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Java Solution for LeetCode algorithm problems, continually updating
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?