题目描述

Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.

In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix

Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.

Example 1:

Input: "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.

Example 2:

Input: "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.

Example 3:

Input: "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.

Example 4:

Input: "/a/./b/../../c/"
Output: "/c"

Example 5:

Input: "/a/../../b/../c//.//"
Output: "/c"

Example 6:

Input: "/a//b////c/d//././/.."
Output: "/a/b/c"

题意：给一个Unix文件路径，返回其最简路径

算法1

题解：我们可以首先获取所有反斜杠之间的内容，那么总共有四种情况.，..，file_name，我们在使用一个栈来维护当前路径。

• 如果读到的是空格或者.，那么不做操作
• 如果读到的是文件名，入栈
• 如果读到的是..，如果栈不为空就退栈。

最后栈中的元素就是当前文件路径，注意是否为空。

C++ 代码

    string simplifyPath(string path) {
        int n = path.length();
        vector<string> res;
        stack<string> st;
        for(int i = 0 ;i < n ;)
        {   
            string cur = "";
            int j = i + 1;
            while(j < n && path[j] != '/')
                cur += path[j++];
            i = j;
            res.push_back(cur);
        }
        for(int i = 0 ; i < res.size() ;i ++)
        {
            if(res[i] == "" || res[i] == ".")continue;
            else if(res[i] == "..") {if(!st.empty()) st.pop();}
            else st.push(res[i]);
        }
        string ans = "";
        while(!st.empty())
        {
            ans = "/" + st.top() + ans;
            st.pop();
        }
        return ans == "" ? "/":ans;
    }