找出n个元素中最小的k个数(编程珠玑第二章)

思想：用最大堆操作来减少时间复杂度，最终的时间复杂度为：O(k)+(n-k)log(k)，其中O(k)是建堆时间，log(k)是每次调整的时间，最多进行n-k次调整。

代码如下：

#include<iostream>

#include<vector>

using namespace std;

void adjust(int k,int m,vector<int>&num)

{

         intj=k;

         while(j<=(m-2)>>1)

         {

                  if(num[j]>num[2*j+1]&& num[j]>num[2*j+2])

                          break;       

                  elseif(num[2*j+1]<=num[2*j+2])

                  {

                          swap(num[j],num[2*j+2]);

                          j=2*j+2;

                  }

                  else

                  {

                          swap(num[j],num[2*j+1]);

                          j=2*j+1;

                  }

         }

}

void makeHeap(intm,vector<int> &num)

{

         for(int i=(m-2)>>1;i>=0;i--)

                  adjust(i,m,num);

}

void partialSort(intm,vector<int> &num)//找出最小的m个元素

{

         makeHeap(m,num);//建最大堆

         for(int i=m;i<num.size();i++)

         {

                  if(num[i]<num[0])

                  {

                          swap(num[i],num[0]);

                          adjust(0,m,num);//不断的把最大元素换出，并且重新调整堆

                  }

         }

}

int main()

{

         intArr[]={3,9,6,8,-10,7,-11,19,30,12,23};

         vector<int>num(Arr,Arr+11);

         partialSort(5,num);

         for(int i=0;i<num.size();i++)

                  cout<<num[i]<<"       ";

         cout<<endl;

         system("pause");

         return0;

}