Given a robot cleaner in a room modeled as a grid.

Each cell in the grid can be empty or blocked.

The robot cleaner with 4 given APIs can move forward, turn left or turn right. Each turn it made is 90 degrees.

When it tries to move into a blocked cell, its bumper sensor detects the obstacle and it stays on the current cell.

Design an algorithm to clean the entire room using only the 4 given APIs shown below.

interface Robot {
  // returns true if next cell is open and robot moves into the cell.
  // returns false if next cell is obstacle and robot stays on the current cell.
  boolean move();

  // Robot will stay on the same cell after calling turnLeft/turnRight.
  // Each turn will be 90 degrees.
  void turnLeft();
  void turnRight();

  // Clean the current cell.
  void clean();
}

Example:

Input:
room = [
  [1,1,1,1,1,0,1,1],
  [1,1,1,1,1,0,1,1],
  [1,0,1,1,1,1,1,1],
  [0,0,0,1,0,0,0,0],
  [1,1,1,1,1,1,1,1]
],
row = 1,
col = 3

Explanation:
All grids in the room are marked by either 0 or 1.
0 means the cell is blocked, while 1 means the cell is accessible.
The robot initially starts at the position of row=1, col=3.
From the top left corner, its position is one row below and three columns right.

Notes:

1. The input is only given to initialize the room and the robot's position internally. You must solve this problem "blindfolded". In other words, you must control the robot using only the mentioned 4 APIs, without knowing the room layout and the initial robot's position.
2. The robot's initial position will always be in an accessible cell.
3. The initial direction of the robot will be facing up.
4. All accessible cells are connected, which means the all cells marked as 1 will be accessible by the robot.
5. Assume all four edges of the grid are all surrounded by wall.

dfs + backtracking

初始從(0, 0)開始，用dirs陣列表示上下左右，把走過的位置資訊編碼成string，用set記錄，用visited陣列標記當前位置是否被訪問過

在dfs函式裡：

因為初始位置永遠是可達的，先呼叫clean()，然後把當前位置加入visited。

遍歷dirs中的四個方向（迴圈4次，每次遞迴函式傳進來的dir是上一次的方向，dir + i是當前方向，為防止越界，對dir + i 取餘，即 (dir + i ) % 4），對於每個方向，先判斷一下這個新位置是否被訪問過，以及是否能到達（用clean()），如果都滿足，則呼叫遞迴函式。每次遍歷完一個方向，robot turn right。

呼叫完遞迴函式後，注意還要把方向轉回原來的方向。即先轉180度，前進一步move()，再轉180度回到原來方向。

time: O(m * n), space: O(m * n)　　-- m, n: row, col of grid

/**
 * // This is the robot's control interface.
 * // You should not implement it, or speculate about its implementation
 * interface Robot {
 *     // Returns true if the cell in front is open and robot moves into the cell.
 *     // Returns false if the cell in front is blocked and robot stays in the current cell.
 *     public boolean move();
 *
 *     // Robot will stay in the same cell after calling turnLeft/turnRight.
 *     // Each turn will be 90 degrees.
 *     public void turnLeft();
 *     public void turnRight();
 *
 *     // Clean the current cell.
 *     public void clean();
 * }
 */
class Solution {
    int[][] dirs = new int[][]{{-1,0},{0,1},{1,0},{0,-1}};  // top, right, down, left
    
    public void cleanRoom(Robot robot) {
        Set<String> visited = new HashSet<>();
        dfs(robot, visited, 0, 0, 0);
    }
    
    private void dfs(Robot robot, Set<String> visited, int cur_dir, int row, int col) {
        robot.clean();
        StringBuilder sb = new StringBuilder();
        sb.append(row);
        sb.append("->");
        sb.append(col);
        visited.add(sb.toString());
        
        for(int i = 0; i < 4; i++) {
            int next_dir = (cur_dir + i) % 4, next_row = row + dirs[next_dir][0], next_col = col + dirs[next_dir][1];
            StringBuilder next = new StringBuilder();
            next.append(next_row);
            next.append("->");
            next.append(next_col);
            if(!visited.contains(next.toString()) && robot.move()) {
                dfs(robot, visited, next_dir, next_row, next_col);
                robot.turnLeft();
                robot.turnLeft();
                robot.move();
                robot.turnLeft();
                robot.turnLeft();
            }
            robot.turnRight();
        }
    }
}