347. Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

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.

// Hashmap + Max heap
vector<int> topKFrequent(vector<int>& nums, int k) { // time: O(nlogn); space: O(n)
    unordered_map<int, int> m;
    priority_queue<pair<int, int> > pq;
    vector<int> res;
    for (int num : nums) ++m[num];
    for (auto& it : m) pq.push({it.second, it.first});
    for (int i = 0; i < k; ++i) {
        res.push_back(pq.top().second);
        pq.pop();
    }
    return res;
}

優化空間使用，只維護一個大小為k的priority_queue，連帶的time complexity可以降至nlogk。

// Hashmap + Max heap
vector<int> topKFrequent(vector<int>& nums, int k) { // time: O(nlogk); space: O(n)
    unordered_map<int, int> m;
    priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pq;
    vector<int> res;
    for (int num : nums) ++m[num];
    for (auto& it : m) {
        pq.push({it.second, it.first});
        if (pq.size() > k) pq.pop();
    }
    while (!pq.empty()) {
        res.push_back(pq.top().second);
        pq.pop();
    }
    return res;
}

// Bucket Sort
vector<int> topKFrequent(vector<int>& nums, int k) { // time: O(nlogn); space: O(n)
    unordered_map<int, int> m;
    vector<vector<int> > bucket(nums.size() + 1);
    vector<int> res;
    for (int num : nums) ++m[num];
    for (auto& it : m) bucket[it.second].push_back(it.first);
    for (int i = nums.size(); i >= 0; --i) {
        for (int j = 0; j < bucket[i].size(); ++j) {
            res.push_back(bucket[i][j]);
            if (res.size() == k) return res;
        }
    }
    return res;
}