LeetCode Simplify Path (JavaScript)

링크 : https://leetcode.com/problems/simplify-path/description/

문제 설명

You are given an absolute path for a Unix-style file system, which always begins with a slash '/'. Your task is to transform this absolute path into its simplified canonical path.

The rules of a Unix-style file system are as follows:

• A single period '.' represents the current directory.
• A double period '..' represents the previous/parent directory.
• Multiple consecutive slashes such as '//' and '///' are treated as a single slash '/'.
• Any sequence of periods that does not match the rules above should be treated as a valid directory or file name. For example, '...' and '....' are valid directory or file names.

The simplified canonical path should follow these rules:

• The path must start with a single slash '/'.
• Directories within the path must be separated by exactly one slash '/'.
• The path must not end with a slash '/', unless it is the root directory.
• The path must not have any single or double periods ('.' and '..') used to denote current or parent directories.

Return the simplified canonical path.

Example 1:

Input: path = "/home/"

Output: "/home"

Explanation:

The trailing slash should be removed.

Example 2:

Input: path = "/home//foo/"

Output: "/home/foo"

Explanation:

Multiple consecutive slashes are replaced by a single one.

Example 3:

Input: path = "/home/user/Documents/../Pictures"

Output: "/home/user/Pictures"

Explanation:

A double period ".." refers to the directory up a level (the parent directory).

Example 4:

Input: path = "/../"

Output: "/"

Explanation:

Going one level up from the root directory is not possible.

Example 5:

Input: path = "/.../a/../b/c/../d/./"

Output: "/.../b/d"

Explanation:

"..." is a valid name for a directory in this problem.

Constraints:

• 1 <= path.length <= 3000
• path consists of English letters, digits, period '.', slash '/' or '_'.
• path is a valid absolute Unix path.

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문제 풀이

'/' 를 기준으로 문자열을 나눈 뒤, 나눈 문자열을 가지고 최종 path를 생성

1. 빈 문자열이거나 .(현재 디렉토리) 인 경우 처리가 필요없으므로 continue

2. ..(이전 디렉토리)인 경우 answer(현재 경로)의 길이가 1이상일 경우 맨 위의 원소를 뺌

3. 아닌 경우 answer에 경로 추가

최종적으로 answer를 '/'로 이어 붙이면 됨

/**
 * @param {string} path
 * @return {string}
 */
var simplifyPath = function(path) {
    const p = path.split('/');
    const answer = [];

    for (let i = 0; i < p.length; i++) {
        if (p[i] === '' || p[i] === '.') continue;
        
        if (p[i] === '..') {
            if (answer.length > 0) {
                answer.pop();
            }
        } else {
            answer.push(p[i]);
        }
    }

    return '/' + answer.join('/'); 
};

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결론

stack을 이용한 구현 문제