机器人能否回到原点 - 简单

ElseWhereR2025-03-28 23:55

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C++

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inspect the topic very first.

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It is letters to decide which side the robot moves. And my thought is quite sample. Assumeing the robot can move back to the origin, then the numbers of L have to be the same with the numbers of R. So are U and D.

Then the code can be easy to write. Travel through the string and count the letters.

cpp 复制代码

class Solution {
public:
    bool judgeCircle(string moves) {
        
        int n = moves.size(); // to figure out the size at first
        int countR = 0;
        int countL = 0;
        int countU = 0;
        int countD = 0;

        for (int i = 0; i < n; i++)
        {
            if (moves[i] == 'R')
            {
                countR = countR + 1;
            }
            else if (moves[i] == 'L')
            {
                countL = countL + 1;
            }
            else if (moves[i] == 'U')
            {
                countU = countU + 1;
            }
            else
            {
                countD++;
            }
        }

        if (countR == countL && countD == countU)
        {
            return true;
        }
        else
        {
            return false;
        }
    }
};

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However, I donot think it is a elegent code. If you gays finish the Nine-year compulsory education，you will know coordinate system.

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↑ y = y + 1;

↓ y = y - 1；

← x = x - 1；

→ x = x + 1；

cpp 复制代码

class Solution {
public:
    bool judgeCircle(string moves) {
        int x = 0;
        int y = 0;

        for (char c : moves) 
        {
            if (c == 'U') y++;
            else if (c == 'D') y--;
            else if (c == 'R') x++;
            else if (c == 'L') x--;
        }
        
        return x == 0 && y == 0;
    }
};

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It is do elegent.

Maybe apple would change the design of ios in the near future. I donot care whether the ios system is fluently or not, I just want to see what desigh looks.

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