题目描述：

LeetCode 692. Top K Frequent Words

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.
2. Can you solve it in O(n) time with only O(k) extra space?

题目大意：

给定非空单词列表，返回出现次数最多的k个单词。

解题思路：

哈希表 + 优先队列（堆）

时间复杂度O(n * log k), 空间复杂度O(n)

利用哈希表统计单词的个数，记为cnt

维护长度为k的优先队列pq（出现频率低、字典序大的单词优先弹出）

Java代码：

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        HashMap<String, Integer> cnt = new HashMap<>();
        for (String word : words) {
            cnt.put(word, cnt.getOrDefault(word, 0) + 1);
        }
        PriorityQueue<String> pq = new PriorityQueue<>(
                (a, b) -> cnt.get(a) == cnt.get(b) ? b.compareTo(a) : cnt.get(a) - cnt.get(b));
        for (String word : cnt.keySet()) {
            pq.add(word);
            if (pq.size() > k) {
                pq.poll();
            }
        }
        LinkedList<String> list = new LinkedList<>();
        while (!pq.isEmpty()) {
            list.addFirst(pq.poll());
        }
        return list;
    }
}

 

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