LeetCode 题解（58）: 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.

题解：

罗马数字共有7个，即I（1）、V（5）、X（10）、L（50）、C（100）、D（500）和M（1000）。按照下述的规则可以表示任意正整数。需要注意的是罗马数字中没有“0”，与进位制无关。一般认为罗马数字只用来记数，而不作演算。

• 重复数次：一个罗马数字重复几次，就表示这个数的几倍。

• 右加左减：
  • 在较大的罗马数字的右边记上较小的罗马数字，表示大数字加小数字。
  • 在较大的罗马数字的左边记上较小的罗马数字，表示大数字减小数字。
  • 左减的数字有限制，仅限于I、X、C。比如45不可以写成VL，只能是XLV
  • 但是，左减时不可跨越一个位数。比如，99不可以用IC（100−1）表示，而是用XCIX（[100−10]+[10−1]）表示。（等同于阿拉伯数字每位数字分别表示。）
  • 左减数字必须为一位，比如8写成VIII，而非IIX。
  • 右加数字不可连续超过三位，比如14写成XIV，而非XIIII。（见下方“数码限制”一项。）

C++版：

class Solution {public:    string intToRoman(int num) {        string result = "";        for(int i = 0; i < num / 1000; i++)            result += "M";        num = num % 1000;                string low, mid, high;        if(num / 100) {            low = "C"; mid = "D"; high = "M";            if(num / 100 == 4) {                result += low;                result += mid;            } else if(num / 100 == 9) {                result += low;                result += high;            } else if(num / 100 < 4)                for(int k = 0; k < num / 100; k ++)                    result += low;            else {                result += mid;                for(int l = 5; l < num / 100; l++)                    result += low;            }            num %= 100;        }        if(num / 10) {            low = "X"; mid = "L"; high = "C";            if(num / 10 == 4) {                result += low;                result += mid;            } else if(num / 10 == 9) {                result += low;                result += high;            } else if(num / 10 < 4)                for(int k = 0; k < num / 10; k ++)                    result += low;            else {                result += mid;                for(int l = 5; l < num / 10; l++)                    result += low;            }            num %= 10;        }                low = "I";mid = "V"; high = "X";        if(num  == 4) {            result += low;            result += mid;        } else if(num  == 9) {            result += low;            result += high;        } else if(num  < 4)            for(int k = 0; k < num; k ++)                result += low;        else {            result += mid;            for(int l = 5; l < num; l++)                result += low;        }        return result;    }};

Java版：

public class Solution {    public String intToRoman(int num) {        String result = new String("");                for(int i = 0; i < num / 1000; i++)            result = result.concat("M");                num %= 1000;        String low, mid, high = new String();        if(num / 100 != 0) {            low = "C"; mid="D"; high="M";            if(num / 100 == 4)                 result = result.concat(low).concat(mid);            else if(num / 100 == 9)                result = result.concat(low).concat(high);            else if(num / 100 < 4)                for(int j = 0; j < num / 100; j++)                    result = result.concat(low);            else {                result = result.concat(mid);                for(int k = 5; k < num / 100; k++)                    result = result.concat(low);            }            num %= 100;        }                if(num / 10 != 0) {            low = "X"; mid="L"; high="C";            if(num / 10 == 4)                 result = result.concat(low).concat(mid);            else if(num / 10 == 9)                result = result.concat(low).concat(high);            else if(num / 10 < 4)                for(int j = 0; j < num / 10; j++)                    result = result.concat(low);            else {                result = result.concat(mid);                for(int k = 5; k < num / 10; k++)                    result = result.concat(low);            }            num %= 10;        }                low = "I"; mid="V"; high="X";        if(num == 4)             result = result.concat(low).concat(mid);        else if(num == 9)            result = result.concat(low).concat(high);        else if(num < 4)            for(int j = 0; j < num; j++)                result = result.concat(low);        else {            result = result.concat(mid);            for(int k = 5; k < num; k++)                result = result.concat(low);        }        return result;    }}

Python版：

class Solution:    # @return a string    def intToRoman(self, num):        result = ""        for i in range(0, num / 1000):            result += "M"                    num %= 1000                if num / 100 != 0:            if num / 100 == 4:                result += "C"                result += "D"            elif num / 100 == 9:                result += "C"                result += "M"            elif num / 100 < 4:                for i in range(num / 100):                    result += "C"            else:                result += "D"                for i in range(5, num / 100):                    result += "C"            num %= 100                   if num / 10 != 0:            if num / 10 == 4:                result += "X"                result += "L"            elif num / 10 == 9:                result += "X"                result += "C"            elif num / 10 < 4:                for i in range(num / 10):                    result += "X"            else:                result += "L"                for i in range(5, num / 10):                    result += "X"            num %= 10                     if num == 4:            result += "I"            result += "V"        elif num == 9:            result += "I"            result += "X"        elif num < 4:            for i in range(num):                result += "I"        else:            result += "V"            for i in range(5, num):                result += "I"                       return result