[LeetCode] 937. Reorder Data in Log Files

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:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

重新排列日志文件。分两种日志，一种是字母日志，一种是数字日志。每个日志的第一个单词决定了他到底是字母日志还是数字日志，请按如下规则对日志排序。

• 所有 字母日志 都排在 数字日志 之前。
• 字母日志 在内容不同时，忽略标识符后，按内容字母顺序排序；在内容相同时，按标识符排序；
• 数字日志 应该按原来的顺序排列。

思路是需要自己写comparator函数。

时间O(nlogn)

空间O(n)

Java实现

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        List<String> llogs = new ArrayList<>();
        List<String> dlogs = new ArrayList<>();
        for (String log : logs) {
            int i = log.indexOf(" ");
            char ch = log.charAt(i + 1);
            if (ch >= '0' && ch <= '9') {
                dlogs.add(log);
            } else {
                llogs.add(log);
            }
        }
        Collections.sort(llogs, new Comparator<String>() {
            @Override
            public int compare(String s1, String s2) {
                int index1 = s1.indexOf(" ");
                String id1 = s1.substring(0, index1);
                String letter1 = s1.substring(index1 + 1);
                int index2 = s2.indexOf(" ");
                String id2 = s2.substring(0, index2);
                String letter2 = s2.substring(index2 + 1);
                int v1 = letter1.compareTo(letter2);
                if (v1 != 0) {
                    return v1;
                }
                int v2 = id1.compareTo(id2);
                return v2;
            }
        });
        String[] res = new String[llogs.size() + dlogs.size()];
        int i = 0;
        for (String s : llogs) {
            res[i++] = s;
        }
        for (String s : dlogs) {
            res[i++] = s;
        }
        return res;
    }
}

LeetCode 题目总结