传送门：

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.

 

大意：给你一些日志，包括英文日志和数字日志，每个日志又包括日志头和日志内容，日志头是第一个单词，日志内容全数字的是数字日志，全英文的是英文日志。要求排序后输出。

排序规则：对英文的日志，去掉日志头，按日志内容字典序排序。对数字的日志，按输入顺序。

                 总体上，英文日志在前，数字日志在后。

思路：对每个字符串的最后一位进行判断之后，分类到两个向量里，对英语日志用stringstream进行分割，然后sort排序。

代码：

class Solution {public:    static bool cmp(string s1,string s2){        string news1="",news2="";        stringstream ss1(s1);        string s;        int k = 0;        while(ss1>>s){            if(k > 0){                news1 = news1 + s +" ";             }            k++;        }        k = 0;        stringstream ss2(s2);        while(ss2>>s){            if(k > 0){                news2 = news2 + s +" ";             }            k++;        }        if(news1
      
        reorderLogFiles(vector
       
        & logs) {        vector
        
         p1,p2,ans;        for(int i = 0 ; i < logs.size() ; i++){            string s = logs[i];            if(s[s.size()-1]<='9' && s[s.size()-1] >= '0'){                p2.push_back(s);            }            else{                p1.push_back(s);            }        }        sort(p1.begin(),p1.end(),cmp);        for(int i = 0 ; i < p1.size() ; i ++){            ans.push_back(p1[i]);        }        for(int i = 0 ; i < p2.size() ; i ++){            ans.push_back(p2[i]);        }        return ans;    }};