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 = 2Output: ["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 = 4Output: ["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.

java

class Word {    int val;    String word;    public Word(String word, int val) {        this.word = word;        this.val = val;    } }public class Solution {    public List<String> topKFrequent(String[] words, int k) {        if (words == null || words.length == 0) {            return null;        }        Map<String, Integer> map = new HashMap<>();        for (String str : words) {            if (map.containsKey(str)) {                map.put(str, map.get(str) + 1);            } else {                map.put(str, 1);            }        }        Comparator<Word> cmp = new Comparator<Word>() {            public int compare(Word a, Word b) {                int diff = a.val - b.val;                if (diff == 0) {                    return b.word.compareTo(a.word);                } else {                    return diff;                }            }        };        Queue<Word> queue = new PriorityQueue<Word>(k + 1, cmp);        for (String str : map.keySet()) {            queue.offer(new Word(str, map.get(str)));            if (queue.size() > k) {                queue.poll();            }        }        List<String> list = new ArrayList<>();        while (!queue.isEmpty()) {            list.add(queue.poll().word);        }        Collections.reverse(list);        return list;    }}