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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"]
Constraints:
0 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i] is guaranteed to have an identifier, and a word after the identifier.
这道题让给日志排序,每条日志是由空格隔开的一些字符串,第一个字符串是标识符,可能由字母和数字组成,后面的是日志的内容,只有两种形式的,要么都是数字的,要么都是字母的。排序的规则是对于内容是字母的日志,按照字母顺序进行排序,假如内容相同,则按照标识符的字母顺序排。而对于内容的是数字的日志,放到最后面,且其顺序相对于原顺序保持不变。博主感觉这道题似曾相识啊,貌似之前在很多 OA 中见过,最后还是被 LeetCode 收入囊中了。其实这道题就是个比较复杂的排序的问题,两种日志需要分开处理,对于数字日志,不需要排序,但要记录其原始顺序。这里就可以用一个数组专门来保存数字日志,这样最后加到结果 res 后面,就可以保持其原来顺序。关键是要对字母型日志进行排序,同时还要把标识符提取出来,这样在遍历日志的时候,先找到第一空格的位置,这样前面的部分就是标识符了,后面的内容就是日志内容了,此时判断紧跟空格位置的字符,假如是数字的话,说明当前日志是数字型的,加入数组 digitLogs 中,并继续循环。如果不是的话,将两部分分开,存入到一个二维数组 data 中。之后要对 data 数组进行排序,并需要重写排序规则,要根据日志内容排序,若日志内容相等,则根据标识符排序。最后把排序好的日志按顺序合并,存入结果 res 中,最后别忘了把数字型日志也加入 res, 参见代码如下:
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:
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:
Constraints:
0 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i]
is guaranteed to have an identifier, and a word after the identifier.这道题让给日志排序,每条日志是由空格隔开的一些字符串,第一个字符串是标识符,可能由字母和数字组成,后面的是日志的内容,只有两种形式的,要么都是数字的,要么都是字母的。排序的规则是对于内容是字母的日志,按照字母顺序进行排序,假如内容相同,则按照标识符的字母顺序排。而对于内容的是数字的日志,放到最后面,且其顺序相对于原顺序保持不变。博主感觉这道题似曾相识啊,貌似之前在很多 OA 中见过,最后还是被 LeetCode 收入囊中了。其实这道题就是个比较复杂的排序的问题,两种日志需要分开处理,对于数字日志,不需要排序,但要记录其原始顺序。这里就可以用一个数组专门来保存数字日志,这样最后加到结果 res 后面,就可以保持其原来顺序。关键是要对字母型日志进行排序,同时还要把标识符提取出来,这样在遍历日志的时候,先找到第一空格的位置,这样前面的部分就是标识符了,后面的内容就是日志内容了,此时判断紧跟空格位置的字符,假如是数字的话,说明当前日志是数字型的,加入数组 digitLogs 中,并继续循环。如果不是的话,将两部分分开,存入到一个二维数组 data 中。之后要对 data 数组进行排序,并需要重写排序规则,要根据日志内容排序,若日志内容相等,则根据标识符排序。最后把排序好的日志按顺序合并,存入结果 res 中,最后别忘了把数字型日志也加入 res, 参见代码如下:
参考资料:
https://leetcode.com/problems/reorder-data-in-log-files/
https://leetcode.com/problems/reorder-data-in-log-files/discuss/192438/C%2B%2B-O(NlogN)-Time-O(N)-Space
https://leetcode.com/problems/reorder-data-in-log-files/discuss/193656/C%2B%2B-stable_sort-easy-to-understand
https://leetcode.com/problems/reorder-data-in-log-files/discuss/193872/Java-Nothing-Fancy-15-lines-2ms-all-clear.
LeetCode All in One 题目讲解汇总(持续更新中...)
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