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 class Solution { public List<Integer> sequentialDigits(int low, int high) { List<Integer> result = new ArrayList(); String str = "123456789"; int i, j, length = str.length(), num; for( i = 2; i <= length; i++) { for( j = 0; j <= length - i; j++) { num = Integer.parseInt(str.substring(j, j + i)); if(num > high) return result; if(num >= low) result.add(num); } } return result; } } Try it on Le