[LeetCode347]Top K Frequent Elements

题目：

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.

分类：Hash Table Heap

代码：

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int,int> hash;
        /*
        for(int i = 0; i < nums.size(); ++i)
        {
             auto ret = hash.insert({nums[i],1});
             if(!ret.second)//如果已经存在
             {
                 ++ret.first->second;
             }
        }
        */
        for(auto num : nums)
            ++hash[num];
        
        //利用堆排序实现
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        for(auto m : hash)
        {
            pq.push({m.second,m.first});
            if(pq.size() > k)
                pq.pop();
        }
        
        //现在根据出现次数元素已经在堆中排好序了
        vector<int> res;
        while(!pq.empty())
        {
            res.push_back(pq.top().second);
            pq.pop();
        }
        reverse(res.begin(),res.end());
        return res;
    }
};