### LeetCode  • ㊗️
• 大家
• offer
• 多多！

## Problem

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

For example, path = “/home/”, => “/home” path = “/a/./b/../../c/”, => “/c” path = “/a/../../b/../c//.//”, => “/c” path = “/a//b////c/d//././/..”, => “/a/b/c”

In a UNIX-style file system, a period (‘.’) refers to the current directory, so it can be ignored in a simplified path. Additionally, a double period (“..”) moves up a directory, so it cancels out whatever the last directory was. For more information, look here: https://en.wikipedia.org/wiki/Path_(computing)#Unix_style

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”.

## Code

class Solution {
public String simplifyPath(String path) {
Stack<String> stack = new Stack<>();
String[] arr = path.split("/");

for(int i = 0; i < arr.length; i++){
String curr = arr[i].trim();
if(curr == null || curr.length() == 0 || curr.equals(".")) continue;

if(curr.equals("..")){
if(!stack.isEmpty()){
stack.pop();
}
} else {
stack.push(curr);
}
}

String res = "";

while(!stack.isEmpty()){
res = "/" + stack.pop() + res;
}

return res.length() == 0 ? "/" : res;
}
}