937 Reorder Data in Log Files

1 题目

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"]

2 尝试解

2.1 分析

给定一组记录，每条记录都是由识别符和若干单词组成。一条记录中的所有单词要么全部字母，要么全部为数字。现在要对记录重新排序，将所有单词为字母的记录按照由单词组成的字符串的顺序排列，如果单词全部相同，则比较识别符。所有单词为数字的记录则保持原有相对顺序，并且放在所有单词为字母的记录之后。

使用一个数组digit和一个小顶堆letter来分别存储数字记录和字母记录。对于每一条记录log，找到识别符之后的由单词组成的字符串words，如果是数字记录，按照相对顺序存入digit末尾，如果是字母记录，则以<words,log>的形式存入letter中。最后将字母记录从letter中依序取出，再与digit合并即可。

2.2 代码

class Solution {
public:
    struct cmp{
        bool operator() (pair<string,string>&A, pair<string,string>&B){
            if(A.first > B.first) return true;
            if(A.first == B.first && A.second > B.second) return true;
            return false;
        }
    };
    vector<string> reorderLogFiles(vector<string>& logs) {
        vector<string> digit;
        vector<string> result;
        priority_queue<pair<string,string>,vector<pair<string,string>>,cmp> letter;
        for(auto log : logs){
            int index = 0;
            while(log[index]!=' ') index++;
            string word = log.substr(index+1);
            if(isdigit(word[0]))
                digit.push_back(log);
            else{
                letter.push(make_pair(word,log));
            }
        }
        while(!letter.empty()){
            result.push_back(letter.top().second);
            letter.pop();
        }
        result.insert(result.end(),digit.begin(),digit.end());
        return result;
    }
};

3 标准解

class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        vector<string> digitLogs, ans;
        vector<pair<string, string>> letterLogs;
        for (string &s : logs) {
            int i = 0;
            while (s[i] != ' ') ++i;
            if (isalpha(s[i + 1])) letterLogs.emplace_back(s.substr(0, i), s.substr(i + 1));
            else digitLogs.push_back(s);
        }
        sort(letterLogs.begin(), letterLogs.end(), [&](auto& a, auto& b) {
            return a.second == b.second ? a.first < b.first : a.second < b.second;
        });
        for (auto &p : letterLogs) ans.push_back(p.first + " " + p.second);
        for (string &s : digitLogs) ans.push_back(s);
        return ans;
    }
};