Question

937. Reorder Data in Log Files

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Description

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:

1. The letter-logs come before all digit-logs.
2. The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
3. 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.

Solutions

Solution 1

class Solution:
  def reorderLogFiles(self, logs: List[str]) -> List[str]:
    def cmp(x):
      a, b = x.split(' ', 1)
      return (0, b, a) if b[0].isalpha() else (1,)

    return sorted(logs, key=cmp)

class Solution {
  public String[] reorderLogFiles(String[] logs) {
    Arrays.sort(logs, this::cmp);
    return logs;
  }

  private int cmp(String a, String b) {
    String[] t1 = a.split(" ", 2);
    String[] t2 = b.split(" ", 2);
    boolean d1 = Character.isDigit(t1[1].charAt(0));
    boolean d2 = Character.isDigit(t2[1].charAt(0));
    if (!d1 && !d2) {
      int v = t1[1].compareTo(t2[1]);
      return v == 0 ? t1[0].compareTo(t2[0]) : v;
    }
    if (d1 && d2) {
      return 0;
    }
    return d1 ? 1 : -1;
  }
}

function reorderLogFiles(logs: string[]): string[] {
  const isDigit = (c: string) => c >= '0' && c <= '9';
  return logs.sort((a, b) => {
    const end1 = a[a.length - 1];
    const end2 = b[b.length - 1];
    if (isDigit(end1) && isDigit(end2)) {
      return 0;
    }
    if (isDigit(end1)) {
      return 1;
    }
    if (isDigit(end2)) {
      return -1;
    }
    const content1 = a.split(' ').slice(1).join(' ');
    const content2 = b.split(' ').slice(1).join(' ');
    if (content1 === content2) {
      return a < b ? -1 : 1;
    }
    return content1 < content2 ? -1 : 1;
  });
}

impl Solution {
  pub fn reorder_log_files(mut logs: Vec<String>) -> Vec<String> {
    logs.sort_by(|s1, s2| {
      let (start1, content1) = s1.split_once(' ').unwrap();
      let (start2, content2) = s2.split_once(' ').unwrap();
      match
        (
          content1.chars().nth(0).unwrap().is_digit(10),
          content2.chars().nth(0).unwrap().is_digit(10),
        )
      {
        (true, true) => std::cmp::Ordering::Equal,
        (true, false) => std::cmp::Ordering::Greater,
        (false, true) => std::cmp::Ordering::Less,
        (false, false) => content1.cmp(&content2).then(start1.cmp(&start2)),
      }
    });
    logs
  }
}