Solution - O(n)
vector<int> topKFrequent(vector<int>& nums, int k)
{
// 统计每个数字的频率
unordered_map<int, int> rec;
for(int i = 0; i < nums.size(); i++)
rec[nums[i]]++;
// 按频率存放数字
vector<vector<int>> bucket(nums.size());
for (auto it : rec)
bucket[it.second - 1].push_back(it.first);
vector<int> res;
for (int i = bucket.size() - 1; i >= 0 && k > 0; i--)
{
for (int temp : bucket[i])
res.push_back(temp);
k -= bucket[i].size();
}
return res;
}