//方案一： O(N)? 不适合海量数据
int* maxK(int A[] , const int& n, const int& k){  if(n<k||n<=0||k<=0||A==NULL) cerr<<"Invalid Input"<<endl;  int* output = new int[k]();  if(n==k){         for(int i = 0; i < n; ++i)            output[i]=A[i];          return output;     }  int r = partition(A,0,n);  while(r!=k){       if(r<k)       r = partition(A,r+1,n);       else        r = partition(A,0,r-1);   }      for(int i = 0; i < n; ++i)            output[i]=A[i];       return output;  }

int partition(int A[], int begin, int end){
　　--end;
   int p = A[end];
   while(true){
      while(A[begin]<p) ++begin;
      while(A[end]>p) --end;
      if(begin>=end) return begin;
      swap(A[begin],A[end]);
      ++begin;
      --end;
   }
}

multiset<int>* maxK(const vector<int> &data, const int& k){        multiset<int,greater<int>> rst;        vector<int>::const_iterator iter = data.begin();        for(;iter!=data.end();++iter){               if(rst.size()<=k)                     rst.insert(*iter);              else{                 int maxData = http://www.mamicode.com/*rst.begin();                 if(maxData>*iter){                      rst.erase(maxData);                      rst.insert(*iter);                   }               }        }        return rst;}