# 937.Reorder Log Files

**937.Reorder Log Files**

难度:Easy

> 你有一个日志数组 logs。每条日志都是以空格分隔的字串。

对于每条日志，其第一个字为字母数字标识符。然后，要么：

标识符后面的每个字将仅由小写字母组成，或； 标识符后面的每个字将仅由数字组成。 我们将这两种日志分别称为字母日志和数字日志。保证每个日志在其标识符后面至少有一个字。

将日志重新排序，使得所有字母日志都排在数字日志之前。字母日志按内容字母顺序排序，忽略标识符；在内容相同时，按标识符排序。数字日志应该按原来的顺序排列。

返回日志的最终顺序。

示例 ：

```
输入：["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
输出：["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]
```

提示：

0 <= logs.length <= 100 3 <= logs\[i].length <= 100 logs\[i] 保证有一个标识符，并且标识符后面有一个字。

先取出内容部分，然后将数字部分移到末尾，然后对于字母部分排序，冒泡排序即可。

```
class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        vector<string>content;
        for(auto l : logs)
        {
            int i=0;
            while(l[i]!=' ')
                i++;
            string tmp=l.substr(i+1);
            content.push_back(tmp);
        }
       
        int second=logs.size()-1;
         int start=second-1;
        
        while(start>=0)
        {
            if(content[second][0] >='a' && content[second][0]<= 'z')         
            {
                if(content[start][0]>='0' && content[start][0] <= '9' )
                {
                    string t=content[start];
                    content[start]=content[second];
                    content[second]=t;
                     t=logs[start];
                    logs[start]=logs[second];
                    logs[second]=t;
                }
                else
                    start--;
                
            }
            else 
            {
                second--;
                if(second ==start)
                    start--;
            }
            
        }
        
        for(int i=0;i<content.size();i++)
            
        {
            if(content[i][0]>='0' && content[i][0]<='9' ) 
            {
                start=i;
                break;
            }
            }
        for(int i=0;i<start;i++)
            for(int j=i+1;j<start;j++)
            {
                if(content[i]>content[j])
                {
                    string t=content[i];
                content[i]=content[j];
                content[j] = t; 
                t=logs[i];
                logs[i]=logs[j];
                logs[j] = t; 
                }
                else if(content[i]==content[j])
                {
                    if(logs[i]>logs[j])
                    {
               string t=content[i];
                content[i]=content[j];
                content[j] = t; 
                t=logs[i];
                logs[i]=logs[j];
                logs[j] = t; 
                    }
                }
            }
        return logs;
    }
};
```

> 执行用时 : 20 ms, 在Reorder Log Files的C++提交中击败了96.79% 的用户\
> 内存消耗 : 16.3 MB, 在Reorder Log Files的C++提交中击败了7.50% 的用户


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