leetcode - 1166. Design File System

Description

You are asked to design a file system that allows you to create new paths and associate them with different values.

The format of a path is one or more concatenated strings of the form: / followed by one or more lowercase English letters. For example, "/leetcode" and "/leetcode/problems" are valid paths while an empty string "" and "/" are not.

Implement the FileSystem class:

  • bool createPath(string path, int value) Creates a new path and associates a value to it if possible and returns true. Returns false if the path already exists or its parent path doesn't exist.
  • int get(string path) Returns the value associated with path or returns -1 if the path doesn't exist.

Example 1:

复制代码
Input: 
["FileSystem","createPath","get"]
[[],["/a",1],["/a"]]
Output: 
[null,true,1]
Explanation: 
FileSystem fileSystem = new FileSystem();

fileSystem.createPath("/a", 1); // return true
fileSystem.get("/a"); // return 1

Example 2:

复制代码
Input: 
["FileSystem","createPath","createPath","get","createPath","get"]
[[],["/leet",1],["/leet/code",2],["/leet/code"],["/c/d",1],["/c"]]
Output: 
[null,true,true,2,false,-1]
Explanation: 
FileSystem fileSystem = new FileSystem();

fileSystem.createPath("/leet", 1); // return true
fileSystem.createPath("/leet/code", 2); // return true
fileSystem.get("/leet/code"); // return 2
fileSystem.createPath("/c/d", 1); // return false because the parent path "/c" doesn't exist.
fileSystem.get("/c"); // return -1 because this path doesn't exist.

Constraints:

复制代码
2 <= path.length <= 100
1 <= value <= 10^9
Each path is valid and consists of lowercase English letters and '/'.
At most 104 calls in total will be made to createPath and get.

Solution

Trie tree

Code

python3 复制代码
class TrieNode:
    def __init__(self, val: int):
        self.child = {}
        self.val = val

class FileSystem:

    def __init__(self):
        self.root = TrieNode(-1)

    def createPath(self, path: str, value: int) -> bool:
        paths = path.split('/')[1:]
        node = self.root
        for each_path in paths[:-1]:
            if each_path not in node.child:
                return False
            node = node.child[each_path]
        if paths[-1] in node.child:
            return False
        # create the last directory
        node.child[paths[-1]] = TrieNode(value)
        return True
        

    def get(self, path: str) -> int:
        node = self.root
        paths = path.split('/')[1:]
        for each_path in paths:
            if each_path in node.child:
                node = node.child[each_path]
            else:
                return -1
        return node.val


# Your FileSystem object will be instantiated and called as such:
# obj = FileSystem()
# param_1 = obj.createPath(path,value)
# param_2 = obj.get(path)
相关推荐
律宏阔2 小时前
WSL Docker 端口明明空闲,但就是绑不上端口
linux·windows
律宏阔2 小时前
WSL 突然断网,无法 ping 内网或外网
linux·windows
穷人小水滴2 小时前
用容器编译 VirtualBox 虚拟机软件 (ArchLinux, podman)
linux·容器·virtualbox
乱码三千3 小时前
如何优雅地直连无公网 IP 的远程 GPU 服务器
linux·人工智能·程序员
yunwei373 小时前
使用 eBPF 跟踪 Nginx 请求
linux·后端·性能优化
yunwei373 小时前
使用 eBPF 跟踪 MySQL 查询
linux·后端·性能优化
yunwei373 小时前
eBPF 示例教程:使用 XDP 捕获 TCP 信息
linux·后端·性能优化
GeW3 小时前
制造业高端化,为什么必须重估Linux和数据库?
linux
百万蹄蹄向前冲3 小时前
双端同步!云服务器装最新Node.js v26.10全过程追踪
服务器·人工智能·node.js
GeW3 小时前
工业场景中的Linux与数据库选型:高端化转型的技术底座
linux