困于环中的机器人

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c++

topic: 1041. 困于环中的机器人 - 力扣（LeetCode）

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Inspect the topic first.

And it looks really familiar with another robot topic.

657. 机器人能否返回原点 - 力扣（LeetCode）https://leetcode.cn/problems/robot-return-to-origin/description/?envType=study-plan-v2&envId=programming-skills

You gays have found what are the differences? That is direction.

Here is the code without considering the direction.

class Solution {
public:
    bool isRobotBounded(string instructions) {
        
        int x = 0;
        int y = 0;
 
        for (char c : instructions) 
        {
            if (c == 'U') y++;
            else if (c == 'D') y--;
            else if (c == 'R') x++;
            else if (c == 'L') x--;
        }
        
        return x == 0 && y == 0;
    }
};

To see if there is any spark from the basic code to solve this problem? In the topic, the robot need to change the direction.

• L --> turn left
• R --> turn right
• G --> go

So what is to describe the direction? Maybe the coordinate axis system, like (x, y)? coordinate axis system is necessary. And besides that, need to add direction.

direction = 0, 向 ↑	y++;
direction = 1, 向 →	x++;
direction = 2, 向 ↓	y--;
direction = 3, 向 ←	x--;

Good.

class Solution {
public:
    bool isRobotBounded(string instructions) {
        int x = 0, y = 0;
        int dir = 0; // 0: North, 1: East, 2: South, 3: West
        
        for (char c : instructions) 
        {
            if (c == 'G') 
            {
                if (dir == 0) 
                {
                    y++; // Move North
                } 
                else if (dir == 1) 
                {
                    x++; // Move East
                } 
                else if (dir == 2) 
                {
                    y--; // Move South
                } 
                else if (dir == 3) 
                {
                    x--; // Move West
                }
            } 
            else if (c == 'L') 
            {
                if (dir == 0) 
                {
                    dir = 3; // North -> West
                } 
                else if (dir == 1) 
                {
                    dir = 0; // East -> North
                } 
                else if (dir == 2) 
                {
                    dir = 1; // South -> East
                } else if (dir == 3) 
                {
                    dir = 2; // West -> South
                }
            } 
            else if (c == 'R') 
            {
                if (dir == 0) 
                {
                    dir = 1; // North -> East
                } 
                else if (dir == 1) 
                {
                    dir = 2; // East -> South
                } 
                else if (dir == 2) 
                {
                    dir = 3; // South -> West
                } else if (dir == 3) 
                {
                    dir = 0; // West -> North
                }
            }
        }
        
        // Robot is in a circle if:
        // 1. It returns to (0, 0), or
        // 2. It's not facing North (dir != 0)
        return (x == 0 && y == 0) || (dir != 0);
    }
};

too many if.

The original position is dir = 0; (0, 0);

if turn left, dir = 3; (0, 0);

if continue turn left, dir = 2; (0, 0);

if continue turn left, dir = 1; (0, 0);

So you can see the pired of turns around is 4 steps. %4 maybe works.

L -> turn left -> dir - 1. but if dir == 0 and dir - 1 is negative, that unsatisfied. so add a period:

L -> turn left -> (dir - 1 + 4) % 4;

R -> turn right -> (dir + 1) % 4;

• Let's take an example.
• the position is (0, 0) and direction is 2(↓）：
• if L：turn left, should be 1(↑）， dir = ( 2 - 1 + 4) % 4 = 1;
• if R: turn right, should be 3(←）， dir = ( 2 + 1) % 4 = 3;

so lets change a little code.

class Solution {
public:
    bool isRobotBounded(string instructions) {
        
        int x = 0;
        int y = 0;
        int dir = 0; // 0: North, 1: East, 2: South, 3: West

        for (char c: instructions)
        {
            if (c == 'G')
            {
                if (dir == 0)
                {
                    y++;
                }
                else if (dir == 1)
                {
                    x++;
                }
                else if (dir == 2)
                {
                    y--;
                }
                else if (dir == 3)
                {
                    x--;
                }
            }

            else if (c == 'L')
            {
                dir = (dir - 1 + 4) % 4;
            }
            else if (c == 'R')
            {
                dir = (dir + 1) % 4;
            }
        }

        return (x == 0 && y == 0) || (dir != 0);
    }
};

and the  code here looks unreadable. So introoduce switch algorhythm in c++.

switch (expression) {
    case value1:
        // 当 expression == value1 时执行的代码
        break;
    case value2:
        // 当 expression == value2 时执行的代码
        break;
    // 可以有任意数量的 case
    default:
        // 当 expression 不匹配任何 case 时执行的代码
}

for a little example:

#include <iostream>
using namespace std;

int main() {
    int choice = 2;

    switch (choice) {
        case 1:
            cout << "选择 1" << endl;
            break;
        case 2:
            cout << "选择 2" << endl;
            break;
        case 3:
            cout << "选择 3" << endl;
            break;
        default:
            cout << "无效选择" << endl;
    }

    return 0;
}

//输出2

Then use switch in the program.

class Solution {
public:
    bool isRobotBounded(string instructions) {

        int x = 0;
        int y = 0;
        int dir = 0; // 0: North, 1: East, 2: South, 3: West

        for (char c : instructions) 
        {
            if (c == 'G') 
            {
                switch (dir) 
                {
                case 0:
                    y++;
                    break;
                case 1:
                    x++;
                    break;
                case 2:
                    y--;
                    break;
                case 3:
                    x--;
                    break;
                }
            }

            else if (c == 'L') 
            {
                dir = (dir - 1 + 4) % 4;
            } 
            else if (c == 'R') 
            {
                dir = (dir + 1) % 4;
            }
        }

        return (x == 0 && y == 0) || (dir != 0);
    }
};