LeetCode 937. Reorder Data in Log Files

原题链接在这里：https://leetcode.com/problems/reorder-data-in-log-files/

题目：

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:

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.

题解：

Sort the logs.

When both are letters, sort based on the second part.

If need to flip, lamda return 1.

Note: split("\s+", 2), 2 could specify only split into 2 parts. If there are multiple "\s+", only split on the first one.

Time Complexity: O(nlogn*m). n = logs.length. m = average length.

Space: O(1).

AC Java:

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        if(logs == null || logs.length == 0){
            return logs;
        }
        
        Arrays.sort(logs, (a, b) -> {
            String [] sa = a.split("\s+", 2);
            String [] sb = b.split("\s+", 2);
            
            boolean isDigita = Character.isDigit(sa[1].charAt(0));
            boolean isDigitb = Character.isDigit(sb[1].charAt(0));
            
            if(!isDigita && !isDigitb){
                int com = sa[1].compareTo(sb[1]);
                if(com == 0){
                    return sa[0].compareTo(sb[0]);
                }
                
                return com;
            }else if(isDigita && isDigitb){
                return 0;
            }else if(isDigita && !isDigitb){
                // flip when a is digit and b is letter
                return 1;
            }else{
                return -1;
            }
        });
        
        return logs;
    }
}