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: Inpu…

[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 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: Inpu…

https://leetcode.com/problems/top-k-frequent-words/ 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 l…

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] class Solution { public List<Integer> topKFrequent(int[] nums, int…

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: Inpu…

这个题的排序是用的PriorityQueue实现自动排列,优先队列用的是堆排序,堆排序请看:http://www.cnblogs.com/stAr-1/p/7569706.html 自定义了优先队列的排序方法,lambda表达式该复习下了... public List<String> topKFrequent(String[] words, int k) { /* 1.建立字符串和出现频率的映射 2.用优先队列PriorityQueue实现排序(相同频率用compareTo方法) 4.挑选出前…

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…

［抄题］: 给一个单词列表,求出这个列表中出现频次最高的K个单词. ［思维问题］: 以为已经放进pq里就不能改了.其实可以改,利用每次取出的都是顶上的最小值就行了.(性质) 不知道怎么处理k个之外的数:先把peak, newpair拿出来 ［一句话思路］: ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 比较之前要新建对象.right.key.compareTo(left.key)是反的,而且需要单独写出来,因为比较的是k…

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…