描述: Given an array of integersnumssorted in ascending order, find the starting and ending position of a giventargetvalue. Your algorithm's runtime complexity must be in the order ofO(logn). If the target is not found in the array, return[-1, -1]. Example 1: Input: nums = [5,7,...
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]. Example 1: Input: nums = [...
LeetCode刷题系列—34. Find First and Last Position of Element in Sorted Array 1.题目描述 英文版: 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 ...
LeetCode 34. Find First and Last Position of Element in Sorted Array 题目描述: 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 ...
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1. Problem Descriptions:Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.If target is not found in the array, return [-1…
- LeetCodeleetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ 解题思路 1. 二分找元素,双指针找区间 class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> res(2, -1); int pos = bsearch(nums, target, 0, nums.size...
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 is not found in the array, return[-1, -1]. ...
Problem 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]. ...
Leetcode - 34. Find First and Last Position of Element in Sorted Array Train of thought Using binary search, if find target, contiguous loop from found position to left or right. classSolution{publicint[]searchRange(int[]nums,inttarget){inttIndex=Arrays.binarySearch(nums,0,nums.length,target...