需要维护一个包含 O(k) 个数字的数组 代码(Python3) class Solution: def topKFrequent(self, nums: List[int], k: int) -> List[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) num_to_cnt = Counter(nums) #将 num_to_cnt 中的数字及其出现次数收集到...
Top K Frequent Elements - LeetCode https://leetcode.com/problems/top-k-frequent-elements/solution/ Python O(n) solution without sort, without heap, without quickselect - LeetCode Discuss https://leetcode.com/problems/top-k-frequent-elements/discuss/81697/Python-O(n)-solution-without-sort-with...
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. Your algorithm's time complexitymust bebetter than O(nlogn), wherenis the...
选择第k个大元素复杂度大概是O(n), 所以一共是O(kn)。 funcquickSort(nums[]int,mymapmap[int]int,sint,eint){ifs>=e-1{return}m:=sfori:=s+1;i<e;i++{ifmymap[nums[i]]>mymap[nums[m]]{fmt.Print(m)nums[i],nums[m+1]=nums[m+1],nums[i]nums[m],nums[m+1]=nums[m+1],nums...
Loading...leetcode.com/problems/top-k-frequent-elements/ #include <utility> // pair的头文件 #include<iostream> #include <vector> #include <queue> // priority_queue 在queueli里面 #include<unordered_map> #include<functional> // greater在 functional里面 using namespace std; class Solution ...
https://leetcode.com/problems/top-k-frequent-elements/ 3种方法 https://leetcode.com/discuss/100713/3-ways-to-solve-this-problem 普通方法heap O(nlogn): classSolution{public:vector<int>topKFrequent(vector<int>&nums,intk){unordered_map<int,int>mp;for(auto&i:nums)mp[i]++;priority_queue<...
347. Top K Frequent Elements 前 K 个高频元素,给定一个非空的整数数组,返回其中出现频率前 k 高的元素。 示例1:输入:nums=[1,1,1,2,2,3],k=2输出:[1,2]示例2:输入:nums=[1],k=1输出:[1] 提示:你可以假设给定的 k 总是合理的,且1≤k≤数组中不相同的元素的个数。你的算
347. Top K Frequent Elements刷题笔记 [问题描述](347. Top K Frequent Elements) 只能说,调库真香,两行代码解决问题 import heapq import collections class Solution: def topKFrequent(self, nums, k): num_counter = collections.Counter(nums) return [x[0] for x in num_counter.most_common(k)]...
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 ...
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...