LeetCode第十二题-将数字转化为罗马数字

Integer to Roman

问题简介：将输入的int类型数字转化为罗马数字
问题详解：罗马数字由七个不同的符号表示：I，V，X，L，C，D和M
符号-数值
I - 1
V - 5
X -10
L - 50
C - 100
D - 500
M - 1000
例如,2用罗马数字写成II,只有两个I加在一起,十二写为XII,解释为X + II, 二十七写成XXVII,即XX + V + II,
罗马数字通常从左到右从最大到最小,但是,四个数字不是IIII,相反,第四个写为IV,因为一个在五个之前,我们减去四个,同样的原则适用于九,即IX,有六个使用减法的实例：
我可以放在V（5）和X（10）之前做4和9,
X可以放在L（50）和C（100）之前,以产生40和90,
C可以放在D（500）和M（1000）之前，以产生400和900,
给定一个整数,将其转换为罗马数字。,输入保证在1到3999的范围内.

举例：
1:
输入: 3
输出: “III”

2:
输入: 4
输出: “IV”

3:
输入: 9
输出: “IX”

4:
输入: 58
输出: “LVIII”
解释: L = 50, V = 5, III = 3.

5:
输入: 1994
输出: “MCMXCIV”
解释: M = 1000, CM = 900, XC = 90 and IV = 4.

解法一：
我这可能算穷举法。。。
因为输入1-3999的数字，也就是把每位的数字做处理

        // 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

class Solution {
    public String intToRoman(int num) {
        if(num < 0 ||num >3999)return "";
        String one = "";
        String two = "";
        String three = "";
        String four = "";
        int x = num%10;
        num = num/10;
        switch(x){
            case 1:
                one = new String("I");
                break;
            case 2:
                one = new String("II");
                break;
            case 3:
                one = new String("III");
                break;
            case 4:
                one = new String("IV");
                break;
            case 5:
                one = new String("V");
                break;
            case 6:
                one = new String("VI");
                break;
            case 7:
                one = new String("VII");
                break;
            case 8:
                one = new String("VIII");
                break;
            case 9:
                one = new String("IX");
                break;
        }
        x = num%10;
        num = num/10;
        switch(x){
            case 0:
                break;                
            case 1:
                two = new String("X");
                break;
            case 2:
                two = new String("XX");
                break;
            case 3:
                two = new String("XXX");
                break;
            case 4:
                two = new String("XL");
                break;
            case 5:
                two = new String("L");
                break;
            case 6:
                two = new String("LX");
                break;
            case 7:
                two = new String("LXX");
                break;
            case 8:
                two = new String("LXXX");
                break;
            case 9:
                two = new String("XC");
                break;
        }
        x = num%10;
        num = num/10;
        switch(x){
            case 0:
                break;                
            case 1:
                three = new String("C");
                break;
            case 2:
                three = new String("CC");
                break;
            case 3:
                three = new String("CCC");
                break;
            case 4:
                three = new String("CD");
                break;
            case 5:
                three = new String("D");
                break;
            case 6:
                three = new String("DC");
                break;
            case 7:
                three = new String("DCC");
                break;
            case 8:
                three = new String("DCCC");
                break;
            case 9:
                three = new String("CM");
                break;
        }
        x = num%10;
        num = num/10;
        switch(x){
            case 0:
                break;                
            case 1:
                four = new String("M");
                break;
            case 2:
                four = new String("MM");
                break;
            case 3:
                four = new String("MMM");
                break;

        }
        String result = four + three + two + one;
        return result;
    }
}

小白刷题之路,请多指教— — 要么大器晚成,要么石沉大海