LeetCode - Simplify Path

1 minute read

Problem description

description

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.

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:

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

Example 2:

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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:

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

Example 4:

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

Example 5:

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

Example 6:

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

Analysis

It’s a stack related problem.

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public String simplifyPath(String path) {
        //path = path.replaceAll("/+", "/");
        String[] array = path.split("/");

        List<String> list = new ArrayList<>();
        for (String s:array){
            if (s.equals(".")){
                continue;
            }
            
            if (s.equals("..")){
                if (list.size() != 0){
                    list.remove(list.size() - 1);                    
                }
            }
            else if (!s.isEmpty()){
                list.add(s);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        
        for(String s:list){
            sb.append("/");
            sb.append(s);
        }
        
        if (sb.length() == 0){
            sb.append("/");            
        }
        
        return sb.toString();
    }

What to improve

  • regex use a lot of time, we should not use it if possible.