LeetCode 12："Integer to Roman"

Thinking

使用迭代的思想，簡化程序

Code

class Solution {
public:
    string intToRoman(int num) {
        // 初始化
        string roman = "";

        if (num == 1000)
        {
            roman = "M";
        }
        if (num == 100)
        {
            roman = "C";
        }
        if (num == 10)
        {
            roman = "X";
        }

        // 如果數字大於1000
        if (num > 1000)
        {
            int indexFourth = num / 1000;
            for (int i = 0; i < indexFourth; i++)
            {
                roman += "M";
            }
            roman += intToRoman(num-indexFourth*1000);
        }

        // 大於100小於100
        if(num > 100 && num < 1000)
        {
            roman = range(num, 100, "C", "D", "M");
        }
        // 大於10小於100
        if (num > 10 && num < 100)
        {
            roman = range(num, 10, "X", "L", "C");
        }
        // 大於0小於10
        if (num > 0 && num < 10)
        {
            roman = range(num, 1, "I", "V", "X");
        }

        return roman;
    }

    // 根據num值，以及區間、當前區間的羅馬標識符，表示當前數字
    string range(int num, int begin, string a, string b, string c)
    {
        string roman = "";

        int index = num / begin;
        switch (index)
        {
        case 1:
        case 2:
        case 3:
            for (int i = 0; i < index; i++)
            {
                roman += a;
            }
            break;
        case 4:
            roman += a;
            roman += b;
            break;
        case 5:
            roman += b;
            break;
        case 6:
        case 7:
        case 8:
            roman += b;
            for (int i = 0; i < index - 5; i++)
            {
                roman += a;
            }
            break;
        case 9:
            roman += a;
            roman += c;
            break;
        }
        roman += intToRoman(num-index*begin);

        return roman;
    }
};