LeetCode(12)Integer to Roman

题目

Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.

分析

该题目要求将给定的1~3999之间的整型数字转换为罗马数字并输出。
解这道题我们必须了解罗马字母与整数之间的对应:
对应规则
对照举例如下:
对照举例

AC代码

class Solution {
public:
    string intToRoman(int num) {
        //存储罗马数字
        string str;
        if (num == 0)
            return "";

        //(1)首先处理最高位千位数字
        if (num >= 1000)
        {
            int count = num / 1000;
            for (int i = 0; i < count; i++)
                str += RomanLeter(1000);
           //得到百位数
            num %= 1000;
           //链接其余三位数字对应的罗马序列
            str += intToRoman(num);
        }//else if
        else if (num >= 100)
        {
            if (num >= 900)
            {
                str = str + RomanLeter(100) + RomanLeter(1000);
                num %= 100;
            }//if               
            else if (num >= 500)
            {
                str += RomanLeter(500);
                num -= 500;

            }//else if
            else if (num >= 400){
                str = str + RomanLeter(100) + RomanLeter(500);
                num -= 400;
            }
            else{
                while (num >= 100)
                {
                    str += RomanLeter(100);
                    num -= 100;
                }//while
            }
            str += intToRoman(num);
        }//else if
        else if (num >= 10)
        {
            if (num >= 90)
            {
                str = str + RomanLeter(10) + RomanLeter(100);
                num %= 10;
            }//if
            else if (num >= 50)
            {
                str += RomanLeter(50);
                num -= 50;              
            }
            else if (num >= 40){
                str = str + RomanLeter(10) + RomanLeter(50);
                num -= 40;
            }
            else{
                while (num >= 10)
                {
                    str += RomanLeter(10);
                    num -= 10;
                }
            }
            str += intToRoman(num);
        }
        else if (num >= 1)
        {
            if (num == 9)
            {
                str = str + RomanLeter(1) + RomanLeter(10);
                num /= 10;
            }
            else if (num >= 5)
            {
                str += RomanLeter(5);
                num -= 5;               
            }
            else if (num >= 4){
                str = str + RomanLeter(1) + RomanLeter(5);
                num -= 4;
            }
            else{
                while (num >= 1)
                {
                    str += RomanLeter(1);
                    num -= 1;
                }
            }
            str += intToRoman(num);
        }
        else
            str += "";
        return str;
    }

    string RomanLeter(int n)
    {
        switch (n)
        {
        case 1:
            return "I"; break;
        case 5:
            return "V"; break;
        case 10:
            return "X"; break;
        case 50:
            return "L"; break;
        case 100:
            return "C"; break;
        case 500:
            return "D"; break;
        case 1000:
            return "M"; break;
        default:
            return ""; break;
        }
    }
};

Git测试程序代码

原文地址:https://www.cnblogs.com/shine-yr/p/5214929.html