我最先想到的解法是先用字典来储存单词出现的个数,再对字典排序,最后拿出前K个,如: #!/usr/bin/env python # -*- coding: utf-8 -*- class Solution(object): def topKFrequent(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ data, res = {}, [] ...
Many big data applications today require querying highly dynamic and large-scale data streams for top-k frequent items in the most recent window of any specified size at any time. This is a challenging problem. We show that our novel solution is not only accurate, but it also one to two ...
之后,我们返回 items 列表中的前 k个元组的key即可。 3 Python 解题代码 其实小根堆也可以结合 Counter 字典(官方解法一:哈希表+排序),所以我们先演示 Counter。 3.1 Counter - 哈希表+排序 解题代码 ## LeetCode 692E - The K Frequent Word - Counter,字典数据结构 from typing import List from collections...
Top- k frequent itemsItem frequency trackingSliding windowMany big data applications today require querying highly dynamic and large-scale data streams to find the top-k most frequent items in the most recent window of a specified size at a specific time. This is a challenging problem. We ...
classSolution{publicList<Integer>topKFrequent(int[] nums,intk){// 统计元素的频率Map<Integer, Integer> map =newHashMap<>(16);for(intnum : nums) { map.put(num, map.getOrDefault(num,0) +1); }// 遍历map,用最小堆保存频率最大的k个元素PriorityQueue<Integer> pq =newPriorityQueue<>(newCo...
Data streams with high volume and complicated items become more and more common, and typical algorithms of finding top-k frequent items on streams, such as counter-based algorithms and sketch algorithms, are gradually not keeping up with efficiency requirements. Our paper focuses on finding top-k...
the number of frequent features of them is often more than data itself, and the retrieval of frequent features by above table-based index structure still needs a high cost. Which makes it challenging to retrieve the top-k frequent items from STBD, especially in the scenario of multi textual ...
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.
Traditional frequent sequential pattern mining only considers the time point-based item or event in the patterns. However, in many application, the events may span over multiple time points and the relations among events are also important. Time interval-based pattern mining is proposed to mine the...
class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: # 统计 nums 中每个数字出现的次数, # 时间复杂度为 O(n) ,空间复杂度为 O(n) num_to_cnt = Counter(nums) #将 num_to_cnt 中的数字及其出现次数收集到数组中, # 时间复杂度为 O(n) ,空间复杂度为 O(n)...