0071. Simplify Path (M)

时间:2021-02-06 11:56:45   收藏:0   阅读:0

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:

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:


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 : "/";
    }
}
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