[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;
}

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