1 class Solution {

 2 public:

 3     vector<int> topKFrequent(vector<int>& nums, int k) {

 4         assert( k > 0 );

 5

 6         unordered_map<int ,int > freq;

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

 8             freq[nums[i]] ++;

 9         assert( k <= freq.size() );

10

11         // 优先队列中，数据对是（频率，元素）的形式

12         priority_queue< pair<int,int> , vector<pair<int,int>>, greater<pair<int,int>>> pq;

13         for( unordered_map<int,int>::iterator iter = freq.begin() ;

14             iter != freq.end() ; iter ++){

15             if(pq.size() == k ){

16                 if( iter->second > pq.top().first){

17                     pq.pop();

18                     pq.push( make_pair( iter->second, iter->first ) );

19                 }

20             }

21             else

22                 pq.push( make_pair( iter->second, iter->first ) );

23         }

24

25         vector<int> res;

26         while( !pq.empty() ){

27             res.push_back( pq.top().second );

28             pq.pop();

29         }

30         return res;

31        }

32 };