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### 题目描述

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.

### 例子

#### 例子 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.

#### 例子 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.

#### 例子 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.

• Could you implement a solution using only `O(1)` extra space complexity and `O(n)` runtime complexity?

### Constraints

• `n == nums.length`
• `1 <= n <= 10^4`
• `0 <= nums[i] <= n`
• All the numbers of `nums` are unique.

### 解题思路

• 时间复杂度: `O(n)`
• 空间复杂度: `O(1)`

GitHub 代码同步地址： 268.MissingNumber.cpp

GitHub: Leetcode-C++-Solution