[LeetCode] Simplify Path（可以不用看）

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

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

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".

注意以下幾點：input"/.", output"/"

                    input"/..", output"/"

                    input"/...", output"/..."即“...”、“....”等是有效檔案名稱

class Solution {public:    string simplifyPath(string path) {         if(path == "/" || path=="")            return path;        string result(1,path[0]);        int len = path.size();        if(path[len-1]!=‘/‘){            path.push_back(‘/‘);            len++;        }        int j=0,start;        stack<int> sstart;        for(int i = 1;i<len;){            if(path[i] != ‘/‘ && path[i] != ‘.‘){//if(1)                                while(i<len && path[i]!=‘/‘){                    result.push_back(path[i++]);                    j++;                }                int flag = j;                while(flag>=0 && result[flag]!=‘/‘){                    flag--;                }                sstart.push(flag+1);                if(i<len-1 && path[i]==‘/‘){                    result.push_back(path[i++]);                    j++;                }else if(i == len-1 && path[i]==‘/‘){                    return result;                 }                }else{                if(path[i]==‘/‘ && result[j]==‘/‘)                    i++;                else if(i<len-1 && path[i]==‘.‘ && path[i+1]==‘/‘){                    i=i+2;                }else if(i<len-2 && path[i]==‘.‘ && path[i+1]==‘.‘&& path[i+2]==‘/‘){                    i = i+3;                    if(result.size() == 1)                        continue;                    else{                        if(result[j]==‘/‘){                            start = sstart.top();                            sstart.pop();                            result.erase(result.begin()+start,result.end());                            j = start-1;                        }else{      //  "/.../""output"/.../"                            int flag = j;                            while(flag>=0 && result[flag]!=‘/‘){                                flag--;                            }                            sstart.push(flag+1);                            result.push_back(path[i-3]);                            result.push_back(path[i-2]);                            if(i-1<len-1 && path[i-1]==‘/‘){                                result.push_back(path[i-1]);                                j+=3;                            }else if(i-1 == len-1 && path[i-1]==‘/‘){                                return result;                             }                            }                      }                }else{                    result.push_back(path[i++]);                    j++;                }                        }//end if(1)        }//end for       while(result[j]==‘/‘ && result.size()!=1){             result = result.substr(0,j);             j--;         }        return result;              }//end func};