Given asortedinteger arrayarr, two integerskandx, return thekclosest integers toxin the array. The result should also be sorted in ascending order. An integerais closer toxthan an integerbif: |a - x| < |b - x|, or |a - x| == |b - x|anda < b Example 1: Input:arr = [1,2...
原题链接:https://leetcode.com/problems/find-k-closest-elements/description/ 题目描述:大概意思是给定一个数组[1,2,3,4,5]和两个常数k,x然后在数组中找到与x最近的的k个元素,找到后的元素依旧按照升序排列。 Given a sorted array, two inte......
[LeetCode] 658. Find K Closest Elements Given a sorted arrayarr, two integerskandx, find thekclosest elements toxin the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred. Example 1: Input: arr = [1,2,3,4,5], ...
[Leetcode]658.Find K Closest Elements 链接:LeetCode658给定一个排序好的数组,两个整数 k 和 x,从数组中找到最靠近 x(两数之差最小)的 k 个数。返回的结果必须要是按升序排好的。如果有两个数与 x 的差值一样,优先选择数值较小的那个数。示例1:输入: [1,2,3,4,5],k=4,x=3[1,2,3,4,5]...
[Leetcode] 658. Find K Closest Elements 解题报告 题目: Given a sorted array, two integerskandx, find thekclosest elements toxin the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred....
英文coding面试练习 day2-1 | Leetcode658 Find K Closest Elements, 视频播放量 23、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 Rollwithlife, 作者简介 To remember,相关视频:英文coding面试练习 day3-2 | Leetcode907 Sum of Subarray Minim
Find K Closest Elements https://leetcode.com/problems/find-k-closest-elements/ 给定一个已排序的数组,以及整数x和整数k,找到数组中最接近x的k个元素存在一个列表中返回。 当有两个元素跟x的差距一样时,优先选择比较小的元素。 一、问题分析 测试用例: 从第一个例子可以看到,尽管5 - 3 = 3 - 1,...
题目地址: https://leetcode.com/problems/find-k-closest-elements/description/ 题目描述: Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements...
来源:LeetCode第658题 难度:中等 给定一个排序好的数组arr,两个整数k和x,从数组中找到最靠近x(两数之差最小)的k个数。返回的结果必须要是按升序排好的。 整数a比整数b更接近x需要满足: |a - x| < |b - x| 或者 |a - x| == |b - x| 且 a < b ...
class Solution { public: vector<int> findClosestElements(vector<int>& arr, int k, int x) { int n = arr.size(); int l = 0, r = n-1; while (l < r) { int m = (l + r) / 2; if (arr[m] < x) l = m + 1; else r = m; } r = min(n-1, l+k-1); l = ma...