How can you convert an integer into a Roman numeral efficiently?

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题目：将整数转换为罗马数字class Solution { public int[] v={1, 5, 10, 50, 100, 500, 1000}; public String[] s={I, V, X, L, C, D, M}; public String intToRoman(int num) { String ans=; int x=num; while (num > 0) { int pos=-1; for (int i=0; i =v[i]) { pos=i; break; } } ans +=s[pos]; num -=v[pos]; } return ans; }}

​​题目​​

class Solution {
public:
int v[7]={1,5,10,50,100,500,1000};
string s[7]={"I","V","X","L","C","D","M"};
string intToRoman(int num) {
string ans="";
int x=num;
while(num){


int pos=-1;
for(int i=0;i<7;i++)
{
if(num<v[i])
{
pos=i-1;
break;
}
}
if(pos==-1)
pos=6;

int y=fun2(num);
if(y!=-1)
{
ans+=fun(y);

num=num-y;
}else{
int x=num/v[pos];

for(int i=0;i<x;i++)
ans+=s[pos];

num=num%v[pos];
}

}
return ans;

}

int fun2(int x)
{
int x2=x;
int i=0;
while(x)
{
x/=10;
i++;
}
int y;
if(i==1)
y=x2;
else
y=(x2/(int)pow(10.0,(i-1)))*((int)pow(10.0,i-1));

if(i==1)
{
if(y==4)
return y;
if(y==9)
return y;
}
else
{
if(y==40)
return y;
if(y==90)
return y;
if(y==400)
return y;
if(y==900)
return y;
}
return -1;
}

string fun(int x)
{
if(x==4)
return "IV";
else if(x==9)
return "IX";
else if(x==40)
return "XL";
else if(x==90)
return "XC";
else if(x==400)
return "CD";
else if(x==900)
return "CM";

return "-1";
}
};