LeetCode-Simplify Path

最新推荐文章于 2020-07-25 17:20:55 发布

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最新推荐文章于 2020-07-25 17:20:55 发布

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分类专栏： LeetCode 文章标签： LeetCode Stack

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本文介绍了一个简化Unix风格文件系统的绝对路径的算法。该算法通过使用栈来处理输入路径中的符号.和..，并返回最短的有效路径。文章详细解释了算法的工作原理，并提供了一个Java实现示例。

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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. For more information, see: Absolute path vs relative path in Linux/Unix

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:

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

Example 2:

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:

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

Example 4:

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

Example 5:

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

Example 6:

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

题意：将一个Unix下的绝对路径进行简化；例如简化包含的当前位置"."、上一个目录位置".."等；

解法：因为需要考虑到当前路径或者上一个路径，这里利用栈来实现；首先将字符串按照"/"进行分割后，遍历字符串数组，向栈中压入路径

如果路径为空，则跳过
如果路径为当前位置"."，则跳过
如果路径为前一个位置，且栈不为空，则将栈顶路径弹出
直到遍历完字符串数组

最后，遍历栈中元素，得到完整的路径，这里需要注意的是最后得到的结果可能为空，需要添加一个根路径；

Java

class Solution {
    public String simplifyPath(String path) {
        if ("".equals(path)) {
            return path;
        }
        String[] paths = path.split("/");
        LinkedList<String> stack = new LinkedList<>();
        for (String p : paths) {
            if ("".equals(p) || ".".equals(p)) {
                continue;
            } else if ("..".equals(p)) {
                if (!stack.isEmpty()) {
                    stack.pop();
                }
            } else {
                stack.push(p);
            }
        }
        StringBuilder result = new StringBuilder();
        while (!stack.isEmpty()) {
            result.insert(0, "/" + stack.pop());
        }

        return "".equals(result.toString()) ? "/" : result.toString();
    }
}