题目

Given an integer, convert it to a roman numeral.

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

分析

首先查了下罗马数字的书写规则
https://www.douban.com/note/335254352/
发现可以从大到小依次书写，比较简单，不多说直接上代码。

实现

class Solution {
public:
    string intToRoman(int num) {
        string ans;
        while(num>=900){
            if(num>=1000){
                num -= 1000;
                ans += "M";
            }
            else{
                num -= 900;
                ans += "CM";
            }
        }
        while(num>=400){
            if(num>=500){
                num -= 500;
                ans += "D";
            }
            else{
                num -= 400;
                ans += "CD";
            }
        }
        while(num>=90){
            if(num>=100){
                num -= 100;
                ans += "C";
            }
            else{
                num -= 90;
                ans += "XC";
            }
        }
        while(num>=40){
            if(num>=50){
                num -= 50;
                ans += "L";
            }
            else{
                num -= 40;
                ans += "XL";
            }
        }
        while(num>=9){
            if(num>=10){
                num -= 10;
                ans += "X";
            }
            else{
                num -= 9;
                ans += "IX";
            }
        }
        while(num>=4){
            if(num>=5){
                num -= 5;
                ans += "V";
            }
            else{
                num -= 4;
                ans += "IV";
            }
        }
        while(num>0){
            num--;
            ans += "I";
        }
        return ans;
    }
};

思考

这里我比较懒，就直接用了若干循环完成了。
其实，还可以将这些循环两两分为一组，放入一个大循环内。
或者直接将两个字母的组合例如“IX”等变为新的一个单元，这样更简单。
另外更讨巧的方法可以是把个十百千位的各种情况都列出来，这样可以直接输出，代码如下。

class Solution {
public:
    string intToRoman(int num) {
        char* digit[10] = {"","I","II","III","IV","V","VI","VII","VIII","IX"};
        char* ten[10] = {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
        char* hundreds[10] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        char* thousand[4] = {"","M","MM","MMM"};
        
        string answer = "";
        answer = answer + thousand[num/1000] + hundreds[(num/100)%10] + ten[(num/10)%10] + digit[(num)%10];
        return answer;
    }
};