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.
思路：这题比较难想的是，怎么找下一个方向，不能random的找，否则会陷入死循环，方法就是四个方向按照一定顺序，也就是clockwise的找，下一层的时候，收集可以move到的点，然后再返回，如果不行，机器人就turnRight(),也就是找到下一个pos，跟dx dy是一致的方向；
/**
* // 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 {
//
private int[] dx = {-1,0,1,0};
private int[] dy = {0,1,0,-1};
public void cleanRoom(Robot robot) {
HashSet<String> visited = new HashSet<>();
dfs(robot, visited, 0, 0, 0);
}
private void dfs(Robot robot, HashSet<String> visited,
int x, int y, int dir) {
String pos = x + "-" + y;
robot.clean();
visited.add(pos);
for(int k = 0; k < 4; k++) {
int newdir = (dir + k) % 4;
int nx = x + dx[newdir];
int ny = y + dy[newdir];
String newpos = nx + "-" + ny;
if(!visited.contains(newpos) && robot.move()) {
dfs(robot, visited, nx, ny, newdir);
// back tracking;
robot.turnLeft();
robot.turnLeft();
robot.move();
robot.turnRight();
robot.turnRight();
}
// 因为dx，dy是顺时针的，那么向右转，其实是面对下一个方向；
robot.turnRight();
}
}
}