出现次数最多的k个数 Top K Frequent Words

问题：

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.

解决：

①  用map，建立每个单词和其出现次数的映射，然后借助PriorityQueue进行自定义的排序，

class Solution {//108ms
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        Map<String,Integer> map = new HashMap<>();//单词与出现次数的映射
        for (String word : words){
            map.put(word,map.getOrDefault(word,0) + 1);
        }
        PriorityQueue<Map.Entry<String,Integer>> priorityQueue =
                new PriorityQueue<>((a,b) -> (a.getValue() == b.getValue() ? b.getKey().compareTo(a.getKey()) : a.getValue() - b.getValue()));//首先将单词按照出现次数由大到小排序，如果次数相同，则按字典序排序
        for (Map.Entry<String,Integer> entry : map.entrySet()){
            priorityQueue.offer(entry);
            if (priorityQueue.size() > k){
                priorityQueue.poll();
            }
        }
        while (! priorityQueue.isEmpty()){
            res.add(0,priorityQueue.poll().getKey());
        }
        return res;
    }
}

② 根据单词出现的次数进行桶排序。

class Solution { //24ms
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        Map<String,Integer> map = new HashMap<>();
        List<String>[] bucket = new List[words.length + 1];
        for (String word : words){
            map.put(word,map.getOrDefault(word,0) + 1);
        }
        for (String key : map.keySet()){
            int count = map.get(key);
            if (bucket[count] == null){
                bucket[count] = new ArrayList<>();
            }
            bucket[count].add(key);
        }
        for (int j = bucket.length - 1;j >= 0 && res.size() < k;j --){
            if (bucket[j] != null){
                Collections.sort(bucket[j]);
                res.addAll(bucket[j]);
            }
        }
        return res.subList(0,k);//
    }
}