LeetCode 692. Top K Frequent Words

原题链接在这里：https://leetcode.com/problems/top-k-frequent-words/description/

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.

题解：

利用HashMap<String, Integer> hm储存word和对应的frequency.

把Map.Entry<String, Integer> entry加到heap中. 

当heap的size() 大于k 就poll()出一个element.

最后把heap的所有element的 key string 加进 res中.

Time Complexity: O(nlogk). n = words.length.

Space: O(n).

AC Java:

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<String>();
        HashMap<String, Integer> hm = new HashMap<String, Integer>();
        for(String s : words){
            hm.put(s, hm.getOrDefault(s, 0)+1);
        }
        
        PriorityQueue<Map.Entry<String, Integer>> heap = new PriorityQueue<Map.Entry<String, Integer>>(
            (a,b) -> a.getValue() == b.getValue() ? b.getKey().compareTo(a.getKey()) : a.getValue() - b.getValue()
        );
        
        for(Map.Entry<String, Integer> entry : hm.entrySet()){
            heap.add(entry);
            if(heap.size() > k){
                heap.poll();
            }
        }
        
        while(!heap.isEmpty()){
            res.add(0, heap.poll().getKey());
        }
        return res;
    }
}

类似Top K Frequent Elements.