forked from havanagrawal/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_692.java
70 lines (60 loc) · 2.34 KB
/
_692.java
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
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
package com.fishercoder.solutions;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.SortedSet;
import java.util.TreeSet;
/**
* 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 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:
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.
Can you solve it in O(n) time with only O(k) extra space?
*/
public class _692 {
public static class Solution1 {
/**
* O(n) extra space
* O(nlogk) time
* */
public List<String> topKFrequent(String[] words, int k) {
Map<String, Integer> map = new HashMap<>();
for (String word : words) {
map.put(word, map.getOrDefault(word, 0) + 1);
}
SortedSet<Map.Entry<String, Integer>> sortedset = new TreeSet<>(
(e1, e2) -> {
if (e1.getValue() != e2.getValue()) {
return e2.getValue() - e1.getValue();
} else {
return e1.getKey().compareToIgnoreCase(e2.getKey());
}
});
sortedset.addAll(map.entrySet());
List<String> result = new ArrayList<>();
Iterator<Map.Entry<String, Integer>> iterator = sortedset.iterator();
while (iterator.hasNext() && k-- > 0) {
result.add(iterator.next().getKey());
}
return result;
}
}
}