README_EN.md

1041. Robot Bounded In Circle

中文文档

Description

On an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:

• "G": go straight 1 unit;
• "L": turn 90 degrees to the left;
• "R": turn 90 degrees to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

 

Example 1:

Input: instructions = "GGLLGG"
Output: true
Explanation: The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).
When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.

Example 2:

Input: instructions = "GG"
Output: false
Explanation: The robot moves north indefinitely.

Example 3:

Input: instructions = "GL"
Output: true
Explanation: The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...

 

Constraints:

• 1 <= instructions.length <= 100
• instructions[i] is 'G', 'L' or, 'R'.

Solutions

Python3

class Solution:
    def isRobotBounded(self, instructions: str) -> bool:
        cur, direction = 0, [0] * 4
        for ins in instructions:
            if ins == 'L':
                cur = (cur + 1) % 4
            elif ins == 'R':
                cur = (cur + 3) % 4
            else:
                direction[cur] += 1
        return cur != 0 or (direction[0] == direction[2] and direction[1] == direction[3])

Java

class Solution {
    public boolean isRobotBounded(String instructions) {
        int[] direction = new int[4];
        int cur = 0;
        for (char c : instructions.toCharArray()) {
            if (c == 'L') {
                cur = (cur + 1) % 4;
            } else if (c == 'R') {
                cur = (cur + 3) % 4;
            } else {
                ++direction[cur];
            }
        }
        return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3]);
    }
}

C++

class Solution {
public:
    bool isRobotBounded(string instructions) {
        vector<int> direction(4);
        int cur = 0;
        for (char c : instructions)
        {
            if (c == 'L') cur = (cur + 1) % 4;
            else if (c == 'R') cur = (cur + 3) % 4;
            else ++direction[cur];
        }
        return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3]);
    }
};

Go

func isRobotBounded(instructions string) bool {
	direction := make([]int, 4)
	cur := 0
	for _, ins := range instructions {
		if ins == 'L' {
			cur = (cur + 1) % 4
		} else if ins == 'R' {
			cur = (cur + 3) % 4
		} else {
			direction[cur]++
		}
	}
	return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3])
}

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