出現次數最多的k個數 Top K Frequent Words

問題：

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.

解決：

①  用map，創建每一個單詞和其出現次數的映射，而後藉助PriorityQueue進行自定義的排序，

class Solution {//108ms
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        Map<String,Integer> map = new HashMap<>();//單詞與出現次數的映射
        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()));//首先將單詞按照出現次數由大到小排序，若是次數相同，則按字典序排序
        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 根據單詞出現的次數進行桶排序。

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);//     } }