# 937. Reorder Log Files

**<font color=red>难度: Easy</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/reorder-log-files/

> 内容描述

```
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: ["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"]
 

Note:

0 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i] is guaranteed to have an identifier, and a word after the identifier.
```

## 解题方案

> 思路 1
******- 时间复杂度: O(N^2)******- 空间复杂度: O(1)******



sb题没什么好说的

就稍微注意一下下面的例子就懂了

```
Input:
["j mo", "5 m w", "g 07", "o 2 0", "t q h"]
Output:
["j mo","5 m w","t q h","g 07","o 2 0"]
Expected:
["5 m w","j mo","t q h","g 07","o 2 0"]
```

beats 100%

```python
class Solution:
    def reorderLogFiles(self, logs):
        """
        :type logs: List[str]
        :rtype: List[str]
        """
        digits_log_idx = []
        letter_log = {}
        letter_logs = []
        
        def allDigit(word):
            return all(w.isdigit() for w in word)
        
        for idx, log in enumerate(logs):
            if allDigit(log.split(' ')[-1]):
                digits_log_idx.append(idx)
            else:
                letters = tuple(log.split(' ')[1:]) # identifier后面的所有letters
                letter_log[letters] = idx 
                letter_logs.append(letters)
                
        letter_logs.sort()
        res = []
        for log in letter_logs:
            res.append(logs[letter_log[log]])
        for idx in digits_log_idx:
            res.append(logs[idx])
        return res
```




























