Int to Roman

来源：互联网发布：工程造价有什么软件编辑：程序博客网时间：2024/05/07 08:55
阿拉伯数字转罗马数字（1~3999）

思路：

直接从个位数开始转起，分别用10,100,1000,10000取余数，不同的余数采用不同的解决办法，最后返回字符串

代码

class Solution {public:    string intToRoman(int num){    int d = 10;    int n = num;    string rslt("");    while (n > 0)    {        //给10取余数        int single = n % 10;        switch (single)        {        case 0:            break;        case 4:            rslt += "IV" ; break;        case 5:            rslt += "V" ; break;            case 1: case 2:case 3:        {            int tmp = single;            while (tmp > 0)            {                rslt += "I";                    --tmp;            }            break;              }        case 6:case 7: case 8:        {            int tmp = single;            rslt += "V";            while (tmp > 5)            {                rslt += "I";                --tmp;            }            break;        }        case 9:            rslt += "IX"; break;        }        //减去        n = n - single;        if (n == 0)            return rslt;        //给100取余数        int decade = n % 100;        switch (decade)        {        case 40:            rslt = "XL"+rslt; break;        case 50:            rslt = "L" + rslt; break;        case 0:            break;        case 10: case 20:case 30:        {            int tmp = decade;            while (tmp >= 10)            {                rslt = "X" + rslt;                    tmp -= 10;            }            break;        }        case 60:case 70: case 80:        {            int tmp = decade;            string tmpRslt = "L";            while (tmp > 50)            {                tmpRslt += "X";                tmp -= 10;            }            rslt = tmpRslt + rslt;            break;        }        case 90:        {            rslt = "XC" + rslt;            break;        }        }        //减去        n = n - decade;        if (n == 0)            return rslt;        //给1000取余数        int hunderd = n % 1000;        switch (hunderd)        {        case 0:            break;        case 400:            rslt = "CD" + rslt; break;        case 500:            rslt = "D" +rslt; break;        case 100: case 200:case 300:        {            int tmp = hunderd;            while (tmp >= 100)            {                rslt = "C" + rslt;                    tmp -= 100;            }            break;        }        case 600:case 700: case 800:        {            int tmp = hunderd;            string tmpRslt = "D";            while (tmp > 500)            {                tmpRslt += "C";                tmp-=100;            }            rslt = tmpRslt + rslt;            break;        }        case 900:         {            rslt = "CM" + rslt;            break;        }        }        //减去        n = n - hunderd;        if (n == 0)            return rslt;        //给10000取余数        int thousand = n % 10000;        switch (thousand)        {        case 1000:            rslt = "M" + rslt; break;        case 2000:            rslt = "MM" + rslt; break;        case 3000:rslt = "MMM" + rslt; break;        }        n = n - thousand;    }    return rslt;}};
0 0