在 10 X 10 网格中移动机器人

Question

我正在应对这个代码挑战：

Given a 2D bot/robot which can only move in four directions, move forward which is UP(U), move backward which is DOWN(D), LEFT(L), RIGHT(R) in a 10x10 grid. The robot can't go beyond the 10x10 area.

Given a string consisting of instructions to move.

Output the coordinates of a robot after executing the instructions. Initial position of robot is at origin(0, 0).

Example:
Input : move = “UDDLRL” 
Output : (-1, -1)
Explanation:
Move U : (0, 0)–(0, 1)
Move D : (0, 1)–(0, 0)
Move D : (0, 0)–(0, -1)
Move L : (0, -1)–(-1, -1)
Move R : (-1, -1)–(0, -1)
Move L : (0, -1)–(-1, -1)
Therefore final position after the complete
movement is: (-1, -1)

我在不使用 10x10 网格信息的情况下使代码正常工作。如何以 OOP 方式将 10x10 网格信息合并到我的解决方案中？我的解决方案不遵循 OOP 原则。

# function to find final position of
# robot after the complete movement

def finalPosition(move): 
    l = len(move)
    countUp, countDown = 0, 0
    countLeft, countRight = 0, 0
 
    # traverse the instruction string 'move'
    for i in range(l): 
        # for each movement increment its respective counter
        if (move[i] == 'U'):
            countUp += 1
        elif(move[i] == 'D'):
            countDown += 1
        elif(move[i] == 'L'):
            countLeft += 1
        elif(move[i] == 'R'):
            countRight += 1
 
    # required final position of robot
    print("Final Position: (", (countRight - countLeft),
          ", ", (countUp - countDown), ")")
 
 
# Driver code
if __name__ == '__main__':
    move = "UDDLLRUUUDUURUDDUULLDRRRR"
    finalPosition(move)

Answer 1

您使用计数器的方式使得检测到您将击中 10x10 网格的边界变得不那么简单。在不改变太多的情况下，您可以将 countUp 和 countDown 变量替换为一个 countVertical 变量，并在上升时向其添加 -1，在下降时向其添加 1。然后忽略一个移动，如果它会使计数器为负或大于 9。显然你会对水平移动做同样的事情。

[编辑： 对您的问题进行编辑后，发现您希望 Y 坐标与我上面假设的相反。所以我改变了Y坐标更新的符号（+1，-1）。]

原来如此。

现在为了使它更面向对象，您可以定义一个 Robot class，它将保持其 x 和 y 坐标。无论如何，最好从您的函数中删除 print 调用，这样该函数只处理移动，而不处理报告（关注点分离）。

这是它的工作原理：

class Robot:
    def __init__(self, x=0, y=0):
        self.position(x, y)
 
    def position(self, x, y):
        self.x = min(9, max(0, x))
        self.y = min(9, max(0, y))

    def move(self, moves):
        for move in moves:
            if move == 'U': 
                self.position(self.x, self.y + 1)
            elif move == 'D':
                self.position(self.x, self.y - 1)
            elif move == 'L':
                self.position(self.x - 1, self.y)
            elif move == 'R':
                self.position(self.x + 1, self.y)
            else:
                raise ValueError(f"Invalid direction '{move}'")
  
if __name__ == '__main__':
    moves = "UDDLLRUUUDUURUDDUULLDRRRR"
    robot = Robot(0, 0)
    robot.move(moves)
    print(f"Final position: {robot.x}, {robot.y}")

Answer 2

这修复了它：

class Robot:

    class Mover:
        def __init__(self, x, y):
            self.x, self.y = x, y

        def new_pos(self, x, y):
            new_x = x + self.x
            new_y = y + self.y
            if (new_x > 9 or new_y > 9):
                raise ValueError("Box dimensions are greater than 10 X 10")
            return new_x, new_y

    WALKS = dict(U=Mover(0, 1), D=Mover(0, -1),
                 L=Mover(-1, 0), R=Mover(1, 0))

    def move(self, moves):
        x = y = 0
        for id in moves:
            x, y = self.WALKS[id].new_pos(x, y)
        return (x,y)

if __name__ == '__main__':
    moves2 = "UDDLLRUUUDUURUDDUULLDRRRR"
    robot = Robot()
    print(robot.move(moves2))

输出：

(2,3)

在 10 X 10 网格中移动机器人

Move Robot in a 10 X 10 grid

oop

robot

python-3.x