字节跳动2-1 算法二轮面试202203-29

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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"};//1000,2000,3000String[] hundreds  = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}; //100,200,300,400,500,600,700,800.900String[] tens      = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}; //10,20,30,40,50,60,70,80,90String[] ones      = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};//123456789public 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/%10private  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。

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