Leetcode
2020.04.25 12:17

# 937. Reorder Data in Log Files

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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.

```class Solution {
public String[] reorderLogFiles(String[] logs) {
Arrays.sort(logs, (str1, str2)->{
String[] sp1 = str1.split(" ", 2);
String[] sp2 = str2.split(" ", 2);

boolean digit1 = Character.isDigit(sp1[1].charAt(0));
boolean digit2 = Character.isDigit(sp2[1].charAt(0));

if(digit1 && digit2){
return 0;
}else if(digit1 || digit2){
return digit1 ? 1 : -1;
}else{
int compare = sp1[1].compareTo(sp2[1]);
if(compare == 0){
return sp1[0].compareTo(sp2[0]);
}
return compare;
}
/*
if(!digit1 && !digit2){
if(sp1[1].compareTo(sp2[1]) == 0){
return sp1[0].compareTo(sp2[0]);
}
return sp1[1].compareTo(sp2[1]);
}
return digit1 ? (digit2 ? 0 : 1) : -1;
*/
});

return logs;
}
}```

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