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

Solution

Time complexity : O(1) (at most 64 time)
Space complexity : O(1) (at most 36 nums)

class Solution {
public:
    vector<int> sequentialDigits(int low, int high) {
        string sample = "123456789";
        vector<int> ans;
        int low_digits = to_string(low).length();
        int high_digits = to_string(high).length();
        
        for (int len = low_digits; len <= high_digits; ++len) {
            for (int start=0; start<=(9-len); ++start) {
                int num = stoi(sample.substr(start, len));
                if (low <= num && num <= high)
                    ans.push_back(num);
            }
        }
        
        return ans;
    }
};