leetcode489 - 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.
1.递归出口：如果扫过了就走。
2.本状态处理：扫地，给memo加记录。
3.递归，四方向：转身试挪动，递归，回溯到原位置原方向。

细节：
1.本题挪动时控制两个对象：一个是用机器人的API让它真的往四个方向试着挪动；一个是用传统的dxdy计算挪动坐标用于去重。
2.坐标用set去重有小trick，可以以规范化的string形式放入set。
3.本题就是回溯的时候因为API不方便，和一般的dfs写得不太一样而已，别的没什么大变化。

实现：

/**
 * // 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 {
    public void cleanRoom(Robot robot) {
        dfs(robot, new HashSet<String>(), 0, 0, 0);
    }

    private void dfs(Robot robot, Set<String> visited, int x, int y, int crtDir) {
        if (visited.contains(x + "->" + y)) {
            return;
        }

        int[] dx = {-1, 0, 1, 0};
        int[] dy = {0, -1, 0, 1};

        robot.clean();
        visited.add(x + "->" + y);

        for (int i = 0; i < 4; i++) {
            // P1: 4方向循环的时候每次不但要控制自己算虚拟坐标，还要控制机器人的API让它真的转了个90度。
            int dir = (crtDir + i + 1) % 4;
            robot.turnRight();
            if (!robot.move()) {
                continue;
            }

            int nx = x + dx[dir];
            int ny = y + dy[dir];
            dfs(robot, visited, nx, ny, dir);
            // go back. Because you executed "move()", backtrack for it.
            robot.turnLeft();
            robot.turnLeft();
            robot.move();
            robot.turnRight();
            robot.turnRight();
        }
    }
}

原文地址：https://www.cnblogs.com/jasminemzy/p/9739743.html