Description：
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.
题意：要求对一个日志数组logs进行排序，logs中的元素为一个字符串(格式为：identifier content)，排序的要求如下
- content为字母的排在数字前
- content为数字的相对位置不变
- 当content相同且不为数字时，根据identifier来排序
解法：本题需要我们做的就是实现一个排序算法使其满足上述的排序要求即可；我们可以按照字符串的格式，对每个字符串分割为两个部分(identifier + content)，之后，按照排序要求进行比较两个字符串即可；
Java
class Solution {
public String[] reorderLogFiles(String[] logs) {
Arrays.sort(logs, (str1, str2) -> {
String[] s1 = str1.split(" ", 2);
String[] s2 = str2.split(" ", 2);
boolean digit1 = s1[1].charAt(0) <= '9';
boolean digit2 = s2[1].charAt(0) <= '9';
if (!digit1 && !digit2) {
return s1[1].compareTo(s2[1]) != 0 ?
s1[1].compareTo(s2[1]) : s1[0].compareTo(s2[0]);
}
return digit1 ? (digit2 ? 0 : 1) : -1;
});
return logs;
}
}