题目：

Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
    Note that "i" comes before "love" due to a lower alphabetical order.

 

Example 2:

Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.

 

Note:

1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
2. Input words contain only lowercase letters.

 

Follow up:

1. Try to solve it in O(n log k) time and O(n) extra space.

分析：

利用HashMap统计出每个字符串出现的次数。维护一个容量为k的最小堆，将字符串和出现的频率加入到堆中，然后再从堆中遍历出结果集。

注意本题当字符串频率相同时，小的字符串要出现在前面。

程序：

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        HashMap<String, Integer> map = new HashMap<>();
        for(String word:words){
            int num = map.getOrDefault(word, 0);
            map.put(word, num + 1);
        }
        PriorityQueue<Pair<String, Integer>> queue = new PriorityQueue<>((a, b) ->{
                return a.getValue().equals(b.getValue()) ? b.getKey().compareTo(a.getKey()) : a.getValue() - b.getValue();
            }
        );
        for(String word:map.keySet()){
            int time = map.get(word);
                queue.offer(new Pair<>(word, time));
            if(queue.size() > k)
                queue.poll();
        }
        List<String> res = new ArrayList<>();
        while(!queue.isEmpty())
            res.add(queue.poll().getKey());
        Collections.reverse(res);
        return res;
    }
}