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...
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] Example 2: Input:nums = [1], k = 1Output:[1] Constraints: 1 <= nums.length <= 105 -104<= num...
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 better than O(n log ...
代码(Go) func topKFrequent(nums []int, k int) []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) numToCnt := make(map[int]int) for _, num := range nums { // num 如果不在 num_to_cnt 中,则初始化为 0 , // 然后对 num 的出现次数加 ...
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]
lee jiayou ・ 2021.05.24 Top K Frequent Elements 解题思路先建立字典,记录每个数字对应的出现频率,并将频率用列表存储。之后对该列表进行排序,得到前k大的频率对应数值,再遍历字典,从中取出大于该值对应的数字。代码 Python3 Python 4 1.2K 0不喝甜粥 ・ 2020.02.03 Java排序Map中的Value,快速返回所需List...
第C++实现LeetCode(347.前K个高频元素)[LeetCode]347.TopKFrequentElements前K个高频元素 Givenanon-emptyarrayofintegers,returnthekmostfrequentelements. Example1: Input:nums=[1,1,1,2,2,3],k=2 Output:[1,2] Example2: Input:nums=[1],k=1 Output:[1] Note: Youmayassumekisalwaysvalid,1≤k≤...
public int[] topKFrequent(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); PriorityQueue<Map.Entry<Integer, Integer>> maxheap = new PriorityQueue<>((a,b) -> b.getValue() - a.getValue()); int[] result = new int[k]; ...
用Java刷LeetCode,总结下每道题的难度、解决关键点,分享java版本答案 Link:347. 前 K 个高频元素 给出一个数组,求得次数最高的topK个元素 容易想到用map来记录个数 // 实用map去统计次数 HashMap<Integer, Integer> map = new HashMap<>(); for (int num : nums) { map.put(num, map.getOrDefault...