# 268. Missing Number

## 1. Question

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.

## 2. Examples

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.


## 3. 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?

## 5. Solutions

### 5.1. 排序遍历

class Solution {
public int missingNumber(int[] nums) {
Arrays.sort(nums);
int ans = 0;

for(int i = 0; i < nums.length; i++) {
if (nums[i] != i) {
return i;
}
}

return nums.length;
}
}


### 5.2. Set排序遍历

class Solution {
public int missingNumber(int[] nums) {
HashSet<Integer> set = new HashSet<>();

int ans = nums.length;

for(int i : nums) {
}

for(int i = 0; i < nums.length; i++) {
if (!set.contains(i)) {
ans = i;
break;
}
}

return ans;
}
}


### 5.3. 数学法

class Solution {
public int missingNumber(int[] nums) {
int len = nums.length;

// 等差数列
int sum = len * (len + 1) / 2;

// 求出数组中所有数字的和
int res = Arrays.stream(nums).sum();

return sum - res;

}
}