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"]
class Solution { public String[] reorderLogFiles(String[] logs) { Arrays.sort(logs, new Comparator<String>() { @Override public int compare(String str1, String str2) { String[] strArr1 = str1.split(" ", 2); String[] strArr2 = str2.split(" ", 2); boolean isOneDigit = Character.isDigit(strArr1[1].charAt(0)); boolean isTwoDigit = Character.isDigit(strArr2[1].charAt(0)); if (!isOneDigit && !isTwoDigit) { int comNum = strArr1[1].compareTo(strArr2[1]); // ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo","a2 act car"] if (comNum == 0) { return strArr1[0].compareTo(strArr2[0]); } return comNum; } else if (isOneDigit && !isTwoDigit) { return 1; } else if (!isOneDigit && isTwoDigit) { return -1; } return 0; } }); return logs; } }