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
class Solution:
    def SequentialDigits(self, low: int, high: int) -> List[int]:
        res = []
        for i in range(1, 10):
            for j in range(i+1, 10):
                i = i*10 +j
                if i > high:
                    break
                if i >= low:
                    res.append(i)
        return sorted(res)
    def sequentialDigits(self, low, high):
        out = []
        queue = deque(range(1,10))
        while queue and queue[0] <= high:
            elem = queue.popleft()
            if low <= elem:
                out.append(elem)
            last = elem % 10
            if last < 9: queue.append(elem*10 + last + 1)
                    
        return out
class Solution:
    def SequentialDigits(self, low: int, high: int) -> List[int]:
        res = []
        for i in range(1, 10):
            for j in range(i+1, 10):
                i = i*10 +j
                if i > high:
                    break
                if i >= low:
                    res.append(i)
        return sorted(res)
    def sequentialDigits(self, low, high):
        out = []
        queue = deque(range(1,10))
        while queue and queue[0] <= high:
            elem = queue.popleft()
            if low <= elem:
                out.append(elem)
            last = elem % 10
            if last < 9: queue.append(elem*10 + last + 1)
                    
        return out