字节跳动2-1 算法二轮面试202203-29_字节研发2-1

罗马数字包含以下七种字符： I， V， X， L，C，D 和 M

I             1
V             5
X             10
L             50
C             100
D             500
M             1000

这道题对应的是leetcode 中的12.整数转罗马数字

package example;
 
public class AlaboJiSuan {
 
 
    //组合数据左减右加
    String[] thousands = {"", "M", "MM", "MMM"};//1000,2000,3000
    String[] hundreds  = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}; //100,200,300,400,500,600,700,800.900
    String[] tens      = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}; //10,20,30,40,50,60,70,80,90
    String[] ones      = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};//123456789
 
    public static void main(String[] args) {
 
        int remark = 11;
 
 
 
 
        AlaboJiSuan alaboJiSuan = new AlaboJiSuan();
 
        String numberStr = alaboJiSuan.getAlboNumberStr(remark);
 
        System.out.println(numberStr);
 
        System.out.println(alaboJiSuan.getAlboNumberStr(3999));
    }
 
 
    //  11
    //  11/%10
    private  String getAlboNumberStr(int remark) {
        StringBuilder stringBuilder = new StringBuilder();
        stringBuilder.append(thousands[remark/1000]);
        stringBuilder.append(hundreds[remark%1000/100]);
        stringBuilder.append(tens[remark%100/10]);
        stringBuilder.append(ones[remark%10]);
        return stringBuilder.toString();
    }
}

解决问题的思路其实是转换 0-9 转换成罗马数字的个十百千位数据。基本方法还是获取到每一位的数字。