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Simplify Path (M)

题目

Given a string path, which is an absolute path (starting with a slash ‘/‘) to a file or directory in a Unix-style file system, convert it to the simplified canonical path.

In a Unix-style file system, a period ‘.‘ refers to the current directory, a double period ‘..‘ refers to the directory up a level, and any multiple consecutive slashes (i.e. ‘//‘) are treated as a single slash ‘/‘. For this problem, any other format of periods such as ‘...‘ are treated as file/directory names.

The canonical path should have the following format:

• The path starts with a single slash ‘/‘.
• Any two directories are separated by a single slash ‘/‘.
• The path does not end with a trailing ‘/‘.
• The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period ‘.‘ or double period ‘..‘)

Return the simplified canonical path.

Example 1:

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

Example 2:

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

Example 4:

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

Constraints:

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

题意

将一个绝对路径转化为标准格式。

思路

先将字符串按照"/"分割，遍历所有子串，维护一个栈，如果是"."，不做处理；如果是".."，若栈非空则出栈一个元素；如果是其他非空字符串，则压入栈中。最后再将栈中的字符串用"/"连接起来。

代码实现

Java

class Solution {
    public String simplifyPath(String path) {
        String[] parsed = path.split("/");
        Deque<String> stack = new ArrayDeque<>();
        for (int i = 0; i < parsed.length; i++) {
            if (parsed[i].equals(".")) {
                continue;
            } else if (parsed[i].equals("..")) {
                if (!stack.isEmpty()) stack.pop();
            } else if (!parsed[i].isEmpty()) {
                stack.push(parsed[i]);
            }
        }

        String ans = "";
        while (!stack.isEmpty()) {
            ans = "/" + stack.pop() + ans;
        }
        return !ans.isEmpty() ? ans : "/";
    }
}

0071. Simplify Path (M)

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原文地址：https://www.cnblogs.com/mapoos/p/14379407.html