[LeetCode] 937. Reorder Data in Log Files

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:

1. The letter-logs come before all digit-logs.
2. The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
3. The digit-logs maintain their relative ordering.

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
All the tokens of logs[i] are separated by a single space.
logs[i] is guaranteed to have an identifier and at least one word after the identifier.

My Solution

class Solution:
    def reorderLogFiles(self, logs: List[str]) -> List[str]:
        digits = []
        letters = []
        for log in logs:
            if log.split()[1].isdigit(): # 1
                digits.append(log)
            else:
                letters.append(log)
        
        letters.sort(key=lambda x: (x.split()[1:], x.split()[0])) # 2
        
        return letters + digits

#1

log를 split했을 때 앞부분은 identifier, 뒷부분은 content. 뒷부분이 숫자인지 아닌지 isdigit() 함수를 써서 판별. digits와 letters로 구분해서 append.

#2

• lambda식을 활용하여 2가지 key를 사용해서 정렬.
• 첫번째 lambda식은 identifier를 제외한 나머지, 즉 content를 기준으로 정렬.
• 그 후 두번째 key를 사용하여 content가 동일할 경우 identifier 기준으로 재정렬.