# 【Leetcode完整解答】268. Missing Number 丟失的數字

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 = [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.

``````   j = j ^ i ^ nums[i]      j ^ i
0: 2 = 0 ^ 0 ^ 3            2 ^ 0 = 2
1: 3 = 2 ^ 1 ^ 0            3 ^ 1 = 2
=     3 ^ 0
2: 0 = 3 ^ 2 ^ 1            0 ^ 2 = 2
=     1 ^ 1``````

3 ^ 1 = 2
11
01
___________
10

3 ^ 2 = 1
11
10
___________
01

a^b^b =a，兩個相同的數字

a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.
In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.