[LeetCode]347. Top K Frequent Elements

347. Top K Frequent Elements

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.

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int,int> hash;
for(int num : nums){
hash[num]++;
}

vector<int> result;
// pair<first, second>: first->value,  second->key
// 最大顶堆
priority_queue<pair<int,int>> pq;

for(auto it = hash.begin(); it != hash.end(); it++){
pq.push(make_pair(it->second, it->first));
// 如果优先队列长度 > hash的长度-k,则现在的堆顶元素肯定是结果之一
if(pq.size() > hash.size() - k){
result.push_back(pq.top().second);
pq.pop();
}
}
return result;
}
};