机器人能否回到原点 - 简单

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C++

topic：657. 机器人能否返回原点 - 力扣（LeetCode）

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inspect the topic very first.

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.

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;
        }
    }
};

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

↑ y = y + 1;

↓ y = y - 1；

← x = x - 1；

→ x = x + 1；

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;
    }
};

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.