［抄题］：
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:
- 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.
- The robot's initial position will always be in an accessible cell.
- The initial direction of the robot will be facing up.
- All accessible cells are connected, which means the all cells marked as 1 will be accessible by the robot.
- Assume all four edges of the grid are all surrounded by wall.
[暴力解法]：
时间分析：
空间分析：
[优化后]：
时间分析：
空间分析：
［奇葩输出条件］：
［奇葩corner case］：
［思维问题］：
想不出还要回到原来的位置，而且如果不move的话要向右转。
［英文数据结构或算法，为什么不用别的数据结构或算法］：
［一句话思路］：
start cleaning然后向四周扩展，能move就move，否则向右转。
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
- 判断方向是否重复，要把现有方向添加到SET中来判断
robot.clean();
set.add(temp);
- 方向的变化是X Y的变化
［二刷］：
This is actually the standard backtracking method. Here is an example, in this problem, suppose the robot starts at (0, 0), and all four directions are accessible. You can assume the robots starts at a crossroad so (0, 0) is the only point to connect other roads. First, it goes up to (-1, 0) and continues from that cell. After all the upper cells are cleaned, it should return to (0, 0), then go right to (0, 1), and so on. Those lines of code make the robot go back to (0, 0) from (-1, 0).
［三刷］：
［四刷］：
［五刷］：
[五分钟肉眼debug的结果]：
［总结］：
backtracing的入口在for循环之前
［复杂度］：Time complexity: O(n) Space complexity: O(n)
［算法思想：迭代/递归/分治/贪心］：
［关键模板化代码］：
［其他解法］：
［Follow Up］：
［LC给出的题目变变变］：
[代码风格] ：
[是否头一次写此类driver funcion的代码] ：
[潜台词] ：