Top K Frequent WOrds

Link: https://leetcode.com/problems/top-k-frequent-words/

Code

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> wordFrequencyMap = new HashMap<>();
        for (String w : words) {
            if (!wordFrequencyMap.containsKey(w)) {
                wordFrequencyMap.put(w, 1);
            } else {
                wordFrequencyMap.put(w, wordFrequencyMap.get(w) + 1);
            }
        }
        
        PriorityQueue<Element> pq = new PriorityQueue<>((e1, e2) -> {
            if (e1.count != e2.count) {
                return e1.count - e2.count;
            }
            
            return e2.word.compareTo(e1.word);
        });
        
        for (String word : wordFrequencyMap.keySet()) {
            Element element = new Element(word, wordFrequencyMap.get(word));
            pq.offer(element);
            while (pq.size() > k) {
                pq.poll();
            }
        }
        
        List<String> results = new LinkedList<>();
        while (!pq.isEmpty()) {
            Element element = pq.poll();
            results.add(element.word);
        }
        
        Collections.reverse(results);
        return results;
    }
    
    private class Element {
        private String word;
        private int count;
        
        Element(String word, int count) {
            this.word = word;
            this.count = count;
        }
    }
}

• Time: O(nlogk).
• Space: O(n).

Reference: https://leetcode.com/problems/top-k-frequent-words/discuss/108346/My-simple-Java-solution-using-HashMap-and-PriorityQueue-O(nlogk)-time-and-O(n)-space

标签：count,pq,word,String,Top,Element,WOrds,wordFrequencyMap,Frequent
来源： https://www.cnblogs.com/blackraven25/p/15125315.html