Example 1: Input:words = ["i","love","leetcode","i","love","coding"], k = 2Output:["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:words = ["the","day","is...
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...
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2Output: ["i", "love"] Explanation:"i" and "love"are the two most frequent words. Note that"i" comes before "love"due to a lower alphabetical order. Example2: Input: ["the", "day", "is", "sunny", "th...
"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.
// LeetCode 2020 medium #35 // 692. Top K Frequent Words // Runtime: 20 ms, faster than 43.75% of C++ online submissions for Top K Frequent Words. // Memory Usage: 11.4 MB, less than 94.44% of C++ online submissions for Top K Frequent Words. class Solution { public: vector<stri...
leetcode-java文章分类OpenStack云计算 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. ...
Top K Frequent Words 题:https://leetcode.com/problems/top-k-frequent-words/ 题目大意 对于 string[] words,输出 出现频率前k高的 word,顺序 为 word 出现的频率 由高到低 ,频率相同的 word 按 字符排序。 思路 其实是对words中的所有word进行一个排序。 排序有两个规则: 1.word 在 words中出现的...
必考算法之 Top K 问题 简介:大家好,这里是《齐姐聊算法》系列之 Top K 问题。 Top K 问题是面试中非常常考的算法题。 Leetcode 上这两题大同小异,这里以第一题为例。 题意:给一组词,统计出现频率最高的 k 个。比如说 “I love leetcode, I love coding” 中频率最高的 2 个就是 I 和 love ...
FindHeaderBarSize FindTabBarSize FindBorderBarSize Given an integer arraynumsand an integerk, returnthekmost frequent elements. You may return the answer inany order. Example 1: Input:nums = [1,1,1,2,2,3], k = 2Output:[1,2]
重点是返回前 K 个频率最高的元素, 所以另一种更简单的方法是直接借助 堆(优先队列) 这种数据结构 维护一个 大小为 K 的堆来动态存储前 K 个频率最高的元素, 其时间复杂度为 O(n) 代码: Java: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution { public List<Integer> topKFrequent...