[LeetCode] 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".

思路：

Unix的path规则可以在这里了解：

http://en.wikipedia.org/wiki/Path_(computing)

归下类的话，有四种字符串：

1. "/"：为目录分隔符，用来分隔两个目录。

2. "."：当前目录

3. ".."：上层目录

4. 其他字符串：目录名

简化的核心是要找出所有的目录，并且如果遇到".."，需要删除上一个目录。

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class Solution { public: string simplifyPath(string path) { string ret, curDir; vector<string> allDir; path.push_back('/'); for(int i=0; i<path.size(); i++) { if(path[i]=='/') { if(curDir.empty()) { continue; } else if(curDir==".") { curDir.clear(); } else if(curDir=="..") { if(!allDir.empty()) allDir.pop_back(); curDir.clear(); } else { allDir.push_back(curDir); curDir.clear(); } } else { curDir.push_back(path[i]); } } for(int i=0; i<allDir.size(); i++) ret.append("/"+allDir[i]); if(ret.empty()) ret = "/"; return ret; } };