LeetCode 简单操作 | 12. Integer to Roman

rical730

于 2017-03-02 11:48:29 发布

阅读量230

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/rical730/article/details/59486523

LeetCode 专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一种将整数转换为罗马数字的算法实现，适用于1到3999范围内的整数转换。该算法使用了数组来映射不同位上的数值与对应的罗马数字符号，通过简单的数学运算完成转换。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
 * Leetcode12. Integer to Roman
 * Funtion: Given an integer, convert it to a roman numeral.Input is guaranteed to be within the range from 1 to 3999.
 * Author: LKJ
 * Date:2016/7/27
 * Hint: Related to12. Roman to Integer
*/
#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Solution {
public:
    string intToRoman(int num) {
        char* Roman[4][10] = {
            {"","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"}
        };
        string result;
        result.append(Roman[3][num/1000]);
        result.append(Roman[2][num%1000/100]);
        result.append(Roman[1][num%100/10]);
        result.append(Roman[0][num%10]);
        return result;
    }
};


int main(){
    int myin;
    string myout;
    Solution SA;
    cout << "Please Enter" << endl;
    cin >> myin;
    myout = SA.intToRoman(myin);
    cout << myout << endl;

    return 0;
}