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.

import java.util.*;
class Solution {
    public String[] reorderLogFiles(String[] logs) {
        LinkedList<String> logList = new LinkedList<>();
        List<Map.Entry<String, String>> tempList = new LinkedList<>();
        Map<String, String> map = new HashMap<>();
        for (String log : logs) {
            String value = log.substring(log.split(" ")[0].length()+1, log.length());
            int charVal = value.charAt(0)-'0';
            if (charVal >=0 && charVal <= 9) {
                logList.addLast(log);
            } else {
                map.put(log, value);
            }
        }
        tempList.addAll(map.entrySet());
        tempList.sort(new Comparator<Map.Entry<String, String>>() {
            @Override
            public int compare(Map.Entry<String, String> o1, Map.Entry<String, String> o2) {
                return o2.getValue().compareTo(o1.getValue());
            }
        });
        for (Map.Entry<String, String> e : tempList) {
            logList.addFirst(e.getKey());
        }
        String[] res = new String[logs.length];
        return logList.toArray(res);
    }
}