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: ["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"]

Note:

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.

思路： 將第一小段去掉後，判斷第一個字元是字母還是數字。

           若是數字，直接放到一個vector裡面即可

          若是字母，只保留後面部分，並建立map對映關係

          最後，排序，輸出即可。

程式碼：

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        vector<string> vec1,vec2;
        int n=logs.size();
        map<string,string> mp;
        mp.clear();
        for(int i=0;i<n;i++)
        {
            int j=0;
            while(logs[i][j]!=' ')
            {
                j++;
            }
            while(logs[i][j]==' ')
            {
                j++;
            }
            if(logs[i][j]>='0'&&logs[i][j]<='9')
            {
                vec1.push_back(logs[i]);
            }
            else
            {
                string tmp;
                for(int k=j;k<logs[i].size();k++) tmp.push_back(logs[i][k]);
                mp[tmp] = logs[i];
                vec2.push_back(tmp);
            }
        }
        sort(vec2.begin(),vec2.end());
        vector<string> ans;
        for(auto &t: vec2) {
           ans.push_back(mp[t]);
        }
        for(auto &t : vec1) ans.push_back(t);
        return ans;
    }
};