937 Reorder-Data-in-Log-Files
difficulty: Easy
section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }
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:
0 <= logs.length <= 100
3 <= logs[i].length <= 100
logs[i]
is guaranteed to have an identifier, and a word after the identifier.
Method One
class Solution {
public String[] reorderLogFiles(String[] logs) {
Arrays.sort(logs, (a,b) -> {
// 用空白分隔符给分成两段,第一段是identifier, 第二段是log内容
String[] splitA = a.split(" ",2);
String[] splitB = b.split(" ",2);
//题目的描述不清楚。它的排序规则是,
// 首先我们比较 log 的内容,letter 排在 digit 之前。
// 都是digit 那么不排序,都是 letter 按 lexicographical 排序。
// letter一样再给 identifier 按 lexicalgraphcial 排序
String idenA = splitA[0];
String idenB = splitB[0];
String logA = splitA[1];
String logB = splitB[1];
boolean isADigit = Character.isDigit(logA.charAt(0));
boolean isBDigit = Character.isDigit(logB.charAt(0));
if(isADigit && isBDigit){
return 0;
}
if(!isADigit && !isBDigit){
int cmp = logA.compareTo(logB);
if( cmp != 0){
return cmp;
}
return idenA.compareTo(idenB);
}
return isADigit ? 1 : -1;
} );
return logs;
}
}
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