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
珍珠是蚌的眼泪8 分钟前
LeetCode_位运算
leetcode·位运算·异或·韩明距离·数字的补数
野犬寒鸦10 分钟前
力扣hot100:旋转图像(48)(详细图解以及核心思路剖析)
java·数据结构·后端·算法·leetcode
墨染点香13 分钟前
LeetCode 刷题【61. 旋转链表】
算法·leetcode·职场和发展
岁忧31 分钟前
(LeetCode 面试经典 150 题) 200. 岛屿数量(深度优先搜索dfs || 广度优先搜索bfs)
java·c++·leetcode·面试·go·深度优先
dbdr090132 分钟前
Linux 入门到精通,真的不用背命令!零基础小白靠「场景化学习法」,3 个月拿下运维 offer,第二十六天
linux·运维·服务器·网络·python·学习
一枝小雨39 分钟前
【OJ】C++ vector类OJ题
数据结构·c++·算法·leetcode·oj题
Tisfy1 小时前
LeetCode 3516.找到最近的人:计算绝对值大小
数学·算法·leetcode·题解
自信的小螺丝钉1 小时前
Leetcode 206. 反转链表 迭代/递归
算法·leetcode·链表
大白的编程日记.1 小时前
【Linux学习笔记】信号的深入理解之软件条件产生信号
linux·笔记·学习
群联云防护小杜1 小时前
服务器异常磁盘写排查手册 · 已删除文件句柄篇
运维·服务器·nginx·开源·lua