347. Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

Note: 

• You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
• Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
  
  

---

计算frequency，用map

找K个最值，用priority queue。这里是K个最大值，需要建一个min-heap，维持其大小不超过K。

struct pairCompare{
    bool operator()(pair<int,int>&p1, pair<int,int>&p2){
        return p1.first>p2.first;
    }
};

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int,int> map; //calculate frequency
        for(int num: nums){
            map[num]++;
        }
        vector<int> result;

        //heap in C++ by default using less as a max-heap
        //need a min-heap here (for k largest values)
        priority_queue<pair<int,int>,vector<pair<int,int>>,pairCompare> pq; 
        for(auto const it: map){
            int key = it.first;
            int val = it.second;
            if(pq.size()<k){
                pq.push(make_pair(val,key));
            }else{
                if(val>pq.top().first){
                    pq.pop();
                    pq.push(make_pair(val,key));
                }
            }
        }
        
        //loop all elements in the min-heap
        while(!pq.empty()){
            result.push_back(pq.top().second);
            pq.pop();
        }
        
        return result;
    }
};