268. Missing Number

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.

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

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 = 
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.

268. Missing Number
``````struct Solution;

impl Solution {
fn missing_number(nums: Vec<i32>) -> i32 {
let mut xor: i32 = 0;
let n = nums.len();
for n in 0..=n {
xor ^= n as i32;
}
for n in nums {
xor ^= n;
}
xor
}
}

#[test]
fn test() {
let nums = vec![9, 6, 4, 2, 3, 5, 7, 0, 1];
assert_eq!(Solution::missing_number(nums), 8);
}
``````