给一非空的单词列表，返回前 k 个出现次数最多的单词。

返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。

示例 1：

示例 2：

注意：

1. 假定 k 总为有效值， 1 ≤ k ≤ 集合元素数。
2. 输入的单词均由小写字母组成。

扩展练习：尝试以 O(n log k) 时间复杂度和 O(n) 空间复杂度解决.

(1)There is a pretty simple Brute Force solution - create a map (word->occurance), then put EntrySet to List and sort this list with a custom comparator. The time complexity is O(N log N).

(2)But since we need to return k popular items only - we don't need to sort all items. So, the second idea is to use Heap with limited capacity k. The time complexity is O(N log K)

(3)The third solution is to use Bucket Sort - O(N + K log K)

(1)

   Map<String, Integer> count = new HashMap();
        for (String word: words) {
            count.put(word, count.getOrDefault(word, 0) + 1);
        }
        List<String> candidates = new ArrayList(count.keySet());
        Collections.sort(candidates, (w1, w2) -> count.get(w1).equals(count.get(w2)) ?
                w1.compareTo(w2) : count.get(w2) - count.get(w1));


        return candidates.subList(0, k);

(2)

Map<String, Integer> map = new HashMap<>();
        for(String w : words) map.put(w, map.getOrDefault(w, 0)+1);
    
        PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>((e1, e2)->{
            int res = e1.getValue() - e2.getValue();
            if(res==0) return e2.getKey().compareTo(e1.getKey());
            return res;
        });
        
        for(Map.Entry e : map.entrySet()) {
            pq.add(e);
            if (pq.size() > k) pq.poll();
        }
        
        List<String> res = new ArrayList<>();
        for(int i=0;i<k;i++) res.add(pq.poll().getKey());
        Collections.reverse(res);
        return res; 

(3)public List<String> topKFrequent(String[] words, int k) {