Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…

本人编程小白,如果有写的不对.或者能更完善的地方请个位批评指正! 这个是leetcode的第34题,这道题的tag是数组,需要用到二分搜索法来解答 34. Find First and Last Position of Element in Sorted Array Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target v…

Description Given a sorted array of n integers, find the starting and ending position of a given target value. If the target is not found in the array, return [-1, -1]. Example Given [5, 7, 7, 8, 8, 10] and target value 8,return [3, 4]. Challenge O(l…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…

一.题目说明 题目是34. Find First and Last Position of Element in Sorted Array,查找一个给定值的起止位置,时间复杂度要求是Olog(n).题目的难度是Medium! 二.我的解答 这个题目还是二分查找(折半查找),稍微变化一下.target==nums[mid]后,需要找前面.后面的值是否=target. 一次写出来,bug free,熟能生巧!怎一个爽字了得! #include<iostream> #include<vecto…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 二分查找 日期 题目地址:https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/ 题目描述 Given an array of integers nums sorted in ascending order,…

思路:先二分查找到一个和target相同的元素,然后再左边二分查找左边界,右边二分查找有边界. class Solution { public: , end = -; int ends; int lSearch(int left, int right, vector<int>& nums, int target) { ; ; if(nums[mid] == target){ || (mid > && nums[mid - ] < target)) retur…

题意懒得抄了,大概是:在升序数组中给定整数target,找到第一个和最后一个target的索引,找到返回{index1, index2},否则返回{-1, -1}: 时间复杂度要求:O(logn) 分析:要求对数时间,又是查找,我们不难想到二分查找.但是有一点,怎么查到第一个和最后一个呢?这困扰了我. 算法:来自leetcode caikehe 1. 使用二分变体,查找target作为index1,再找target+1, 作为index2: 2. 对index做判断,如果它未越界且它的对应值等于…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given targetvalue. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. E…

1. 原始题目 给定一个按照升序排列的整数数组 nums,和一个目标值 target.找出给定目标值在数组中的开始位置和结束位置. 你的算法时间复杂度必须是 O(log n) 级别. 如果数组中不存在目标值,返回 [-1, -1]. 示例 1: 输入: nums = [5,7,7,8,8,10], target = 8 输出: [3,4] 示例 2: 输入: nums = [5,7,7,8,8,10], target = 6 输出: [-1,-1] 2. 思路 既然是有序数组,可以2分法.如果存…

原题目:Search for a Range, 现在题目改为: 34. Find First and Last Position of Element in Sorted Array Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity…

乘风破浪:LeetCode真题_034_Find First and Last Position of Element in Sorted Array 一.前言 这次我们还是要改造二分搜索,但是想法却有一点不一样. 二.Find First and Last Position of Element in Sorted Array 2.1 问题 2.2 分析与解决 查找问题,时间复杂度要求对数级别的,我们自然的想到了二分查找,和上一题一样,需求都是有点不一样的,这次是有重复的数字,找出某一特定的重…

目录 题目链接 注意点 解法 小结 题目链接 Find First and Last Position of Element in Sorted Array - LeetCode 注意点 nums可能为空 时间复杂度为O(logn) 解法 解法一:最普通的二分搜索,先找到一个target,然后向两边拓展. class Solution { public: int binarySearch(vector<int>& nums, int target) { int left = 0,rig…

leetcode-algorithms-34 Find First and Last Position of Element in Sorted Array Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in th…

Given a sorted array of integers, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target , -]. For example, Given [, , , , , ] and target value , , ]. Analysi…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given targetvalue.Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. Ex…

描述: Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4970 访问. 给定一个按照升序排列的整数数组 nums,和一个目标值 target.找出给定目标值在数组中的开始位置和结束位置. 你的算法时间复杂度必须是 O(log n) 级别. 如果数组中不存在目标值,返回 [-1, -1]. 输入: nums = [5,7,7,8,8,10], target = 8 输出: [3,4] 输入: nums = [5,7,7,…

题目链接 题目大意:找出一串升序数组中target值的起始下标和结束下标值,如果不存在则返回{-1,-1}. 解法一:用二分查找,找到数组中的target,然后找其左边和右边的target下标值.代码如下(耗时11ms): public int[] searchRange(int[] nums, int target) { if(nums == null || nums.length == 0) { int[] r = {-1, -1}; return r; } int low = 0, hig…

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1].…

二分查找不只是查找,还可以根据需求添加条件进行查找,比如这个题,左端点的条件就是边界点或者小于target,右端点的条件就是!=size()或者大于.根据这个找到查找的条件…

问题:给定一个有序数组和一个目标值,输出目标值在数组中的起始位置和终止位置,如果目标值不在数组中,则输出[-1,-1] 示例: 输入:nums = [1,2,3,5,5,7] target = 5 输出:[3,4] 输入:nums = [1,5,8,9] target = 7 输出:[-1,-1] 解决思路:二分查找直到找到第一个目标值,再对目标值左右进行二分查找 Python代码: class Solution(object): def searchRange(self, nums, targ…

给定一个按照升序排列的整数数组 nums,和一个目标值 target.找出给定目标值在数组中的开始位置和结束位置. 你的算法时间复杂度必须是 O(log n) 级别. 如果数组中不存在目标值,返回 [-1, -1]. 示例 1: 输入: nums = [5,7,7,8,8,10], target = 8 输出: [3,4] 示例 2: 输入: nums = [5,7,7,8,8,10], target = 6 输出: [-1,-1] class Solution { public: vector…

题目来源 https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given…

原题链接在这里:https://leetcode.com/problems/missing-element-in-sorted-array/ 题目: Given a sorted array A of unique numbers, find the K-th missing number starting from the leftmost number of the array. Example 1: Input: A = [4,7,9,10], K = 1 Output: 5 Explan…

108. 将有序数组转换为二叉搜索树 108. Convert Sorted Array to Binary Search Tree 题目描述 将一个按照升序排列的有序数组,转换为一棵高度平衡二叉搜索树. 本题中,一个高度平衡二叉树是指一个二叉树每个节点的左右两个子树的高度差的绝对值不超过 1. 每日一算法2019/5/17Day 14LeetCode108. Convert Sorted Array to Binary Search Tree 示例: 给定有序数组: [-10,-3,0,5,9…

题目来源: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ 题意分析: 给定一个有重复的翻转的数组,找到最小的数. 题目思路: 由于有重复的存在,所以当中间数和两端存在相等的时候就不能用二分的方法来做了,最坏的情况是线性时间复杂度,所以直接线性查找这题也可以做. 代码(python): class Solution(object): def findMin(self, nums): ""&qu…

题目来源: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ 题意分析: 在一个不重复的翻转的数组里面找到最小那个.例如:4 5 6 7 0 1 2,最小是0. 题目思路: 这里可以利用二分的方法去找最小的值. 代码(python): class Solution(object): def findMin(self, nums): """ :type nums: List[int] :rty…