LeetCode | 71. Simplify Path

题外话：最近每天都在反复的思考，何以解忧，唯有刷题^______________^

题目：

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".

题意解析：

输入一个字符串，表示某个文件路径，本题的目标是将该输入路径简化为形式：

                                            simplified path = slash '/' + folder_name + ...

当出现多余的斜杠和反斜杠，需要简化为一个；当出现 '.' 为当前路径，则忽略；当出现 '..' 则返回上一级路径。

理清题意之后是思考如何解这个题目，因为需要考虑返回上一级路径，我的做法是把当前所有扫描合法的folder_name存储到一个数组里，当检测到 ''..'' 则弹出数组尾部的folder_name。

值得注意的坑是：有时候folder_name中可能会包含'.' 和 ".." ，所以每次需要判断当前扫描到点时是否符合预期约束。

代码：

string simplifyPath(string path) {        int len = path.length();        if(len < 1)            return path;        vector<string> subpaths;        string cur = "", res = "";        for(int i = 0; i<len; i++)        {            if(path[i] == '/' || path[i] == '\/')            {                if(cur != "")                {                    subpaths.push_back(cur);                    cur = "";                }                continue;            }            if(path[i] == '.')            {                if(cur == "")                {                    if(i < len - 1 && path[i+1] == '.')                    {if(i+1 == len - 1 || (i+1 < len - 1 && path[i+2] == '/')){i++;if(subpaths.size() > 0)subpaths.pop_back();continue;}                    }else if(i == len - 1 || (i < len - 1 && path[i+1] == '/')){continue;}                }            }            cur += path[i];        }if(cur != "")subpaths.push_back(cur);        for(int i = 0; i<subpaths.size(); i++)        {            res += '/' + subpaths[i];        }        if(res == "")            res += '/';        return res;    }

题外话：

转眼已入秋，树叶慢慢变黄，金黄色的时候一定要去拍照呀！不念过去的想法，不畏将来的困惑，就看眼前的快乐和成长。加油~