我正在努力解决Leetcode问题489。机器人房间清洁器使用回溯。具体来说,我尝试在四个方向中的每一个方向移动机器人,如果四个方向都被阻塞或清理,则返回。
下面的代码不起作用,我正试图用这个简单的示例对其进行调试:
grid = [[1,1,1],
[1,0,1]]
其中1表示机器人可以访问单元,0表示单元被阻止。机器人从这个初始位置开始:
initial_row = 0
initial_col = 1
在我运行代码后,机器人只清洁了网格的右侧部分(c-清洁,d-左脏):
[[d,c,c],
[d,0,c]]
它似乎正确地返回到单元格[0,1],并尝试移动到左上角(单元格[0,0]),但机器人没有移动。move()返回false,函数在清理左上角之前返回。如果您能给我一些建议,我将不胜感激。
代码:
struct hsh {
size_t operator() (const pair<int,int>& p) const {
hash<int> intHash;
return intHash(p.first) ^ intHash(p.second);
}
};
class Solution {
public:
unordered_set<pair<int, int>, hsh> cleaned;
vector<int> directions {0, 1, 2, 3}; // up, down, right, left
void cleanRoom(Robot& robot) {
int row_s = 0; // relative start row
int col_s = 0; // relative start col
clean(row_s, col_s, robot);
}
void clean(int row, int col, Robot& robot) {
cleaned.insert({row, col});
robot.clean();
for (int dir : directions) {
if (!is_cleaned(row, col, dir)) {
turn(robot, dir);
if (robot.move()) {
turn_back(robot, dir);
int row_new = get_new_row(row, dir);
int col_new = get_new_col(col, dir);
clean(row_new, col_new, robot);
} else {
turn_back(robot, dir);
}
}
}
}
bool is_cleaned(int row, int col, int dir) {
int row_new = get_new_row(row, dir);
int col_new = get_new_col(col, dir);
if(cleaned.find({row_new, col_new}) != cleaned.end()) {
return true;
}
return false;
}
void turn(Robot& robot, int dir) {
if (dir == 0) { // up
} else if (dir == 1) { // down
robot.turnLeft();
robot.turnLeft();
} else if (dir == 2) { // right
robot.turnRight();
} else { // left
robot.turnLeft();
}
}
void turn_back(Robot& robot, int dir) {
if (dir == 0) { // back to up from up
} else if (dir == 1) { // back to up from down
robot.turnLeft();
robot.turnLeft();
} else if (dir == 2) { // back to up from right
robot.turnLeft();
} else { // back to up from left
robot.turnRight();
}
}
int get_new_row(int row, int dir) {
if (dir == 0) { // up
row--;
return row;
} else if (dir == 1) { // down
row++;
return row;
}
return row;
}
int get_new_col(int col, int dir) {
if (dir == 2) { // right
col++;
return col;
} else if (dir == 3) { // left
col--;
return col;
}
return col;
}
};
问题在于,机器人类接口在回溯时不会自动将机器人放回原位(机器人对象通过引用传递并保持其位置)。添加了一个特定的功能,在回溯后将机器人移回修复了该问题:
void go_back(Robot& robot, int dir) {
// Keep mind that the robot always ends up facing up after every move
if (dir == 0) { // moved up - returning down
robot.turnLeft();
robot.turnLeft();
robot.move();
robot.turnLeft();
robot.turnLeft();
} else if (dir == 1) { // moved down - returning up
robot.move();
} else if (dir == 2) { // moved right - returning left
robot.turnLeft();
robot.move();
robot.turnRight();
} else { // moved left - returning right
robot.turnRight();
robot.move();
robot.turnLeft();
}
}
然后在递归清洁()函数中调用这个函数:
void clean(int row, int col, Robot& robot) {
cleaned.insert({row, col});
robot.clean();
for (int dir : directions) {
if (!is_cleaned(row, col, dir)) {
turn(robot, dir);
if (robot.move()) {
turn_back(robot, dir);
int row_new = get_new_row(row, dir);
int col_new = get_new_col(col, dir);
clean(row_new, col_new, robot);
go_back(robot, dir); // Move the robot back while backtracking!
} else {
turn_back(robot, dir);
}
}
}
}