Showing posts with label System Design. Show all posts
Showing posts with label System Design. Show all posts

LeetCode 588 - Design In-Memory File System


http://bookshadow.com/weblog/2017/05/21/leetcode-design-in-memory-file-system/
Design an in-memory file system to simulate the following functions:
ls: Given a path in string format. If it is a file path, return a list that only contains this file's name. If it is a directory path, return the list of file and directory names in this directory. Your output (file and directory names together) should in lexicographic order.
mkdir: Given a directory path that does not exist, you should make a new directory according to the path. If the middle directories in the path don't exist either, you should create them as well. This function has void return type.
addContentToFile: Given a file path and file content in string format. If the file doesn't exist, you need to create that file containing given content. If the file already exists, you need to append given content to original content. This function has void return type.
readContentFromFile: Given a file path, return its content in string format.
Example:
Input: 
["FileSystem","ls","mkdir","addContentToFile","ls","readContentFromFile"]
[[],["/"],["/a/b/c"],["/a/b/c/d","hello"],["/"],["/a/b/c/d"]]
Output:
[null,[],null,null,["a"],"hello"]
Explanation:
filesystem
Note:
  1. You can assume all file or directory paths are absolute paths which begin with / and do not end with / except that the path is just "/".
  2. You can assume that all operations will be passed valid parameters and users will not attempt to retrieve file content or list a directory or file that does not exist.
  3. You can assume that all directory names and file names only contain lower-case letters, and same names won't exist in the same directory.

题目大意:

设计内存文件系统模拟实现如下功能:
ls:给定路径字符串,若对应一个目录,则输出其中包含的目录和文件(字典序);若对应一个文件,则只输出该文件名
mkdir:创建目录,若目录不存在,则递归创建缺少的目录
addContentToFile:在文件中追加内容,若文件不存在,则新建
readContentFromFile:从文件中读取内容并返回
注意:
  1. 你可以假设文件和目录的路径均为绝对路径,以/开始,并且结尾不包含/,除非路径就是"/"本身
  2. 你可以假设所有操作均合法,用户不会常识读取一个不存在的文件,或者ls一个不存在的目录
  3. 你可以假设所有目录名称和文件名称都只包含小写字母,并且在同一目录下不会存在同名的目录或者文件
树形结构(Tree)
目录节点Node包含两个字段dirs和files,分别存储其中包含的子目录节点和文件内容
X. Use TreeMap to contain children nodes - so they are sorted.
https://discuss.leetcode.com/topic/90016/file-system-as-a-tree-java
    private class TreeNode implements Comparable<TreeNode> {
        String label;
        boolean isDir;
        Map<String, TreeNode> dirContents = new TreeMap<>();
        String fileContents;
        
        public int compareTo(TreeNode other) {
            return label.compareTo(other.label);
        }
        
        public int hashCode() {
            return Objects.hash(label);
        }
        
        public boolean equals(Object treeNode) {
            TreeNode other = (TreeNode) treeNode;
            return Objects.equals(label, other.label);
        }
    }    
    
    private TreeNode sentinel = new TreeNode();

    public FileSystem() {
        TreeNode root = new TreeNode();
        root.label = "";
        root.isDir = true;
        sentinel.dirContents.put("", root);
    }
    
    public List<String> ls(String path) {
        String[] labels = path.split("/");
        TreeNode curr = (labels.length != 0) ? sentinel : sentinel.dirContents.get("");
        List<String> result = new ArrayList<>();
        for(String label : labels) {
            curr = curr.dirContents.get(label);
            System.out.println(">>" + curr);
        }
        if(curr.isDir) {
            for(String label : curr.dirContents.keySet()) {
                result.add(label);
            }
        }
        else {
            result.add(curr.label);
        }
        
        return result;
    }
    
    public void mkdir(String path) {
        String[] labels = path.split("/");
        TreeNode curr = sentinel;
        int currIndex = 0;
        for(String label : labels) {
            TreeNode next = curr.dirContents.get(label);
            if(next == null) {
                break;
            }
            currIndex++;
            curr = next;
        }
        
        for(int i = currIndex; i < labels.length; i++) {
            TreeNode newDir = new TreeNode();
            newDir.isDir = true;
            newDir.label = labels[i];
            curr.dirContents.put(labels[i], newDir);
            
            curr = newDir;
        }
    }
    
    public void addContentToFile(String filePath, String content) {
        String[] labels = filePath.split("/");
        TreeNode curr = sentinel, prev = null;
        
        for(String label : labels) {
            prev = curr;
            if(curr == null) {
                break;
            }
            curr = curr.dirContents.get(label);
        }
        
        if(curr == null) {
            curr = new TreeNode();
            curr.isDir = false;
            curr.label = labels[labels.length - 1];
            curr.fileContents = content;
            prev.dirContents.put(curr.label, curr);
        }
        else {
            curr.fileContents = curr.fileContents == null ? content : curr.fileContents + content;
        }
    }
    
    public String readContentFromFile(String filePath) {
        String[] labels = filePath.split("/");
        TreeNode curr = sentinel;
        
        for(String label : labels) {
            curr = curr.dirContents.get(label);
        }
        
        return (curr.fileContents == null) ? "" : curr.fileContents;
    }

X. https://discuss.leetcode.com/topic/90000/java-solution-file-class
https://discuss.leetcode.com/topic/90004/file-system-as-graph
    class File {
        boolean isFile = false;
        Map<String, File> children = new HashMap<>();
        String content = "";
    }
    
