Time Cost

11min1s

Implementation

Use min heap to save k-th frequent elements

Code

• Solution

  class Solution {
  public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> counts;
        for (int num: nums) {
            if (counts.count(num)) {
                counts[num]++;
            }else{
                counts[num]=0;
            }
        }
  
        priority_queue<pair<int, int>, 
            vector<pair<int, int>>, 
            greater<pair<int, int>>> minheap;
          
        for (auto const& [key, value]: counts) {
            if (minheap.size() < k) {
                minheap.emplace(value, key);
            }else{
                if (value > minheap.top().first) {
                    minheap.pop();
                    minheap.emplace(value, key);
                }
            }
        }
  
        vector<int> result;
        while (!minheap.empty()) {
            result.push_back(minheap.top().second);
            minheap.pop();
        }
  
        return result;
    }
  };