Top K Frequent Words-优快云博客

本文介绍了一种通过哈希表和优先队列实现的算法，用于筛选给定列表中频率最高的k个单词，并按频率和字母顺序进行排序。

Given a list of words and an integer k, return the top k frequent words in the list.

You should order the words by the frequency of them in the return list, the most frequent one comes first. If two words has the same frequency, the one with lower alphabetical order come first.

[
    "yes", "lint", "code",
    "yes", "code", "baby",
    "you", "baby", "chrome",
    "safari", "lint", "code",
    "body", "lint", "code"
]

for k = 3, return ["code", "lint", "baby"].

for k = 4, return ["code", "lint", "baby", "yes"],

思路：Hash map + PriorityQueue（最小元素在顶部） + 定义 compare

 1 class Pair {
 2     String key;
 3     int value;
 4     
 5     Pair(String key, int value) {
 6         this.key = key;
 7         this.value = value;
 8     }
 9 }
10 
11 public class Solution {
12     /**
13      * @param words an array of string
14      * @param k an integer
15      * @return an array of string
16      */
17      
18     private Comparator<Pair> pairComparator = new Comparator<Pair>() {
19         public int compare(Pair left, Pair right) {
20             if (left.value != right.value) {
21                 return left.value - right.value > 0 ? 1 : -1;
22             }
23             return right.key.compareTo(left.key);
24         }
25     };
26     
27     public String[] topKFrequentWords(String[] words, int k) {
28         if (k == 0) {
29             return new String[0];
30         }
31         
32         HashMap<String, Integer> counter = new HashMap<>();
33         for (String word : words) {
34             if (counter.containsKey(word)) {
35                 counter.put(word, counter.get(word) + 1);
36             } else {
37                 counter.put(word, 1);
38             }
39         }
40         
41         PriorityQueue<Pair> Q = new PriorityQueue<Pair>(k, pairComparator);
42         for (String word : counter.keySet()) {
43             Pair peak = Q.peek();
44             Pair newPair = new Pair(word, counter.get(word));
45             if (Q.size() < k) {
46                 Q.add(newPair);
47             } else if (pairComparator.compare(newPair, peak) > 0) {
48                 Q.poll();
49                 Q.add(new Pair(word, counter.get(word)));
50             }
51         }
52         
53         String[] result = new String[k];
54         int index = 0;
55         while (!Q.isEmpty()) {
56             result[index++] = Q.poll().key;
57         }
58         for (int i = 0; i < result.length / 2; i++) {
59             String tmp = result[i];
60             result[i] = result[result.length - i - 1];
61             result[result.length - i - 1] = tmp;
62         }
63         return result;
64     }
65 }