489. Robot Room Cleaner - Hard

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

初始从(0, 0)开始，用dirs数组表示上下左右，把走过的位置信息编码成string，用set记录，用visited数组标记当前位置是否被访问过

在dfs函数里：

因为初始位置永远是可达的，先调用clean()，然后把当前位置加入visited。

遍历dirs中的四个方向（循环4次，每次递归函数传进来的dir是上一次的方向，dir + i是当前方向，为防止越界，对dir + i 取余，即 (dir + i ) % 4），对于每个方向，先判断一下这个新位置是否被访问过，以及是否能到达（用clean()），如果都满足，则调用递归函数。每次遍历完一个方向，robot turn right。

调用完递归函数后，注意还要把方向转回原来的方向。即先转180度，前进一步move()，再转180度回到原来方向。

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();
        }
    }
}