Top K Frequent Words

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分类专栏： PriorityQueue 文章标签： leetcode

于 2018-12-28 13:59:42 首次发布

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/85320962

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本文介绍了一种使用最大堆实现的高效算法，用于从给定单词列表中找出出现频率最高的k个单词。通过构建一个最大堆并利用优先队列，算法能在O(n log k)的时间复杂度内完成任务，同时保持O(n)的空间复杂度。文章详细解释了如何通过比较单词的频率和字典序来确保输出的正确性和稳定性。

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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:

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

Follow up:

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

思路：注意这个题目需要最大到最小输出，用maxheap，Nlogk;

class Solution {
    class Node {
        public String word;
        public int fre;
        public Node(String word, int fre) {
            this.word = word;
            this.fre = fre;
        }
    }
    
    public List<String> topKFrequent(String[] words, int k) {
        HashMap<String, Integer> countmap = new HashMap<>();
        for(String word: words) {
            countmap.put(word, countmap.getOrDefault(word, 0) + 1);
        }
        PriorityQueue<Node> pq = new PriorityQueue<Node>((a, b) -> {
            if(a.fre != b.fre) {
                return b.fre - a.fre;
            } else {
                return a.word.compareTo(b.word);
            }
        });
        for(String key: countmap.keySet()) {
            pq.offer(new Node(key, countmap.get(key)));
        }
        
        List<String> list = new ArrayList<>();
        int count = 0;
        while(!pq.isEmpty()) {
            Node node = pq.poll();
            if(count < k) {
                list.add(node.word);
                count++;
            } else {
                break;
            }
        }
        return list;
    }
}