leetcode Roman& Integer

关注

原创最新推荐文章于 2026-07-12 13:46:36 发布 249 阅读 AI 写同款· GEO 优化 ›

【HALCON工业视觉应用探索】02. C# + Halcon铸件毛刺机器人打磨引导：点云处理从实验室到工业现场的全流程落地摘要：铸件毛刺的自动化打磨是机械制造领域的重要难题，其随机性（高度0.1-3mm）和复杂工况导致人工打磨效率低（3分钟/件）、工伤率高。本文基于Halcon 24.11与.NET 6，构建了一套点云处理方案：通过线激光扫描仪获取三维点云，经预处理、毛刺检测、路径规划三大模块，引导机器人完成打磨。虚拟实验测试显示，对高度＞1mm的毛刺检出率达100%，定位误差±0.08mm，但工业现场受油污、粉尘影响，性能降至检出率68-75%、处理时间＞45秒。阅读详情

参考：http://blog.youkuaiyun.com/ljiabin/article/details/39968583

1、罗马字母与数值的对应关系,罗马数字的范围0~3999

int toNumber(char ch) {
        switch (ch) {
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            case 'M': return 1000;
        }
        return 0;
    }

2、CDL ==>550

转化方式：从右往左，如果左值比右值小，则减；如果左值比右值大，则加

public int romanToInt(String s) {
        char[] array=s.toCharArray();
        int result=toNumber(array[array.length-1]);
        for(int i=array.length-2;i>=0;i--){
            int t2=toNumber(array[i+1]);
            int t1=toNumber(array[i]);
            if(t1>=t2){
                result+=t1;
            }
            else{
                result-=t1;
            }
        }
        return result;
    }

3、转化方式：数值每个位置的数都表示一个罗马数字，如69（LXIX），其中LX和IX分别表示60和9.

roman的数组，表示{1,2,3，。。。10,20,30.。。。100,200,300.。。}

    public String intToRoman(int num) {
        String result="";
        String[][] roman={
            {"","I","II","III","IV","V","VI","VII","VIII","IX"},
            {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
            {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
            {"","M","MM","MMM"}
        };
        int col=0;
        while(num!=0){
            int remain=num%10;
            result=roman[col][remain]+result;
            col++;
            num=num/10;
        }
        return result;
    }

还有一种暴力方式：

class Solution {  
public:  
    string intToRoman(int num) {  
        string str;    
        string symbol[]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};    
        int value[]=    {1000,900,500,400, 100, 90,  50, 40,  10, 9,   5,  4,   1};   
        for(int i=0;num!=0;++i)  
        {  
            while(num>=value[i])  
            {  
                num-=value[i];  
                str+=symbol[i];  
            }  
        }  
        return str;  
    }  
};

收录于