human
Give me a the Python code you requested given the following:
I found an interesting problem on https://leetcode.com/problems/sequential-digits/:
An integer has sequential digits if and only if each digit in the number is one more than the previous digit.
Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits.
 
Example 1:
Input: low = 100, high = 300
Output: [123,234]
Example 2:
Input: low = 1000, high = 13000
Output: [1234,2345,3456,4567,5678,6789,12345]

 
Constraints:

10 <= low <= high <= 10^9

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        nl, nh = len(str(low)), len(str(high))
        ans = []
        sequential = '123456789'
        for n in range(nl, min(nh+1,10)):
            start, delta = int(sequential[:n]), int('1'*n)
            k_min = 0 if n > nl else max(-(-(low-start) // delta), 0)
            k_max = 9 - n if n < nh else (min(high, int(sequential[-n:]))-start) // delta
            ans += [start + k*delta for k in range(k_min, k_max + 1)]

        return ans