974. Reorder Data in Log Files

• Difficulty: Easy

• Topics: String

• Similar Questions:

Problem:

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.

Solutions:

class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        auto comparator = [this](const string& log1, const string& log2) {
            int pos1 = 0;
            int pos2 = 0;
            if (isLetterLog(log1, pos1) && isLetterLog(log2, pos2)) {
                string logContent1 = log1.substr(pos1);
                string logContent2 = log2.substr(pos2);
                if (logContent1 != logContent2) {
                    return logContent1 < logContent2;
                }

                return log1.substr(0, pos1 - 1) < log2.substr(0, pos2 - 1);
            } else if (isLetterLog(log1, pos1) && !isLetterLog(log2, pos2)) {
                return true;
            } else if (!isLetterLog(log1, pos1) && isLetterLog(log2, pos2)) {
                return false;
            } else {
                return false;
            }
        };

        stable_sort(logs.begin(), logs.end(), comparator);
        return logs;
    }

private:
    bool isLetterLog(const string& log, int& pos) {
        pos = 0;
        while (log[pos] != ' ') ++pos;
        ++pos; // eat empty space
        return !isdigit(log[pos]);
    } 

};