[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:

//复杂度 O(n*log(n-k)#include <queue>  //priority_queue在这个库里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){  //取出频次高的k个数，剩下的数放在堆里res.push_back(pq.top().second);pq.pop();}}return res;}