397. Reorder Data in Log Files
Problem
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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:
Letter-logs: All words (except the identifier) consist of lowercase English letters.
Digit-logs: All words (except the identifier) consist of digits.
Reorder these logs so that:
The letter-logs come before all digit-logs.
The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
The digit-logs maintain their relative ordering.
Return the final order of the logs.
Questions before reading example
// Q) // 1. identifier with empty is available? // => nope // 2. all words have identifier ? // => yeap // 3. what is a identifier? // => the word whicch has a specific rule // 4. any other exceptional case? uppercase words, other words.. // => maybe
Example
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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".
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"]
Solution
- Discussion 을 참고한 나의 풀이
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class Solution {
public String[] reorderLogFiles(String[] logs) {
List<String> letterLogs = new ArrayList<>();
List<String> digitLogs = new ArrayList<>();
for (String log : logs) {
if (isDigit(log.charAt(log.length() - 1))) {
digitLogs.add(log);
} else {
letterLogs.add(log);
}
}
List<String> sortedLetterLogs = letterLogs.stream()
.sorted(Comparator.comparing((String log) -> {
String identifier = log.split(" ")[0];
return log.substring(identifier.length());
}).thenComparing((String log) -> {
String identifier = log.split(" ")[0];
return identifier;
}))
.collect(Collectors.toList());
sortedLetterLogs.addAll(digitLogs);
return sortedLetterLogs.toArray(String[]::new);
}
private static boolean isDigit(Character character) {
return (int) character >= 48 && (int) character <= 57;
}
}
Spent time
- 1시간 30분..?
Review
- 문제를 제대로 읽지 못해서 한참 헤매였다.
- 코너 케이스를 제대로 파악하지 못하였다.
- 처음에는 Map 으로 풀려고 했다. (이것도 한참 걸렸다..)
- 그런데 Key (identifier) 가 중복이 되는 CASE 가 있어서 포기했다.
- Discussion 을 보니.. 이것 이외에도 문제를 잘못읽은 부분이 있었다.
- 크게 digits 과 letters 를 분류, (이 과정에서 처음에 input String 을 그대로 관리하면 됐는데, 이를 또 쪼개려 했다)
- letters sorting (이 과정에서 contents 가 같으면 identifier 를 비교해야하는것도 놓쳤다.)
- 두가지만 잘하면 쉽게 풀리는 문제인데.. 너무 헤맸다.
- 아직 많이 부족하다..