示例 1：
输入: ["i", "love", "leetcode", "i", "love", "coding"], k = 2输出: ["i", "love"]解析: "i" 和 "love" 为出现次数最多的两个单词，均为2次。    注意，按字母顺序 "i" 在 "love" 之前。
来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/top-k-frequent-words著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class WordNumber {    String word;    Integer number;
    public String getWord() {        return word;    }
    public void setWord(String word) {        this.word = word;    }
    public Integer getNumber() {        return number;    }
    public void setNumber(Integer number) {        this.number = number;    }}

public class DayCode {    public static void main(String[] args) {        String[] words = new String[]{                "i", "love", "leetcode", "i", "love", "coding"        };        int k = 2;
        String[] words1 = new String[]{                "a", "aa", "aaa"        };        int k1 = 1;        List<String> ans = new DayCode().topKFrequent(words, k);        System.out.println(ans.toString());
        List<String> ans1 = new DayCode().topKFrequent(words1, k1);        System.out.println(ans1.toString());    }
    public List<String> topKFrequent(String[] words, int k) {        Map<String, Integer> map = new HashMap<>();        for (String word : words) {            map.put(word, map.getOrDefault(word, 0) + 1);        }
        PriorityQueue<WordNumber> priorityQueue = new PriorityQueue<>(                new Comparator<WordNumber>() {                    @Override                    public int compare(WordNumber o1, WordNumber o2) {                        if (o1.getNumber().equals(o2.getNumber())) {                            return o1.getWord().compareTo(o2.getWord());                        } else {                            return o2.getNumber() - o1.getNumber();                        }
                    }                }        );
        for (Map.Entry<String, Integer> i : map.entrySet()) {            WordNumber WordNumber = new WordNumber();            WordNumber.setWord(i.getKey());            WordNumber.setNumber(i.getValue());            priorityQueue.add(WordNumber);        }
        List<String> ans = new ArrayList<>();        for (int i = 0; i < k; i++) {            WordNumber temp = priorityQueue.poll();            ans.add(temp.getWord());        }
        return ans;    }}