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.

扫地机器人。

房间（用格栅表示）中有一个扫地机器人。格栅中的每一个格子有空和障碍物两种可能。

扫地机器人提供4个API，可以向前进，向左转或者向右转。每次转弯90度。

当扫地机器人试图进入障碍物格子时，它的碰撞传感器会探测出障碍物，使它停留在原地。

请利用提供的4个API编写让机器人清理整个房间的算法。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/robot-room-cleaner
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是DFS。这道题既然是模拟扫地机器人，那么DFS其实是蛮直观的。可以用BFS做吗？不能。因为BFS做是没法回到原点的。

这道题大部分的实现都还是比较直观的，需要注意以下几点

• 需要一个hashset记录访问过的坐标
• 给机器人定义的移动方向是需要有顺序的，要不然是顺时针要不然是逆时针。我这里给的是顺时针的做法，上右下左。当机器人在某一个方向走到头的时候，需要让它原地180度掉头，回到初始位置，再去扫描下一个位置

时间O(mn) - 遍历的坐标，应该是一个二维的房间吧

空间O(n) - hashset

Java实现

 1 class Solution {
 2     // the order of the directions matters
 3     private int[][] dirs = new int[][] { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
 4 
 5     public void cleanRoom(Robot robot) {
 6         HashSet<String> visited = new HashSet<>();
 7         int[] cur = new int[] { 0, 0 };
 8         dfs(robot, cur, 0, visited);
 9     }
10 
11     private void dfs(Robot robot, int[] cur, int dir, HashSet<String> visited) {
12         robot.clean();
13         visited.add(cur[0] + "-" + cur[1]);
14         for (int i = 0; i < dirs.length; i++) {
15             int nextDir = (dir + i) % 4;
16             int[] next = new int[] { cur[0] + dirs[nextDir][0], cur[1] + dirs[nextDir][1] };
17             if (!visited.contains(next[0] + "-" + next[1]) && robot.move()) {
18                 dfs(robot, next, nextDir, visited);
19                 robot.turnRight();
20                 robot.turnRight();
21                 robot.move();
22                 robot.turnLeft();
23             } else {
24                 robot.turnRight();
25             }
26         }
27     }
28 }

 

flood fill题型总结

LeetCode 题目总结