[leetcode] problem 692 – top k frequent words

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

No.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.

No.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.

Code

public List<String> (String[] words, int k) {
    List<String> result = new LinkedList<>();
    Map<String, Integer> map = new HashMap<>();

    for (String word : words)
        map.put(word, map.getOrDefault(word, 0) + 1);

    PriorityQueue<Map.Entry<String, Integer>> heap = new PriorityQueue<>(new Comparator<Map.Entry<String, Integer>>() {
        
        public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) {
            if (o1.getValue() != o2.getValue())
                return o1.getValue() - o2.getValue();
            else
                return o2.getKey().compareTo(o1.getKey());
        }
    });

    for (Map.Entry entry : map.entrySet()) {
        heap.offer(entry);

        if (heap.size() > k)
            heap.poll();
    }

    while (!heap.isEmpty())
        result.add(0, heap.poll().getKey());

    return result;
}

[leetcode] problem 692 – top k frequent words

Example

No.1

No.2

Note

Follow up

Code

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