Description
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
Example 1:1
2
3
4Input: ["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 2:1
2
3
4Input: ["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.
Solution
O(NlogK)1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27class Solution {
public List<String> topKFrequent(String[] words, int k) {
if (words == null || words.length == 0) return new ArrayList<String>();
HashMap<String, Integer> map = new HashMap<>();
for (String word: words){
map.put(word, map.getOrDefault(word, 0)+1);
}
PriorityQueue<String> pq = new PriorityQueue<>(new Comparator<String>(){
public int compare(String a, String b){
if (map.get(a) == map.get(b)) return b.compareTo(a);
else return map.get(a)-map.get(b);
}
});
for (String word: map.keySet()){
pq.offer(word);
if (pq.size() > k) pq.poll();
}
LinkedList<String> res = new LinkedList<>();
while(!pq.isEmpty()) res.addFirst(pq.poll());
return res;
}
}