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,…
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…
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…
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…
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].…
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].…
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…