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 =......
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 ...
Link:https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ Description# Given an array of integersnumssorted in ascending order, find the starting and ending position of a giventargetvalue. 给定一个升序排序整数数组nums,找给定值target的第一次和最后一次出现的位置。
The loop exits when left > right, and left will point to the first occurrence of the target or the insertion point if the target is not found. findRightmostIndex Function: This function performs a binary search to find the rightmost index of the target. If the target is found, it continu...
- 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...
LeetCode-Find First and Last Position of Element in Sorted Array,LeetCodeJavaC++FindFirstandLastPositionofElementinSortedArray
LeetCode Top 100 Liked Questions 34. Find First and Last Position of Element in Sorted Array(Java版; Medium) 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. ...
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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...
if (nums[start] != nums[mid]) start++; //move start to lower bound (first position) else end--; //move end to higher bound (last position) } } if (nums[start] == target && nums[end] == target) { res[0] = start;