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
该题目要求将给定的1~3999之间的整型数字转换为罗马数字并输出。
解这道题我们必须了解罗马字母与整数之间的对应：
对照举例如下：
AC代码
class Solution {
public:
string intToRoman(int num) {
//存储罗马数字
string str;
if (num == 0)
return "";
//（1）首先处理最高位千位数字
if (num >= 1000)
{
int count = num / 1000;
for (int i = 0; i < count; i++)
str += RomanLeter(1000);
//得到百位数
num %= 1000;
//链接其余三位数字对应的罗马序列
str += intToRoman(num);
}//else if
else if (num >= 100)
{
if (num >= 900)
{
str = str + RomanLeter(100) + RomanLeter(1000);
num %= 100;
}//if
else if (num >= 500)
{
str += RomanLeter(500);
num -= 500;
}//else if
else if (num >= 400){
str = str + RomanLeter(100) + RomanLeter(500);
num -= 400;
}
else{
while (num >= 100)
{
str += RomanLeter(100);
num -= 100;
}//while
}
str += intToRoman(num);
}//else if
else if (num >= 10)
{
if (num >= 90)
{
str = str + RomanLeter(10) + RomanLeter(100);
num %= 10;
}//if
else if (num >= 50)
{
str += RomanLeter(50);
num -= 50;
}
else if (num >= 40){
str = str + RomanLeter(10) + RomanLeter(50);
num -= 40;
}
else{
while (num >= 10)
{
str += RomanLeter(10);
num -= 10;
}
}
str += intToRoman(num);
}
else if (num >= 1)
{
if (num == 9)
{
str = str + RomanLeter(1) + RomanLeter(10);
num /= 10;
}
else if (num >= 5)
{
str += RomanLeter(5);
num -= 5;
}
else if (num >= 4){
str = str + RomanLeter(1) + RomanLeter(5);
num -= 4;
}
else{
while (num >= 1)
{
str += RomanLeter(1);
num -= 1;
}
}
str += intToRoman(num);
}
else
str += " ";
return str;
}
string RomanLeter(int n)
{
switch (n)
{
case 1:
return "I"; break;
case 5:
return "V"; break;
case 10:
return "X"; break;
case 50:
return "L"; break;
case 100:
return "C"; break;
case 500:
return "D"; break;
case 1000:
return "M"; break;
default:
return ""; break;
}
}
};