Problem

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

跟Roman To Integer是對應問題。即將整數轉換成對應的羅馬數字。羅馬數字規則見wiki：羅馬數字規則

Java 實現（個人解法）

package com.coderli.leetcode.algorithms.medium;

/**
 * Given an integer, convert it to a roman numeral.
 * <p>
 * Input is guaranteed to be within the range from 1 to 3999
 *
 * @author li.hzh
 */
 

public class IntegerToRoman {

    public static void main(String[] args) {
        IntegerToRoman integerToRoman = new IntegerToRoman();
        System.out.println(integerToRoman.intToRoman(1));
        System.out.println(integerToRoman.intToRoman(10));
        System.out.println(integerToRoman.intToRoman
 
(90));
        System.out.println(integerToRoman.intToRoman(1437));
    }

    public String intToRoman(int num) {
        int times = 1;
        String result = "";
        while (num > 0) {
            int tempIntValue = num % 10;
            num /= 10;
            String tempStrValue = "";
            if
 
 (tempIntValue == 5) {
                switch (times) {
                    case 1:
                        result = tempStrValue = "V";
                        break;
                    case 2:
                        result = "L" + result;
                        break;
                    case 3:
                        result = "D" + result;
                        break;
                }
            } else if (tempIntValue % 5 <= 3) {
                int length = tempIntValue;
                switch (times) {
                    case 1:
                        if (tempIntValue > 5) {
                            tempStrValue = "V";
                            length = tempIntValue - 5;
                        }
                        for (int i = 0; i < length; i++) {
                            tempStrValue += "I";
                        }
                        result = tempStrValue + result;
                        break;
                    case 2:
                        if (tempIntValue > 5) {
                            tempStrValue = "L";
                            length = tempIntValue - 5;
                        }
                        for (int i = 0; i < length; i++) {
                            tempStrValue += "X";
                        }
                        result = tempStrValue + result;
                        break;
                    case 3:
                        if (tempIntValue > 5) {
                            tempStrValue = "D";
                            length = tempIntValue - 5;
                        }
                        for (int i = 0; i < length; i++) {
                            tempStrValue += "C";
                        }
                        result = tempStrValue + result;
                        break;
                    case 4:
                        for (int i = 0; i < tempIntValue; i++) {
                            tempStrValue += "M";
                        }
                        result = tempStrValue + result;
                        break;
                }
            } else {
                switch (times) {
                    case 1:
                        result = tempIntValue < 5 ? "IV" + result : "IX" + result;
                        break;
                    case 2:
                        result = tempIntValue < 5 ? "XL" + result : "XC" + result;
                        break;
                    case 3:
                        result = tempIntValue < 5 ? "CD" + result : "CM" + result;
                        break;
                }
            }
            times++;
        }
        return result;
    }

}

分析

上面解法其實受到了羅馬轉整數問題的慣性思維。把思維限制到了羅馬數字只能用基礎的I、V、X等元素上。羅馬數字的規則不再介紹了。思路其實跟羅馬轉數字一致，抓住羅馬數字其實也是個位、十位、百位等一位一位組合起來的，因此我們一位一位的生成對應的羅馬數字即可。

正因為羅馬數字也是一位一位對應的，因此網上流傳一個簡便的寫法，程式碼很簡潔：

public static String change(int num) {
		String [][]str = 
			{
		            {"","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 roman = (str[3][num / 1000 % 10])+(str[2][num / 100 % 10])+(str[1][num / 10 % 10])+(str[0][num % 10]);
		
		return roman;

把個十百千各位可能羅馬字元準備好，然後直接計算整數對應位上的值，取出陣列中對應索引位的羅馬值拼到一起就可以了。