Simplify Path

Given an absolute path for a file (Unix-style), simplify it.

For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"

click to show corner cases.

Corner Cases:

• Did you consider the case where path = "/../"?
  In this case, you should return "/".
• Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
  In this case, you should ignore redundant slashes and return "/home/foo".

 

it takes me a lot of time...

 

class Solution {
public:
    string simplifyPath(string path) {
    vector<string> s;
    string name;
    const char *p = path.c_str();
    while (*p != '')
    {
        while (*p == '/')                   //去除多余的/
            p++;
        while (*p != '/' && *p !='')      //分段，用name储存
        {
            name.push_back(*p++);
        }
        if (name[0] == '')                //是否是结尾，结尾则不处理
            ;
        else if (name.compare("..") == 0)   //判断是否有上一级目录，有则返回上一级目录
        {
            if (s.size() > 0)
                s.erase(s.end());
        }
        else if (name.compare(".") != 0)    //当前目录.不用处理，否则加入到vector中
        {
            s.push_back(name);
        }
        name.erase(0);
    }
    string ret;
    if (s.size() == 0)                      //特殊情况/
        return "/";
    vector<string>::iterator it;
    for(it=s.begin(); it!=s.end(); it++){
        ret += "/";
        ret += *it;
    }
    return ret;
    }
};