机器人能否回到原点 - 简单

*************

C++

topic:657. 机器人能否返回原点 - 力扣(LeetCode)

*************

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.

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;
        }
    }
};

|----------------------------------------------------------------------------|
| |

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;

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;
    }
};

|----------------------------------------------------------------------------|
| |

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.

|----------------------------------------------------------------------------|
| |

相关推荐
机器人零零壹3 小时前
技术白皮书:iRobotCAM如何解决机器人激光切割应用中的核心难点,从CAD离线编程到双目视觉定位与轨迹自适应提升复杂空间曲线切割的精度、效率与柔性化水平
人工智能·算法
布莱克6054 小时前
C/C++ 中变量的声明与定义:区别、联系与常见误区
c语言·开发语言·c++·声明·定义
ZhengEnCi6 小时前
大模型的分类到底怎么分:从基础模型到推理模型、从通用到专用的完整图谱
算法
取名真是6 小时前
Python 算法知识点从基础到进阶
开发语言·python·算法
罗西的思考7 小时前
【轻量化复刻大模型 / KIMI K3】KDA 模块交互式演示
人工智能·算法·机器学习
沫璃染墨9 小时前
从零入门计算机网络系列(十二):HTTP长连接与会话管理详解——从Connection到Cookie与Session
linux·服务器·c++·tcp/ip·ubuntu·http·交互
nm_zfyjt9 小时前
信息流广告投放优化实践:线索成本评估模型与转化链路重构(家装建材案例)
算法·重构·内蒙古朋友圈广告服务商·内蒙古朋友圈获客
xiaoye-duck10 小时前
《C++ 项目:手写大模型接入 SDK》DeepSeek接口设计(下):流式响应实现和测试
开发语言·网络·c++·语言模型
w_zero_one10 小时前
算法--构造
java·数据结构·算法
其实防守也摸鱼10 小时前
信创安全竞赛知识总结与题库
大数据·网络·人工智能·算法·安全·php