[Leetcode]937. Reorder Log Files给日志文件排序

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:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

题意：

给定日志文件(规定首单词是认证号)， 要求以字母日志在前，数字日志在后的顺序排序

思路：

重写一个Comparator

代码：

 1 class Solution {
 2       public String[] reorderLogFiles(String[] logs) {
 3        Comparator<String> myComp = new Comparator<String>() {
 4          @Override
 5          public int compare(String s1, String s2) {
 6            String[] split1 = s1.split(" ", 2);
 7            String[] split2 = s2.split(" ", 2);
 8            boolean isDigit1 = Character.isDigit(split1[1].charAt(0));
 9            boolean isDigit2 = Character.isDigit(split2[1].charAt(0));
10            if(!isDigit1 && !isDigit2) {             
11              int comp = split1[1].compareTo(split2[1]);
12              if(comp != 0)
13               return comp;
14               return split1[0].compareTo(split2[0]);
15            }
16              return isDigit1 ? (isDigit2 ? 0 : 1) : -1;
17          }
18        };
19        Arrays.sort(logs, myComp);
20        return logs;
21     }
22 }