Given a non-empty array of integers, return the k most frequent elements. For example, Given [1,1,1,2,2,3] and k = 2, return [1,2]. Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements. Your algorithm’s time complexity must be better than O(n log...
The privacy protection mechanism is independent of the mining process, resulting in the accuracy of the top-k frequent itemsets being affected by the k value. When the algorithm is applied to large-scale frequent itemsets, balance the data availability and privacy security is difficult. In view ...
func topKFrequent(nums []int, k int) []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) numToCnt := make(map[int]int) for _, num := range nums { // num 如果不在 num_to_cnt 中,则初始化为 0 , // 然后对 num 的出现次数加 1 numToCn...
LeetCode 347. Top K Frequent Elements(出现频率最多的 k 个元素) 题意:求出现频率最多的 k 个元素。分析:统计每个元素出现次数,按出现次数将元素分组,然后按频率从高到低取k个元素。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 ...
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...
原题链接 :https://leetcode.com/problems/top-k-frequent-elements/ Given a non-empty array of integers, return thekmost frequent elements. 给定一个不为空的数字数组,返回出现频率最高的k个元素。 Example 1: Input: nums = [1,1,1,2,2,3], k = 2 ...
Given a non-empty array of integers, return the k most frequent elements. Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] 1. 2. Example 2: Input: nums = [1], k = 1 Output: [1] 1. 2. Note: You may assume k is always valid, 1 ≤ k ≤ number of...
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 ...
Top K Frequent Element top k 频率的数字 定义一个新结构,按照结构中频率字段排序即可。这里用了map去纪律每个数字的频率,时间复杂度一样,空间复杂度更低。 另外重写了一下快速排序,也可以用堆排,基数排序,实际都差不多。 有人用快排的思想去选择前k个元素,时间复杂度有点高,做了很多重复工作。选择第k个大...
https://leetcode.com/problems/top-k-frequent-elements/ Given a non-empty array of integers, return thekmost frequent elements. For example, Given[ 1,1,1,2,2,3]and k = 2, return[1,2]. Note: You may assumekis always valid, 1 ≤k≤ number of unique elements. ...