java - 检查机器人的给定移动序列是否在 Java 中是圆形的

Question

我正在处理以下任务:

Given a sequence of moves for a robot, check if the sequence is circular or not. A sequence of moves is circular if first and last positions of robot are same. 
A move can be on of the following.

  G - Go one unit
  L - Turn left
  R - Turn right 

Examples:

Input: path[] = "GLGLGLG"
Output: Given sequence of moves is circular 

Input: path[] = "GLLG"
Output: Given sequence of moves is circular 

The movements described in the input string are repeated for an infinite time. Your task is to find if there exists a circle, whose radius is some positive real number R,
such that the robot never leaves it. If such a circle exists return "YES" otherwise "NO"

我找到了这个任务的解决方案:

public class Circle {


    String check(String commands) {

        int initialX = 0;
        int initialY = 0;

        int x = 0;
        int y = 0;
        String direction = "north";

        for (int i = 0; i < commands.length(); i++) {

            if (direction.equals("north")) {
                if (commands.charAt(i) == 'G') {
                    y++;
                } else if (commands.charAt(i) == 'L') {
                    direction = "west";
                } else if (commands.charAt(i) == 'R') {
                    direction = "east";
                } else {
                    System.out.println("Wrong command");
                }
            } else if (direction.equals("east")) {
                if (commands.charAt(i) == 'G') {
                    x++;
                } else if (commands.charAt(i) == 'L') {
                    direction = "north";
                } else if (commands.charAt(i) == 'R') {
                    direction = "south";
                } else {
                    System.out.println("Wrong command");
                }
            } else if (direction.equals("south")) {
                if (commands.charAt(i) == 'G') {
                    y--;
                } else if (commands.charAt(i) == 'L') {
                    direction = "east";
                } else if (commands.charAt(i) == 'R') {
                    direction = "west";
                } else {
                    System.out.println("Wrong command");
                }
            } else if (direction.equals("west")) {
                if (commands.charAt(i) == 'G') {
                    x--;
                } else if (commands.charAt(i) == 'L') {
                    direction = "south";
                } else if (commands.charAt(i) == 'R') {
                    direction = "north";
                } else {
                    System.out.println("Wrong command");
                }
            }
        }

        if (direction.equals("north") && (((x-initialX)*(x-initialX) + (y-initialY)*(y-initialY)) > 0)) {
            return "NO";
        } else {
            return "YES";
        }
    }
}

一切看起来都很完美，但我无法理解这种情况:

if (direction.equals("north") && (((x-initialX)*(x-initialX) + (y-initialY)*(y-initialY)) > 0))

你能帮我理解为什么我们需要在这种情况下返回 NO 吗，公式 && (((x-initialX)*(x-initialX) + (y -initialY)*(y-initialY)) > 0 表示？为什么条件只检查“北”方向而不检查其他方向。

Answer 1

最后一个条件可以缩短为:

if (direction.equals("north") && (x != 0 || y != 0))

这个条件意味着在给定的步骤序列之后，机器人有初始方向但没有初始位置。在这种情况下，此序列将机器人的位置移动 (x, y)。这意味着在 n 次重复此序列后，机器人将位于 (n*x, n*y) 位置。当 x != 0 || 时，它不受任何半径的约束y != 0。

当x 和y 都为零时，机器人在给定的步骤序列后具有初始方向和初始位置。这是一个循环，答案是"is"。

否则，机器人的方向在步骤序列后发生了变化。有两种可能:

1. 方向已更改为相反(南)。假设在一系列步骤之后，机器人移动到坐标 (x, y)。但是当我们在相反的方向重复这个序列时，坐标将被 (-x, -y) 改变，我们将返回到 (0, 0) 初始坐标方向。

2. 方向已更改为西 或东。在这种情况下，我们将在四个步骤之后以初始方向返回到初始位置。例如，假设在第一个序列之后方向更改为 east 并且位置为 (x, y):

   Direction  |  Coordinates
   -----------+--------------
   north      |  (0, 0)
   east       |  (x, y)
   south      |  (x+y, y-x)
   west       |  (y, -x)
   north      |  (0, 0)