【LeetCode】937、重新排列日志文件

937、Reorder Log Files重新排列日志文件

难度：简单

题目描述

英文：

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.
中文：

你有一个日志数组 logs。每条日志都是以空格分隔的字串。

对于每条日志，其第一个字为字母数字标识符。然后，要么：
- 标识符后面的每个字将仅由小写字母组成，或；
- 标识符后面的每个字将仅由数字组成。
我们将这两种日志分别称为字母日志和数字日志。保证每个日志在其标识符后面至少有一个字。

将日志重新排序，使得所有字母日志都排在数字日志之前。字母日志按字母顺序排序，忽略标识符，标识符仅用于表示关系。数字日志应该按原来的顺序排列。

返回日志的最终顺序。

提示

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

示例

Example 1:

1 2	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"]

解题思路

思路一

先区分数字日志和字母日志，然后字母日志按内容排序，最后合并两部分内容输出。

代码提交

Python2，用时48ms，内存11.1M

class Solution(object):
    def reorderLogFiles(self, logs):
        """
        :type logs: List[str]
        :rtype: List[str]
        """
        letterList = []
        numberList = []
        for log in logs:
            logList = log.split(" ")
            if logList[1].isdigit():
                numberList.append(log)
            else:
                letterList.append((logList[0], " ".join(logList[1:])))
        letterList.sort(key=lambda x:x[1])
        resultList = [" ".join(a) for a in letterList]
        return resultList+numberList

后来看了一下用时短的代码，方法都很简单，但是代码很简洁高效。

用0、1标记字母、数字日志，之后一次排序返回。

Python2，用时36ms，内存11.1M

class Solution(object):
    def reorderLogFiles(self, logs):
        """
        :type logs: List[str]
        :rtype: List[str]
        """
        def f(log):
            id, rest = log.split(' ', 1)
            return (0, rest) if rest[0].isalpha() else (1,)
        
        return sorted(logs, key = f)