LeetCode in Kotlin
2000. Reverse Prefix of Word
Easy
Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.
- For example, if
word = "abcdefd"andch = "d", then you should reverse the segment that starts at0and ends at3(inclusive). The resulting string will be"dcbaefd".
Return the resulting string.
Example 1:
Input: word = “abcdefd”, ch = “d”
Output: “dcbaefd”
Explanation: The first occurrence of “d” is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is “dcbaefd”.
Example 2:
Input: word = “xyxzxe”, ch = “z”
Output: “zxyxxe”
Explanation: The first and only occurrence of “z” is at index 3.
Reverse the part of word from 0 to 3 (inclusive), the resulting string is “zxyxxe”.
Example 3:
Input: word = “abcd”, ch = “z”
Output: “abcd”
Explanation: “z” does not exist in word.
You should not do any reverse operation, the resulting string is “abcd”.
Constraints:
1 <= word.length <= 250wordconsists of lowercase English letters.chis a lowercase English letter.
Solution
class Solution {
fun reversePrefix(word: String, ch: Char): String {
var i = 0
var j = word.indexOf(ch)
val charArr = word.toCharArray()
while (i < j) {
val temp = charArr[i]
charArr[i] = charArr[j]
charArr[j] = temp
i++
j--
}
return String(charArr)
}
}