一、题目大意 标签: 查找 https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array 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。 如果数组中不存在目标值 target,返回 [-1, -1]。 进阶: 你可以设计并实现时间复杂...
- 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....
代码性能: Runtime:12ms, faster than38.75% of C++ online submissions for Find First and Last Position of Element in Sorted Array. Memory Usage:10.4MB, less than70.33% of C++ online submissions for Find First and Last Position of Element in Sorted Array. 三、改进 上一个题目,发现mid = begin...
[Leetcode][python]Find First and Last Position of Element in Sorted Array/在排序数组中查找元素的第一个和最后一个位置 题目大意 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...
34. Find First and Last Position of Element in Sorted Array 二分搜索,34.FindFirstandLastPositionofElementinSortedArray关于二分搜索的精华帖按照其方法,分别编写求上下界的二分搜索函数,求得上下界,代码如下:classSolution{private:intleft_bound(vector<int>&n
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, -1]. You must write an algorithm with O(log n) runtime complexity. 2. Analysis: To solve this problem ...
思路: 写两个函数,分别用于寻找第一个index和最后一个index,搜索方法用二分法查找即可。注意点: 如果查找到nums[mid] == target,要注意分两种情况,...
34. Find First and Last Position of Element in Sorted Array/在排序数组中查找元素的第一个和最后一个位置 Share 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 ...
https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ 题目描述 Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. ...
2. Solution 解析:最容易想到的就是二分查找,只是要进行一些修改,另一个方法是分别从前往后找以及从后往前找,满足条件就退出。 Version 1 代码语言:javascript 复制 classSolution:defsearchRange(self,nums,target):length=len(nums)start=-1end=-1left=0right=length-1whileleft<=right:mid=(left+right)// ...