### 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: ["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.

### 区分开“数字log”和“字母log”，只给字母log排序

#### 怎么比较act car和off key？

["act", "car"]
["off", "key"]


|
act car
off key


   |
aaa
aaaa


#### 另外两个提高效率的地方

1. 不要用lambda表达式，效率比老老实实写Comparator类慢10倍。用lambda表达式32ms，不用2ms
2. String[]效率比List<String>效率高。

#### 代码

class Solution {
public String[] reorderLogFiles(String[] logs) {
String[] letterLogs = new String[100];
String[] digitLogs = new String[100];
int pl = 0, pd = 0;
for (String log : logs) {
if (Character.isDigit(log.charAt(log.indexOf(' ') + 1))) {
digitLogs[pd++] = log;
} else {
letterLogs[pl++] = log;
}
}
Arrays.sort(letterLogs, 0, pl, new Comparator<String>(){
public int compare(String a, String b) {
int pa = a.indexOf(' ') + 1;
int pb = b.indexOf(' ') + 1;
return a.substring(pa).compareTo(b.substring(pb));
}
});
String[] res = new String[logs.length];
int p = 0;
for (int i = 0; i < pl; i++) {
res[p++] = letterLogs[i];
}
for (int i = 0; i < pd; i++) {
res[p++] = digitLogs[i];
}
return res;
}
}