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.

很常見的top-k問題，我一開始的想法就是用哈希表來實(shí)現(xiàn)。
注意priority_queue的使用，排序是first，所以把second放在first。

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> map;
        priority_queue<pair<int, int>> pq;
        vector<int> res;
        int n = nums.size();
        for (int i=0; i<n; i++) {
            map[nums[i]]++;
        }
        for (auto iter=map.begin(); iter!=map.end(); iter++) {
            pq.push(make_pair(iter->second, iter->first));
        }
        for (int i=0; i<k; i++) {
            res.push_back(pq.top().second);
            pq.pop();
        }
        return res;
    }
};