备注:在 LeetCode 中的运行时间也不是特别慢。 Java 实现 importjava.util.Map;importjava.util.HashMap;importjava.util.List;importjava.util.ArrayList;classSolution{publicList<Integer>topKFrequent(int[] nums,intk){// 统计元素的频率Map<Integer, Integer> freqMap =newHashMap<>();for(intnum : nums...
347. Top K Frequent ElementsMedium Topics Companies Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order. Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k ...
LeetCode - Top K Frequent Elements Given a non-empty array of integers,returnthe 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...
Len() // 堆顶已被移动到切片最后,方便删除 x := (*h)[n-1] *h = (*h)[0 : n-1] return x } 题目链接: Top K Frequent Elements : leetcode.com/problems/t 前K 个高频元素: leetcode-cn.com/problem LeetCode 日更第 84 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
https://leetcode.com/problems/top-k-frequent-elements/ 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. ...
[leetcode] 347. Top K Frequent Elements Description Given a non-empty array of integers, return the k most frequent elements. Example 1: Input: AI检测代码解析 nums = [1,1,1,2,2,3], k = 2 1. Output: AI检测代码解析 [1,2]
主体思想:先排序,记录每个数的个数,和每个数的初始下标。分别存在两个列表里,构建字典。将字典排序,输出前K个高频元素class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[int]: if len(nums)==1: return nums
Swap(arr_ptr + largest * SizeOfElements, arr_ptr + index * SizeOfElements, SizeOfElements); AdjustHeap(arr, largest, SizeOfElements, heapSize, CMP); } } int* topKFrequent(int* nums, int numsSize, int k, int* returnSize) { timesHash head = NULL,s,t; ...
## LeetCode 692E - The K Frequent Word - minHeap from typing import List from heapq import nsmallest from collections import Counter class Solution: def findKthLargest(self, words: 'List[str]', k: 'int') -> list: counter = Counter(words) ## 这里 counter 的key是某个单词 ## 这个用...
链接:力扣https://leetcode-cn.com/problems/top-k-frequent-elements 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 class Solution(object): def topKFrequent(self, nums, k): from collections import Counter counter = Counter(nums).most_common(k) ...