LeetCode in Kotlin
1844. Replace All Digits with Characters
Easy
You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.
There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.
- For example,
shift('a', 5) = 'f'andshift('x', 0) = 'x'.
For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).
Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.
Example 1:
Input: s = “a1c1e1”
Output: “abcdef”
Explanation: The digits are replaced as follows:
-
s[1] -> shift(‘a’,1) = ‘b’
-
s[3] -> shift(‘c’,1) = ‘d’
-
s[5] -> shift(‘e’,1) = ‘f’
Example 2:
Input: s = “a1b2c3d4e”
Output: “abbdcfdhe”
Explanation: The digits are replaced as follows:
-
s[1] -> shift(‘a’,1) = ‘b’
-
s[3] -> shift(‘b’,2) = ‘d’
-
s[5] -> shift(‘c’,3) = ‘f’
-
s[7] -> shift(‘d’,4) = ‘h’
Constraints:
1 <= s.length <= 100sconsists only of lowercase English letters and digits.shift(s[i-1], s[i]) <= 'z'for all odd indicesi.
Solution
class Solution {
fun replaceDigits(s: String): String {
val sb = StringBuilder()
for (c in s.toCharArray()) {
if (Character.isAlphabetic(c.code)) {
sb.append(c)
} else {
sb.append((sb[sb.length - 1].code + Character.getNumericValue(c)).toChar())
}
}
return sb.toString()
}
}