937.Reorder-Data-in-Log-Files

937. Reorder Data in Log Files

题目地址

https://leetcode.com/problems/reorder-data-in-log-files/

题目描述

You are given an array of logs. Each log is a space-delimited string of words, where the first word is the identifier.

There are two types of logs:

Letter-logs: All words (except the identifier) consist of lowercase English letters.
Digit-logs: All words (except the identifier) consist of digits.
Reorder these logs so that:

The letter-logs come before all digit-logs.
The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
The digit-logs maintain their relative ordering.
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"]
Explanation:
The letter-log contents are all different, so their ordering is "art can", "art zero", "own kit dig".
The digit-logs have a relative order of "dig1 8 1 5 1", "dig2 3 6".

Example 2:
Input: logs = ["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"]

Constraints:
1 <= logs.length <= 100
3 <= logs[i].length <= 100
All the tokens of logs[i] are separated by a single space.
logs[i] is guaranteed to have an identifier and at least one word after the identifier.

代码

Approach #1

Time: O(1) && Space: O(1)

class Solution {
  public String[] reorderLogFiles(String[] logs) {
      Comparator<String> cmp = new Comparator<String>() {
        public int compare(String log1, String log2) {
          String[] split1 = log1.split(" ", 2);
          String[] split2 = log2.split(" ", 2);

          boolean isDigit1 = Character.isDigit(split1[1].charAt(0));
          boolean isDigit2 = Character.isDigit(split2[1].charAt(0));

          if (!isDigit1 && !isDigit2) {
            int cmp = split1[1].compareTo(split2[1]);
            if (cmp != 0)   return cmp;

            if (cmp == 0) {
                return split1[0].compareTo(split2[0]);
            } else {
                return cmp;
            }
          }

          if (!isDigit1 && isDigit2) {
              return -1;
          } else if (isDigit1 && !isDigit2) {
              return 1;
          } else {
              return 0;
          }

      }
    };


    Arrays.sort(logs, cmp);
    return logs;
  }
}