出現次數較多的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 使用PriorityQueue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()));//構造小頂堆，首先將單詞按照出現次數由小到大排序，若是次數相同，則按字典序由大到小排序
        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;
    }
}

相反的排序方式：

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        List<String> res = new ArrayList<>();
        Map<String,Integer> map = new HashMap<>();
        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() ? a.getKey().compareTo(b.getKey()) : b.getValue() - a.getValue()));
        for (Map.Entry<String,Integer> entry : map.entrySet()){
            priorityQueue.offer(entry);
        }
        int i = 0;
        while(! priorityQueue.isEmpty() && i < k){
            res.add(priorityQueue.poll().getKey());
            i ++;
        }
        return res;
    }
}

② 使用桶排序

class Solution {//24ms     public List<String> topKFrequent(String[] words, int k) {         List<String> res = new ArrayList<>();         Map<String,Integer> map = new HashMap<>();         for (String word : words) {             map.put(word, map.getOrDefault(word, 0) + 1);         }         List<String>[] bucket = new ArrayList[words.length + 1];         for (Map.Entry<String,Integer> entry : map.entrySet()){             int count = entry.getValue();             if (bucket[count] == null){                 bucket[count] = new ArrayList<>();             }             bucket[count].add(entry.getKey());         }         for (int j = bucket.length - 1;j >= 0 && res.size() < k;j --){//按照出現次數加入結果集合             if (bucket[j] != null){//若是出現次數相同，按照字典序排序                 Collections.sort(bucket[j]);                 res.addAll(bucket[j]);             }         }         return res.subList(0,k);//返回前k個值     } }