原题链接在这里：https://leetcode.com/problems/simplify-path/

题目：

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

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

题解：

首先用string.split 把原有path 按照"/"分开 存入 String 数组strArr中.

"."代表当前层. /a/b/. 还是在b folder中.

从前往后扫strArr. 遇到"." 和 " "直接跳过，遇到正常字符就压入栈中，遇到".."时若stack不是空就pop()一次。

得到最后的stack后一个一个出栈加入到StringBuilder中即可。

Note: 注意corner case 如"///", 此时应该返回"/", 但若是用split拆分是返回一个空的strArr, 所以当stack是空的时候，特殊返回"/".

Time Complexity: O(path.length()). Space: O(path.length()).

AC Java:

 1 class Solution {
 2     public String simplifyPath(String path) {
 3         if(path == null || path.length() == 0){
 4             return path;
 5         }
 6         
 7         //split string into array
 8         //when encoutering "." or " ", continue
 9         //when encoutering "..", if stack is not empty, pop stack and return to upper level
10         //when encoutering else, push into stack.
11         String [] parts = path.trim().split("/");
12         Stack<String> stk = new Stack<>();
13         for(String part : parts){
14             if(part.equals(".")){
15                 continue;
16             }
17             
18             if(part.equals("..")){
19                 // Either way, when encountering "..", it needs to skip.
20                 // Thus it can't combine below if with outter one 
21                 if(!stk.isEmpty()){
22                     stk.pop();
23                 }
24                 
25                 continue;
26             }
27             
28             if(part.length() > 0){
29                 stk.push(part);
30             }
31         }
32         
33         StringBuilder sb = new StringBuilder();
34         while(!stk.isEmpty()){
35             sb.insert(0, "/"+stk.pop());
36         }
37         
38         // Corner case like "///", it should return "/".
39         if(sb.length() == 0){
40             sb.insert(0, "/");
41         }
42         
43         return sb.toString();
44     }
45 }