937. Reorder Log Files

#### QUESTION:

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.

#### EXPLANATION:

1.将字符串数组进行拆分，分成数字和字母，数字直接按照原顺序

2.对字母的list进行排序，依据就是每个字符的字典顺序

#### SOLUTION:

``````class Solution {
public String[] reorderLogFiles(String[] logs) {
ArrayList<String> letter = new ArrayList<>();
ArrayList<String> digit = new ArrayList<>();
String[] result = new String[logs.length];
for(int i = 0;i<logs.length;i++){
String log = logs[i];
String[] splits = log.split(" ");
if(splits.charAt(0)>=48 && splits.charAt(0)<=57)
else

}
Collections.sort(letter, (o1, o2) -> {
String[] s1 = o1.split(" ");
String[] s2 = o2.split(" ");
int len1 = s1.length;
int len2 = s2.length;
for (int i = 1; i < Math.min(len1, len2); i++) {
if (!s1[i].equals(s2[i])) {
return s1[i].compareTo(s2[i]);
}
}
return 0;
});
for(int i = 0;i<logs.length;i++){
if(i<letter.size())
result[i] = letter.get(i);
else
result[i] = digit.get(i-letter.size());
}
return result;
}
}
``````
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