classSolution:def_largestNumber(self,nums:List[int])->str:"""利用sorted和cmp_to_key"""fromfunctoolsimportcmp_to_key#先通过map函数将nums转换为字符串temp=list(map(str,nums))temp.sort(key=cmp_to_key(lambdax,y:int(x+y)-int(
演示代码: fromtypingimportListimportrandomdeffindKthLargest(nums:List[int],k:int)->int:# 将问题转化为寻找第n-k个最小元素k=len(nums)-kdefquickSelect(l,r):pivot,p=nums[r],l# 将小于等于pivot的元素移动到左侧foriinrange(l,r):ifnums[i]<=pivot:nums[p],nums[i]=nums[i],nums[p]p+=1...
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.Example 1:Input: [3,2,1,5,6,4] and k = 2 Output: 5 Example 2:Input: [3,2,3,1,2,4,5,5,6] and k = 4 Output: 4 Note:...
/** * 基于快速选择排序 * 时间 O(n) 证明过程可以参考《算法导论》9.2 * 空间 O(nlongn) * * @param nums nums * @param k k * @return res */ public int findKthLargestWithQuickSelectSort(int[] nums, int k) { return quickSelect(nums, 0, nums.length - 1, nums.length - k); } ...
Description Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element. Example 1: Input: [3,2,1,5,6,4] and k = 2 Output: 5 1. 2. Example 2: ...
英文网址:215. Kth Largest Element in an Array。 中文网址:215. 数组中的第K个最大元素。 思路分析 求解关键:这是一个常规问题,使用借用快速排序的 partition 的思想完成。关键在于理解 partition 的返回值,返回值是拉通了整个数组的索引值,这一点是非常重要的,不要把问题想得复杂了。
215. 数组中的第K个最大元素 - 给定整数数组 nums 和整数 k,请返回数组中第 k 个最大的元素。 请注意,你需要找的是数组排序后的第 k 个最大的元素,而不是第 k 个不同的元素。 你必须设计并实现时间复杂度为 O(n) 的算法解决此问题。 示例 1: 输入: [3,2,1,5,6,4]
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element. For example, Given [3,2,1,5,6,4] and k = 2, return 5. Note: You may assume k is always valid, 1 ≤ k ≤ array's length. ...
【题目】Say you have an array for which the i_th element is the price of a given stock on day _i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the followi...
If no element exists, return an empty string "". Challenge: Perform all these in O(1) time complexity. 【解答】实现一个数据结构,加一、减一,获取最大值的 key 和最小值的 key,都是 O(1) 的时间复杂度。 这类题目可以说遇到多次了,有一些通用思路: 要O(1),根据 key 去取 value 的,无非...