top_k问题

来源：互联网发布：键盘按键显示软件编辑：程序博客网时间：2024/06/17 06:03

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;}

0 0