Go语言LeetCode题解937重新排列日志文件

输入：logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"]
输出：["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]
解释：
字母日志的内容都不同，所以顺序为 "art can", "art zero", "own kit dig" 。
数字日志保留原来的相对顺序 "dig1 8 1 5 1", "dig2 3 6" 。

输入：logs = ["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"]

class Solution {    
public:    
    static bool cmp(string& str1, string& str2){
        int idx1=str1.find(' '), idx2=str2.find(' ');
        auto substr1=str1.substr(idx1+1), substr2=str2.substr(idx2+1);//标识符后的子串
        if(substr1 != substr2)return  substr1<substr2;//子串进行字典序排序
        return str1<str2;//子串相同，排序标识符
    }
    vector<string> reorderLogFiles(vector<string>& logs) {     
        vector<string> nums, alphas, res;
         //日志分成两类字母和数字分别存放在alphas 和 nums中，res为最后的结果。    
        for (auto &iter : logs) {    
            int index = iter.find_first_of(' '); //分割tag 和 log 的pos
            int idx_log = index + 1;  //log的pos  
            if (idx_log < iter.size() && iter[idx_log] <= '9' && iter[idx_log] >= 0) {    
                nums.push_back(iter);    
            }    
            else if (idx_log < iter.size()){    
                alphas.push_back(iter);    
            }    
        }    
        sort(alphas.begin(), alphas.end(), cmp);     
        for (auto &iter : alphas) {       
            res.push_back(iter);        
        }     
        for (auto &iter : nums) {    
            res.push_back(iter);    
        }   
    return res;    
    }    
};