## order

This article mainly records the missing numbers of leetcode

## subject

```
Given an array num containing N numbers in [0, n], find the number that does not appear in the array within the range of [0, n].
Advanced:
Can you solve this problem by implementing an algorithm with linear time complexity and only using additional constant space?
Example 1:
Input: num = [3,0,1]
Output: 2
Explanation: n = 3, because there are 3 numbers, all numbers are in the range [0,3]. 2 is the missing number because it does not appear in nums.
Example 2:
Input: num = [0,1]
Output: 2
Explanation: n = 2, because there are 2 numbers, all numbers are in the range [0,2]. 2 is the missing number because it does not appear in nums.
Example 3:
Input: num = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9, because there are 9 numbers, all numbers are in the range [0,9]. 8 is the missing number because it does not appear in nums.
Example 4:
Input: num = [0]
Output: 1
Explanation: n = 1, because there is 1 number, all numbers are in the range [0,1]. 1 is the missing number because it does not appear in nums.
Tips:
n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All numbers in nums are unique
Source: leetcode
Link: https://leetcode-cn.com/problems/missing-number
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```

## Solution

```
class Solution {
public int missingNumber(int[] nums) {
int result = nums.length;
for (int i = 0; i < nums.length; ++i){
result ^= nums[i];
result ^= i;
}
return result;
}
}
```

## Summary

Use here`All numbers in nums are unique`

This condition uses XOR operation to calculate the index and value of the array, and the result of traversal is the missing number.

## doc

- Missing numbers