[Leetcode] #347 Top K Frequent Elements

Discription:

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.

Solution:

//複雜度 O(n*log(n-k)
#include <queue>  //priority_queue在這個庫裏

vector<int> topKFrequent(vector<int>& nums, int k) {
	unordered_map<int, int> map;
	for (int num : nums){
		map[num]++;
	}
	vector<int> res;
	priority_queue<pair<int, int>> pq;
	for (auto it:map){
		pq.push(make_pair(it.second, it.first));
		if (pq.size() > map.size() - k){  //取出頻次高的k個數，剩下的數放在堆裏
			res.push_back(pq.top().second);
			pq.pop();
		}
	}
	return res;
}

GitHub-Leetcode:https://github.com/wenwu313/LeetCode