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# 692.Top-K-Frequent-Words

## 692. Top K Frequent Words

## 题目地址

<https://leetcode.com/problems/top-k-frequent-words/>

## 题目描述

```
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.
```

## 代码

### Approach #1: Sorting

**Intuition and Algorithm**

Count the frequency of each word, and sort the words with a custom ordering relation that uses these frequencies. Then take the best `k` of them.

**Complexity Analysis**

* Time Complexity: &#x4F;*(\_NlogN*), where N is the length of `words`. We count the frequency of each word in &#x4F;*(\_N*) time, then we sort the given words in O\_(\_NlogN) time.
* Space Complexity: O\_(\_N), the space used to store our `candidates`.

```java
class Solution {
    public List<String> topKFrequent(String[] words, int k) {
    Map<String, Integer> count = new HashMap();
    for (String word: words) {
      count.put(word, count.getOrDefault(word, 0) + 1);
    }
    List<String> candidates = new ArrayList(count.keySet());
    Collections.sort(candidates, (w1, w2) -> count.get(w1).equals(count.get(w2))) ? w1.compareTo(w2) : count.get(w2) - count.get(w1));

    return candidates.subList(0, k);
  }

}
```

### Approach #2 Heap

**Intuition and Algorithm**

Count the frequency of each word, then add it to heap that stores the best `k` candidates. Here, "best" is defined with our custom ordering relation, which puts the worst candidates at the top of the heap. At the end, we pop off the heap up to `k` times and reverse the result so that the best candidates are first.

```java
class Solution {
  public List<String> topKFrequent(String[] words, int k) {
    Map<String, Integer> count = new HashMap();
    for (String word : words) {
      count.put(word, count.getOrDefault(word, 0) + 1);
    }
    PriorityQueue<String> heap = new PriorityQueue<String>(
    (w1, w2) -> count.get(w1).equals(count.get(w2)) ? w2.compareTo(w1) : count.get(w1) - count.get(w2));

    for (String word : count.keySet()) {
      heap.offer(word);
      if (heap.size() > k) heap.poll();
    }

    List<String> ans = new ArrayList();
    while (!heap.isEmpty()) {
      ans.add(heap.poll());
    }

    Collections.reverse(ans);

    return ans;
  }
}
```