937. reorder log files

Problems

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:

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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.
Accepted
2,197
Submissions
4,155

Arrays.sort

String.split(“ “, n): you could only split to n part

Comparator:

• both are not digit –> general compareTo

• 1 is digit && 2 is digit: –> 0

• 1 is digit && 2 is not digit –> 1

• 1 is not digit && 2 is digit –> -1

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class Solution {
    public String[] reorderLogFiles(String[] logs) {
        Arrays.sort(logs, (l1, l2) -> {
            String [] sp1 = l1.split(" ", 2);
            String [] sp2 = l2.split(" ", 2);
            boolean isDigit1 = Character.isDigit(sp1[1].charAt(0));
            boolean isDigit2 = Character.isDigit(sp2[1].charAt(0));
            if(!isDigit1 && !isDigit2) {
                return sp1[1].compareTo(sp2[1]);
            }
            // 1 is digit && 2 is digit: --> 0
            // 1 is digit && 2 is not digit --> 1
            // 1 is not digit && 2 is digit --> -1
            return !isDigit1 ? -1 : (isDigit2 ? 0 : 1);
        });
        return logs;
    }
}