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This article will explain in detail how LeetCode solves the problem of finding the first and last location of elements in a sorted array. The editor thinks it is very practical, so I share it for you as a reference. I hope you can get something after reading this article.

Topic description

Given an array of integers nums in ascending order, and a target value target. Find out the start and end positions of the given target value in the array.

The time complexity of your algorithm must be O (log n).

If the target value does not exist in the array, return [- 1,-1].

Example:

Input: nums = [5, 7, 7, 7, 7, 8, 8, 10], target = 8 output: [3, 4] input: nums = [5, 7, 7, 8, 8, 10], target = 6 output: [- 1, 5, 5, 7, 8, 10, 8, 10, 8, 10, 10], 8, 10, 6, 6, output: [- 1] topic resolution.

The problem requires a time complexity of O (log n), so it is clear to use the binary search method.

First, two pointer variables are defined to store the indexes of the left and right positions. First of all, find the leftmost index of the target value, through the loop to prevent the element from being lost, retain the rightmost element each time, and the left pointer moves by + 1. Determine whether the target value is included in the array at the end of the loop, and exit directly if not. The one on the right is the same as the left, but just the opposite.

Code implementation / / 34. Next permutation / / https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/// time complexity: O (n) / / Space complexity: O (1) class Solution {public int [] searchRange (int [] nums, int target) {int [] res = new int [] {- 1,-1}; int left = 0 Int right = nums.length-1; int l = left; int r = right; while (left

< right) { int mid = (left + right) / 2; if (nums[mid] < target) { left = mid + 1; } else { right = mid; } } if (left>

Right | | Num [left]! = target) {return new int [] {- 1mam talk 1};} while (l

< r) { int mid = (l + r) / 2 + 1; if (nums[mid] >

Target) {r = mid-1;} else {l = mid;}} if (left > right | | left > r) {return new int [] {- 1,-1} } else {return new int [] {left, r} } on "LeetCode how to solve the problem of finding the first and last location of elements in a sorted array" is shared here. I hope the above content can be of some help to you, so that you can learn more knowledge. if you think the article is good, please share it for more people to see.

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