    File root = null;
    
    public FileSystem() {
        root = new File();
    }
    
    public List<String> ls(String path) {
        String[] dirs = path.split("/");
        File node = root;
        List<String> result = new ArrayList<>();
        String name = "";
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                return result;
            }
            node = node.children.get(dir);
            name = dir;
        }
        
        if (node.isFile) {
            result.add(name);
        }
        else {
            for (String key : node.children.keySet()) {
                result.add(key);
            }
        }
        
        Collections.sort(result);
        
        return result;
    }
    
    public void mkdir(String path) {
        String[] dirs = path.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }
    }
    
    public void addContentToFile(String filePath, String content) {
        String[] dirs = filePath.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }
        node.isFile = true;
        node.content += content;
    }
    
    public String readContentFromFile(String filePath) {
        String[] dirs = filePath.split("/");
        File node = root;
        for (String dir : dirs) {
            if (dir.length() == 0) continue;
            if (!node.children.containsKey(dir)) {
                File file = new File();
                node.children.put(dir, file);
            }
            node = node.children.get(dir);
        }

        return node.content;
    }

Google – Stop All Way


Google – Stop All Way
一道非常有意思的Design题。Google engineer们的脑洞也是一个比一个大。
题目很简单,就是design data structure来实现美国的stop all way路口,具体就是实现
void arriveCar(Lane lane, Car car);
Car getNextCar();
[Solution]
首先对于Car和Lane的design. 每个Lane都会有一个Queue of Car, 每个Car也会有一个Lane object.
然后对于schedule, 有两种solution。
注意考虑下多线程的情况。
[Solution #1]
第一种是round robin。1,2,3,4,四个路口,比如这次从路口1走了一辆车,下次从路口2开始扫描,看有没有车等着,有就从2走,再下次就从3开始扫。如果路口2没有车,那就看3,扫完一圈回到路口1,要是路口1有车,就走,没有,就说明此时此刻4个路口都没有车。
但是这样的解法看上去就不太合理。因为如果想象一下现实生活中的场景,其实会是一个multithreading的过程,4个路口相当于4个threads不停的有车来,而同一时间只能走一辆车,先来先走。
那么对于round robin的方法,会有bug。比如现在4个路口都没有车,然后getNextCar()和arriveCar(4, car)被两个threads同时call,那按照上面说的从0开始扫,发现没车,到1,结果这时候又有一个thread call arriveCar(3, car), 那么这样的情况扫到路口3的时候发现有车,就会让路口3的车先走。但事实却是路口4的车先到。
[Solution #2]
Maintain一个queue来记录schedule的顺序,在getNextCar()的时候只要poll出queue里的第一辆车就可以了.
当arriveCar(Lane lane, Car car)的时候,如果当前lane里一辆车都没有,把新来的这辆车offer进schedule queue. 否则只需要offer进当前路口自己的queue就可以了。
当getNextCar()之后,将next car所属车道的下一辆车offer进schedule queue。
这样即可以保证先到先走的顺序,用一个BlockingQueue也可以解决multithreading的情况。
class Car {
  int id;
  Lane lane;
   
  public Car(int id, Lane lane) {
    this.id = id;
    this.lane = lane;
  }
   
}
   
class Lane {
  Queue<Car> queue;
  int id;
   
  public Lane(int id) {
    this.id = id;
    this.queue = new LinkedList<>();
  }
}
 
// round robin - not survive multithreading
class Solution {
   
  List<Lane> lanes = new ArrayList<>();
  lanes.add(new Lane(0));
  lanes.add(new Lane(1));
  lanes.add(new Lane(2));
  lanes.add(new Lane(3));
 
  int currLane = 0;
     
  public void arriveCar(Car car, Lane lane) {
    if (car == null || lane == null) {
      return;
    }
    lane.queue.offer(car);
  }
   
  // round robin - not survive multithreading
  // 比如现在所有车道都没有车,currlane在0。
  // 然后getNextCar()和arriveCar(car, 4)同时被两个threads call
  // currlane从0开始循环,循环到1车道的时候又有另个一thread call了arriveCar(car, 3)
  // 结果会变成3车道的车比4车道的车先走,但事实是4车道的车先到。
  public Car getNextCar() {
    Car result = null;
    int tmp = currLane;
    while (lanes.get(tmp).queue.isEmpty()) {
      tmp = (tmp + 1) % 4;
      if (tmp == currLane) {
        break;
      }
    }
    if (!lanes.get(tmp).queue.isEmpty()) {
      result = lanes.get(tmp).poll();
      currLane = (tmp + 1) % 4;
    }
     
    return result;
  }
}
 
 
class Solution2 {
  Lane lane0 = new Lane(0);
  Lane lane1 = new Lane(1);
  Lane lane0 = new Lane(2);
  Lane lane1 = new Lane(3);
   
  Queue<Car> next = new LinkedList<>();
   
   
  public void arriveCar(Car car, Lane lane) {
    if (car == null || lane == null) {
      return;
    }
     
    if (lane.queue.isEmpty()) {
      next.offer(car);
    }
    car.lane = lane;
    lane.queue.offer(car);
  }
   
  public Car getNext() {
    if (next.isEmpty()) {
      return null;
    }
     
    Car result = next.poll();
    result.lane.queue.poll();
    if (!result.lane.queue.isEmpty()) {
      next.offer(result.lane.queue.peek());
    }
     
    return result;
  }
}
Read full article from Google – Stop All Way

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