LeetCode in Kotlin
1291. Sequential Digits
Medium
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
class Solution {
fun sequentialDigits(low: Int, high: Int): List<Int> {
val arr = intArrayOf(
12, 23, 34, 45, 56, 67, 78, 89, 123, 234, 345, 456, 567, 678, 789, 1234, 2345, 3456,
4567, 5678, 6789, 12345, 23456, 34567, 45678, 56789, 123456, 234567, 345678, 456789,
1234567, 2345678, 3456789, 12345678, 23456789, 123456789
)
val result: MutableList<Int> = ArrayList()
for (j in arr) {
// 234 148 234 256
if (j in low..high) {
result.add(j)
}
}
return result
}
}