## Description

https://leetcode.com/problems/self-dividing-numbers/

A *self-dividing number* is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because `128 % 1 == 0`

, `128 % 2 == 0`

, and `128 % 8 == 0`

.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

**Example 1:**

Input:left = 1, right = 22Output:[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

**Note:**The boundaries of each input argument are `1 <= left <= right <= 10000`

.

## Explanation

Iterate numbers and count numbers which can be self divided by its digits.

## Python Solution

```
class Solution:
def selfDividingNumbers(self, left: int, right: int) -> List[int]:
results = []
while left <= right:
is_self_divide = True
for c in str(left):
if int(c) == 0 or int(left) % int(c) != 0:
is_self_divide = False
if is_self_divide:
results.append(int(left))
left += 1
return results
```

- Time Complexity: O(N)
- Space Complexity: O(N)