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) { freqMap.put(num, freqMap.getOrDefault(num,0) +1...
this.elements[i] = x; }; BinMinHeap.prototype.isEmpty = function() { return this.size == 0; }; BinMinHeap.prototype.deleteMin = function() { if (this.isEmpty()) { return this.elements[0]; } var minElement = this.elements[1]; var lastElement = this.elements[this.size]; this....
importheapqclassSolution:deffindKthLargest(self,nums:list,k:int)->int:# Solution: 小顶堆。# Special considerationsnumsLen=len(nums)ifnumsLen==0:returnNone# ParametersminHeap=[]# minHeap = MinHeap(500)foriinrange(0,numsLen):# ake comparision if there are enough elements in the heapiflen(...
TopK的问题,思路就是用堆来解决。 先以前K个元素构建一个大小为K的小顶堆,然后从K个元素之后,遍历从索引在K后面的元素,如果有大于小顶堆的堆顶元素的,那么就交换两个元素并重新构建小顶堆。遍历到最后的小顶堆堆顶就是第K大的元素了。时间复杂度O(n+klogk) classSolution {publicintfindKthLargest(int[] ...
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数组中, # 时间复杂度为 O(n) ,空间复杂度为 O(n) heap: List[Tuple[int, int]] = [ # heapq 默认是最小堆, # 所以这里堆 cnt 取相反数 (-cnt, num) for num, cnt in num_to_cnt....
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]
使用JAVA语言解题十分的方便,因为java语言中有priorityqueue的大根堆与小根堆数据结构,其内部是有完全二叉树进行实现的,固定长度的K,自动进行排序。进行删除堆顶与堆底部的操作。但是由于java语言中的priorityqueue默认是使用 class Solution { public int[] getLeastNumbers(int[] arr, int k) { ...
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≤...
【题目】Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index i to val. Example: 代码语言:javascript 代码运行次数:0 运行 复制 Gi...
【题目】Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Note: Each of the array element will not exceed 100. The array size will not exceed 200. Example 1: ...