共计 1349 个字符，预计需要花费 4 分钟才能阅读完成。
Problem
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 <= 1003 <= logs[i].length <= 100logs[i] is guaranteed to have an identifier, and a word after the identifier.
Solution
class Solution {
public String[] reorderLogFiles(String[] logs) {
Comparator<String> comparator = new Comparator<String>() {
@Override
public int compare(String s1, String s2) {
int i1 = s1.indexOf(‘ ‘);
int i2 = s2.indexOf(‘ ‘);
char c1 = s1.charAt(i1+1);
char c2 = s2.charAt(i2+1);
if (c1 <= ‘9’) {
return c2 <= ‘9’ ? 0 : 1;
} else if (c2 <= ‘9’) {
return -1;
} else {
//all letters
int res = s1.substring(i1+1).compareTo(s2.substring(i2+1));
if (res == 0) return s1.substring(0,i1).compareTo(s2.substring(0,i2));
else return res;
}
}
};
Arrays.sort(logs, comparator);
return logs;
}
}