692. Top K Frequent Words leetcode

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.
遍历一遍，统计各个词出现的频率，放入map。
然后将node放进小堆里，维护一个大小为k的堆，统计出来top k的词汇。
然后，依次poll出堆中的元素，放入result中。
【692. Top K Frequent Words】堆的排序规则为：
按count从小到大排序，count一样，再按word的字典序从大到小排序。与答案要求的正好相反，这样从堆中拿出元素时，要插入reuslt的头部。逆序排列

class Solution { public List> topKFrequent(String[] words, int k) { List> result = new LinkedList<>(); Map, Node> map = new HashMap<>(); for (String word : words) { if (!map.containsKey(word)) { map.put(word, new Node(word, 1)); } else { map.get(word).count += 1; } } PriorityQueue minHeap = new PriorityQueue<>((o1, o2) -> { if (o1.count != o2.count) { return Integer.compare(o1.count, o2.count); } else { return o2.word.compareTo(o1.word); } }); for (String s : map.keySet()) { minHeap.add(map.get(s)); if (minHeap.size() > k) { minHeap.poll(); } } while (!minHeap.isEmpty()) { result.add(0, minHeap.poll().word); } return result; }private class Node { String word; int count; public Node(String word, int count) { this.word = word; this.count = count; } } }