• Home
• com.github.javadev
• leetcode-in-java11
• 1.19
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g0201_0300.s0268_missing_number.readme.md Maven / Gradle / Ivy

268\. Missing Number

Easy

Given an array `nums` containing `n` distinct numbers in the range `[0, n]`, return _the only number in the range that is missing from the array._

**Example 1:**

**Input:** nums = [3,0,1]

**Output:** 2

**Explanation:** n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums. 

**Example 2:**

**Input:** nums = [0,1]

**Output:** 2

**Explanation:** n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums. 

**Example 3:**

**Input:** nums = [9,6,4,2,3,5,7,0,1]

**Output:** 8

**Explanation:** n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums. 

**Example 4:**

**Input:** nums = [0]

**Output:** 1

**Explanation:** n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums. 

**Constraints:**

*   `n == nums.length`
*   1 <= n <= 104
*   `0 <= nums[i] <= n`
*   All the numbers of `nums` are **unique**.

**Follow up:** Could you implement a solution using only `O(1)` extra space complexity and `O(n)` runtime complexity?