原文链接: https://leetcode-cn.com/problems/reorder-data-in-log-files

英文原文

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.

中文题目

给你一个日志数组 logs。每条日志都是以空格分隔的字串，其第一个字为字母与数字混合的 标识符 。

有两种不同类型的日志：

• 字母日志：除标识符之外，所有字均由小写字母组成
• 数字日志：除标识符之外，所有字均由数字组成

请按下述规则将日志重新排序：

• 所有 字母日志 都排在 数字日志 之前。
• 字母日志 在内容不同时，忽略标识符后，按内容字母顺序排序；在内容相同时，按标识符排序。
• 数字日志 应该保留原来的相对顺序。

返回日志的最终顺序。

 

示例 1：

输入：logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"]
输出：["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]
解释：
字母日志的内容都不同，所以顺序为 "art can", "art zero", "own kit dig" 。
数字日志保留原来的相对顺序 "dig1 8 1 5 1", "dig2 3 6" 。

示例 2：

输入：logs = ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
输出：["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]

 

提示：

• 1 <= logs.length <= 100
• 3 <= logs[i].length <= 100
• logs[i] 中，字与字之间都用 单个 空格分隔
• 题目数据保证 logs[i] 都有一个标识符，并且在标识符之后至少存在一个字

通过代码

官方题解

方法：自定义排序

思路和算法

我们按照指定的自定义顺序进行排序，而不是按默认顺序排序。

排序规则如下：

• 字母日志先于数字日志；
• 字母日志按字母数字顺序排列，先按内容排序，再按标识符排序；
• 数字日志的顺序保持不变。

这些想法很容易转化为代码。

[solution1]
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;
    }
}

[solution1]
class Solution(object):
    def reorderLogFiles(self, logs):
        def f(log):
            id_, rest = log.split(" ", 1)
            return (0, rest, id_) if rest[0].isalpha() else (1,)

        return sorted(logs, key = f)

复杂度分析

• 时间复杂度：$O(\mathcal{A}\log \mathcal{A})$，其中 $\mathcal{A}$ 是 logs 的内容总和。

• 空间复杂度：$O(\mathcal{A})$。

统计信息

通过次数	提交次数	AC比率
12722	21647	58.8%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言