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) {
if(nums.size()==0)
return vector<int>();
unordered_map<int, int> nums_count;
for(int i=0; i<nums.size(); i++)
{
if(nums_count.count(nums[i]))
{
nums_count[nums[i]]+=1;
}else
{
nums_count[nums[i]]=1;
}
}
vector<int> res;
priority_queue<pair<int,int>> pq;
for(auto it = nums_count.begin(); it != nums_count.end(); it++)
{
pq.push(make_pair(it->second, it->first));
if(pq.size()>(int)nums_count.size()-k)
{
res.push_back(pq.top().second);
pq.pop();
}
}
return res;
}
};