<p>An&nbsp;integer has <em>sequential digits</em> if and only if each digit in the number is one more than the previous digit.</p>

<p>Return a <strong>sorted</strong> list of all the integers&nbsp;in the range <code>[low, high]</code>&nbsp;inclusive that have sequential digits.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> low = 100, high = 300
<strong>Output:</strong> [123,234]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> low = 1000, high = 13000
<strong>Output:</strong> [1234,2345,3456,4567,5678,6789,12345]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>10 &lt;= low &lt;= high &lt;= 10^9</code></li>
</ul>

class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        result = []
        s = "123456789"
        l2 = [s[i:j+1] for i in range(9) for j in range(i, 9)]

        for num in l2:
            num = int(num)
            if low<=num<=high:
                result.append(num)

        return sorted(result)
16500