【LeetCode】12. Integer to Roman (2 solutions)

Integer to Roman

Given an integer, convert it to a roman numeral.

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

逐区间做处理。

解法一：非递归

class Solution {
public:
string intToRoman(int num) {
string ret;
//M<-->1000
while(num >= 1000)
{
ret += 'M';
num -= 1000;
}
//to here, num < 1000
//CM<-->900
if(num >= 900)
{
ret += "CM";
num -= 900;
}
//to here, num < 900
//D<-->500
if(num >= 500)
{
ret += 'D';
num -= 500;
}
//to here, num < 500
if(num >= 400)
{
ret += "CD";
num -= 400;
}
//to here, num < 400
//C<-->100
while(num >= 100)
{
ret += 'C';
num -= 100;
}
//to here, num < 100
//XC<-->90
if(num >= 90)
{
ret += "XC";
num -= 90;
}
//to here, num < 90
//L<-->50
if(num >= 50)
{
ret += 'L';
num -= 50;
}
//to here, num < 50
//XL<-->40
if(num >= 40)
{
ret += "XL";
num -= 40;
}
//to here, num < 40
//X<-->10
while(num >= 10)
{
ret += 'X';
num -= 10;
}
//to here, num < 10
//IX<-->9
if(num >= 9)
{
ret += "IX";
num -= 9;
}
//to here, num < 9
//V<-->5
if(num >= 5)
{
ret += 'V';
num -= 5;
}
//to here, num < 5
//IV<-->4
if(num >= 4)
{
ret += "IV";
num -= 4;
}
//to here, num < 4
//I<-->1
while(num >= 1)
{
ret += 'I';
num -= 1;
}
return ret;
}
};

解法二：递归

class Solution {
public:
string intToRoman(int num) {
if(num >= 1000)
return "M" + intToRoman(num-1000);
else if(num >= 900)
return "CM" + intToRoman(num-900);
else if(num >= 500)
return "D" + intToRoman(num-500);
else if(num >= 400)
return "CD" + intToRoman(num-400);
else if(num >= 100)
return "C" + intToRoman(num-100);
else if(num >= 90)
return "XC" + intToRoman(num-90);
else if(num >= 50)
return "L" + intToRoman(num-50);
else if(num >= 40)
return "XL" + intToRoman(num-40);
else if(num >= 10)
return "X" + intToRoman(num-10);
else if(num >= 9)
return "IX" + intToRoman(num-9);
else if(num >= 5)
return "V" + intToRoman(num-5);
else if(num >= 4)
return "IV" + intToRoman(num-4);
else if(num >= 1)
return "I" + intToRoman(num-1);
else
return "";
}
};