Leetcode Simplify Path

题目

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<string> 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;
    }
};