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困于环中的机器人

news 来源：原创 2025/7/1 10:19:09

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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: Westfor (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: Westfor (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 casedefault:// 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: Westfor (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);}
};

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