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…

版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. C#版 - Leetcode 347. Top K Frequent Elements - 题解 在线提交: https://leetcode.com/problems/top-k-frequent-elements/ Description Given a non-empty array of integers…

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] Example 2: Input: nums = [1], k = 1 Output: [1] Note: You may assume k is always valid, 1 ≤ k ≤ number of unique ele…

题目地址:https://leetcode.com/problems/top-k-frequent-elements/ 从一个数组中求解出现次数最多的k个元素,本质是top k问题,用堆排序解决. 关于堆排序,其时间复杂度在最好和最坏的场景下都是O(nlogn). 一开始想定义一个结构体,包含元素和元素个数两个成员,后直接用pair存储即可. 解题思路: 1. 分别统计每个数字的个数,建立数字和个数的映射,用pair存储,first=数字个数,second=数字,然后存入集合. 2. 以不同数字…

[LeetCode]692. Top K Frequent Words 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/top-k-frequent-words/description/ 题目描述: Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency f…

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 bet…

1单节点上的topK (1)批量数据 数据结构:HashMap, PriorityQueue 步骤:(1)数据预处理:遍历整个数据集,hash表记录词频 (2)构建最小堆:最小堆只存k个数据. 时间复杂度:O(n +n*lgk) = O(nlgk) 空间复杂度:O(|n|+k) (|n| = number of unique words) lintcode原题:Top K Frequent Words (2)流式数据 数据结构:TreeMap, HashMap 步骤:有新数据到来时,HashMa…

A typical solution is heap based - "top K". Complexity is O(nlgk). typedef pair<int, unsigned> Rec; struct Comp { bool operator()(const Rec &r1, const Rec &r2) { return r1.second > r2.second; } }; class Solution { public: vector…

PageRanking 通过: Input degree of link "Flow" model - 流量判断喜好度 传统的方式又是什么呢? Every term在某个doc中的权重(地位). 公共的terms在Query与Doc中对应的的地位(单位化后)直接相乘,然后全部加起来,构成了cosin相似度. Efficient cosine ranking 传统放入堆的模式:n * log(k) 使用Quick Select:n + k * log(k) : "find to…

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: 347. Top K Frequent ElementsYou may assume k is always valid, 1 ≤ k ≤ number of unique elements. Your algorithm's…