[Leetcode] Integer to Roman

第三次刷这个题了，这次一次A过！总结起来是利用了小于4000的数字的罗马数字表示上的相对独立性，我们可以只针对正数表示的个位、十位、百位、千位来分别设计计算的策略。

0： return “”

1-3：return相应数字个当前符号

4：返回当前符号+高一级的符号

5：返回高一级符号

6-8：返回高一级符号+相应数量的当前符号

9：当前符号+高两级的符号

上代码：

import java.util.*;

public class Solution {
    private String getUnit(int digit, String currentNotation, String largerNotation,String largerlargerNotation){
        String res = "";
        if(digit==0) return "";
        if(digit<=3){
            for(int i=0;i<digit;i++) res+=currentNotation;
        }else if(digit==4){
            res = currentNotation + largerNotation;
        }else if(digit<=8){
            res = largerNotation;
            digit = digit - 5;
            for(int i=0;i<digit;i++) res = res + currentNotation;
        }else{
            res = currentNotation + largerlargerNotation;
        }
        return res;
    }
    public String intToRoman(int num) {
        String res = "";
        int gewei = num %10;
        int shiwei = (num/10)%10;
        int baiwei = (num/10/10)%10;
        int qianwei = (num/10/10/10)%10;
        String qianString = getUnit(qianwei,"M","","");
        res = res + qianString;
        String baiString = getUnit(baiwei,"C","D","M");
        res = res + baiString;
        String shiString = getUnit(shiwei,"X","L","C");
        res = res + shiString;
        String geString = getUnit(gewei,"I","V","X");
        res = res + geString;
        return res;
    }
}