总结

1. 做这种题时需要考虑 corner case, 比如连续几个 slash

2. 状态函数带参数 (int &pos), 主函数带 while 循环 while(pos < len), 我意识到对于 char* 或 string 操作时, 参数带上 lenth 也是蛮好的

3. 这次很快就 A 了, 甚至忘记以前为什么会在这里栽跟头

 

代码

string getFolderName(const string &path, int &i, const int &len)  {	string res;	while(i < len && path[i] != '/')  {		res.push_back(path[i]);		i ++;	}	return res;}class Solution {public:    string simplifyPath(string path) {  		deque
     
       record;  		int i = 0;  		int len = path.size();  		while(i < len)  {  			if(path[i] == '/')  {  				i ++;  			}  else  {  				string folderName = getFolderName(path, i, len);  				if(folderName == ".")  {  					// do nothing  				}  else if(folderName.size() == 2 && folderName == "..")  {  					if(!record.empty())  record.pop_back();  				}  else  {  					record.push_back(folderName);  				}  			}  		}  		string simplifiedPath = "";		while(!record.empty())  {			simplifiedPath.push_back('/');			simplifiedPath.append(record.front());			record.pop_front();		}		if(simplifiedPath.size() == 0)  {			simplifiedPath = "/";		}     		return simplifiedPath;    }};