# 937.Reorder-Data-in-Log-Files

## 937. Reorder Data in Log Files

## 题目地址

<https://leetcode.com/problems/reorder-data-in-log-files/>

## 题目描述

```
You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier.  Then, either:

Each word after the identifier will consist only of lowercase letters, or;
Each word after the identifier will consist only of digits.
We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

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"]
```

## 代码

Approach #1 Custom Sort

**Intuition and Algorithm**

Instead of sorting in the default order, we'll sort in a custom order we specify.

The rules are:

* Letter-logs come before digit-logs;
* Letter-logs are sorted alphanumerically, by content then identifier;
* Digit-logs remain in the same order.

It is straightforward to translate these ideas into code.

```java
class Solution {
    public String[] reorderLogFiles(String[] logs) {
    Arrays.sort(logs, (log1, log2) -> {
      String[] split1 = log1.split(" ", 2);
      String[] split2 = log2.split(" ", 2);
      boolean isDigit1 = Character.isDigit(split1[1].charAt(0));
      boolean isDigit2 = Character.isDigit(split2[1].charAt(0));
      if (!isDigit1 && !isDigit2) {
        int cmp = split1[1].compareTo(split2[1]);
        if (cmp != 0) return cmp;
        return split1[0].compareTo(split2[0]);
      }
      return isDigit1 ? (isDigit2 ? 0 : 1) : -1;
    });

    return logs;
  }

}
```


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