Medium
You are given an array of logs
. Each log is a space-delimited string of words, where the first word is the identifier.
There are two types of logs:
Reorder these logs so that:
Return the final order of the logs.
Example 1:
Input: logs = [“dig1 8 1 5 1”,”let1 art can”,”dig2 3 6”,”let2 own kit dig”,”let3 art zero”]
Output: [“let1 art can”,”let3 art zero”,”let2 own kit dig”,”dig1 8 1 5 1”,”dig2 3 6”]
Explanation:
The letter-log contents are all different, so their ordering is “art can”, “art zero”, “own kit dig”.
The digit-logs have a relative order of “dig1 8 1 5 1”, “dig2 3 6”.
Example 2:
Input: logs = [“a1 9 2 3 1”,”g1 act car”,”zo4 4 7”,”ab1 off key dog”,”a8 act zoo”]
Output: [“g1 act car”,”a8 act zoo”,”ab1 off key dog”,”a1 9 2 3 1”,”zo4 4 7”]
Constraints:
1 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i]
are separated by a single space.logs[i]
is guaranteed to have an identifier and at least one word after the identifier.import java.util.Collections
class Solution {
fun reorderLogFiles(logs: Array<String>): Array<String?> {
val digi: MutableList<String> = ArrayList()
val word: MutableList<String> = ArrayList()
for (s in logs) {
if (Character.isDigit(s[s.length - 1])) digi.add(s) else word.add(s)
}
Collections.sort(
word,
Comparator { s1, s2 ->
val firstSpacePosition = s1.indexOf(" ")
val firstWord = s1.substring(firstSpacePosition, s1.length)
val secondSpacePosition = s2.indexOf(" ")
val secondWord = s2.substring(secondSpacePosition, s2.length)
if (firstWord.compareTo(secondWord) == 0) {
val firstSpacePosition1 = s1.indexOf(" ")
val firstWord1 = s1.substring(0, firstSpacePosition1)
val secondSpacePosition1 = s2.indexOf(" ")
val secondWord1 = s2.substring(0, secondSpacePosition1)
return@Comparator firstWord1.compareTo(secondWord1)
}
firstWord.compareTo(secondWord)
},
)
val result = arrayOfNulls<String>(digi.size + word.size)
var `in` = 0
for (s in word) result[`in`++] = s
for (s in digi) result[`in`++] = s
return result
}
}