很傻的方法,穷举;
用4个int的数组,分别存放num的千、百、十、个位,然后针对每一位列出各种情况;
事实证明,占用内存和耗时 都很高;
class Solution {
public:
string intToRoman(int num)
{
int temp[4] = {0};
string res;
for (int i = 0; i < 4; i++)
{
temp[3-i] = num % 10;
num /= 10;
}
for (int i = 0; i < temp[0]; i++)
{
res += "M";
}
switch (temp[1])
{
case 0:break;
case 1:res += "C"; break;
case 2:res += "C"; res += "C"; break;
case 3:res += "C"; res += "C"; res += "C"; break;
case 4:res += "CD"; break;
case 5:res += "D"; break;
case 6:res += "D"; res += "C"; break;
case 7:res += "D"; res += "C"; res += "C"; break;
case 8:res += "D"; res += "C"; res += "C"; res += "C"; break;
case 9:res += "CM"; break;
}
switch (temp[2])
{
case 0:break;
case 1:res += "X"; break;
case 2:res += "X"; res += "X"; break;
case 3:res += "X"; res += "X"; res += "X"; break;
case 4:res += "XL"; break;
case 5:res += "L"; break;
case 6:res += "L"; res += "X"; break;
case 7:res += "L"; res += "X"; res += "X"; break;
case 8:res += "L"; res += "X"; res += "X"; res += "X"; break;
case 9:res += "XC"; break;
}
switch (temp[3])
{
case 0:break;
case 1:res += "I"; break;
case 2:res += "I"; res += "I"; break;
case 3:res += "I"; res += "I"; res += "I"; break;
case 4:res += "IV"; break;
case 5:res += "V"; break;
case 6:res += "V"; res += "I"; break;
case 7:res += "V"; res += "I"; res += "I"; break;
case 8:res += "V"; res += "I"; res += "I"; res += "I"; break;
case 9:res += "IX"; break;
}
return res;
}
};