[LeetCode] 937. Reorder Data in Log Files

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.

重新排列日志文件。

给你一个日志数组 logs。每条日志都是以空格分隔的字串，其第一个字为字母与数字混合的 标识符 。

有两种不同类型的日志：

字母日志：除标识符之外，所有字均由小写字母组成
数字日志：除标识符之外，所有字均由数字组成
请按下述规则将日志重新排序：

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 应该保留原来的相对顺序。
返回日志的最终顺序。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reorder-data-in-log-files
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是先将字母日志和数字日志分开，然后仅对字母日志排序，数字日志无需排序，最后再合并两种日志。具体步骤如下

• 通过每个字符串的最后一个字符判断当前日志属于字母还是数字，并放到不同的 list 里，这里我分别定义为 letters 和 digits
• 对于字母日志 letters 的排序，因为日志的大致格式是 [标识符 + 日志内容]，所以这里我们两两比较的时候，先看日志内容是否一样，如果内容一样，则按日志整体的字典序排序；如果日志内容不一样，则仅按日志内容的字典序排序
• 最后将排好序的 letters 和 digits 两个 list 再写回结果集

时间O(nlogn)

空间O(nlogn) - 默认的快排是需要一定空间的

JavaScript实现

/**
 * @param {string[]} logs
 * @return {string[]}
 */
var reorderLogFiles = function (logs) {
    var letters = [], nums = [];

    // Separate digit-logs from letter-logs
    logs.forEach(function (log) {
        if (log.split(" ")[1].charAt(0) >= '0' && log.split(" ")[1].charAt(0) <= '9') {
            nums.push(log);
        } else {
            letters.push(log);
        }
    });

    // Sort letter-logs
    letters.sort(function (a, b) {
        var cmp = a.slice(a.indexOf(" ")).localeCompare(b.slice(b.indexOf(" ")));
        return cmp === 0 ? a.localeCompare(b) : cmp;
    });
    return [...letters, ...nums];
};

 

Java实现

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        // corner case
        if (logs == null || logs.length == 0) {
            return logs;
        }

        // normal case
        int n = logs.length;
        List<String> digits = new ArrayList<>();
        List<String> letters = new ArrayList<>();
        for (String log : logs) {
            char last = log.charAt(log.length() - 1);
            if (last >= '0' && last <= '9') {
                digits.add(log);
            } else {
                letters.add(log);
            }
        }

        Collections.sort(letters, new Comparator<>() {
            @Override
            public int compare(String s1, String s2) {
                int index1 = s1.indexOf(" ");
                String letter1 = s1.substring(index1 + 1);
                int index2 = s2.indexOf(" ");
                String letter2 = s2.substring(index2 + 1);
                if (letter1.equals(letter2)) {
                    return s1.compareTo(s2);
                }
                return letter1.compareTo(letter2);
            }
        });

        String[] res = new String[n];
        int i = 0;
        for (String str : letters) {
            res[i++] = str;
        }
        for (String str : digits) {
            res[i++] = str;
        }
        return res;
    }
}

 

LeetCode 题目总结