速排序的思路来求解最小的k个数，时间复杂度为O（n），这个方法需要修改输入的数组，如果先排序在输出时间复杂度最好为O（nlogn）:

#include<iostream>
using namespace std;

int Partion(int a[], int left, int right)
{
	int key = a[left];
	while(left < right)
	{
		while(left < right && key <= a[right]) --right;
		a[left] = a[right];
		while(left < right && key >= a[left]) ++left;
		a[right] = a[left];
	}
	a[left] = key;
	return left;
}

void top_k(int a[], int n, int k)
{
	int left = 0;
	int right = n - 1;
	int index = Partion(a, left, right);
	while(index != k - 1)
	{
		if(index > k - 1)
		{
			right = index - 1;
			index = Partion(a, left, right);
		} else {
		    left = index + 1;
			index = Partion(a ,left, right);
		}
	}

	for(int i = 0; i < k; i++){
		cout << a[i] << endl;
	}
}

int main()
{
	int a[] = {4, 5, 1, 6, 2, 7, 3, 8};
 
	top_k(a, 8, 4);
	return 0;
}

针对海量数据可以用大根堆来保存前k个数据，后面的数跟根节点比较，时间复杂度是O（nlogk）:

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

typedef multiset<int, greater<int>>  Set;
typedef multiset<int, greater<int>>::iterator Iterator;

void top_k(vector<int> &data, Set &least, size_t k)
{
	least.clear();
	vector<int>::const_iterator iter = data.cbegin();
	for(; iter != data.cend(); ++iter)
	{
		if((least.size()) < k)
		  least.insert(*iter);
		else {
			Iterator iterGreatest = least.begin();
			if(*iter < *iterGreatest)
			{
				least.erase(iterGreatest);
				least.insert(*iter);
			}
		}
	}
}

int main()
{
	vector<int> data = {4, 5, 1, 6, 2, 7, 3, 8};
	Set least; 
	top_k(data, least, 4);
	auto iter = least.cbegin();
	while(iter != least.cend())
	  cout << *iter++ << endl;
	return 0;
}