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...
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 1 numToCn...
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 be better than O(n log n), whe...
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...
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 n), where n is the array’s size. 题意:给定一个一维数组nums和一个整数k,返回k个出现次数最多的元素;并且要求实践复杂度为O(nlogn); ...
原题链接 :https://leetcode.com/problems/top-k-frequent-elements/ Given a non-empty array of integers, return thekmost frequent elements. 给定一个不为空的数字数组,返回出现频率最高的k个元素。 Example 1: Input: nums = [1,1,1,2,2,3], k = 2 ...
functopKFrequent(nums[]int,kint)[]int{iflen(nums)<=1{returnnums}ans:=make([]int,0)iflen(nums)==0{returnans}// 记录出现的次数cache:=map[int]int{}fori:=0;i<len(nums);i++{cache[nums[i]]++}// 排序取到前 k 个高频元素tempAry:=make([][]int,len(nums)+1)fork,v:=rangecache...
347. 前 K 个高频元素 - 给你一个整数数组 nums 和一个整数 k ,请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。 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: * 1 <= nums
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<...
Explanation: "the", "is", "sunny" and "day" are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively. Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.