# Problem

Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.

Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000

For example, two is written as `II` in Roman numeral, just two one’s added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:

• `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
• `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
• `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

# Example

Example 1:

Input: 3
Output: “III”

Example 2:

Input: 4
Output: “IV”

Example 3:

Input: 9
Output: “IX”

Example 4:

Input: 58
Output: “LVIII”
Explanation: L = 50, V = 5, III = 3.

Example 5:

Input: 1994
Output: “MCMXCIV”
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

# Solution

``````impl Solution {
pub fn int_to_roman(num: i32) -> String {
let mut ret = String::from("");

let mut temp = num;
if temp >= 1000 {
let ret1 = temp/1000;
for i in 0..ret1 {
ret.push('M');
}
temp -= 1000*ret1;
}

if temp >= 900 {
ret.push('C');
ret.push('M');
temp = temp - 900;
}

if temp >= 500 {
ret.push('D');
temp -= 500;
}

if temp >= 400 {
ret.push('C');
ret.push('D');
temp -= 400;
}

if temp >= 100 {
let ret1 = temp/100;
for i in 0..ret1 {
ret.push('C');
}
temp -= ret1*100;
}

if temp >= 90 {
ret.push('X');
ret.push('C');
temp = temp - 90;
}

if temp >= 50 {
ret.push('L');
temp -= 50;
}

if temp >= 40 {
ret.push('X');
ret.push('L');
temp -= 40;
}

if temp >= 10 {
let ret1 = temp/10;
for i in 0..ret1 {
ret.push('X');
}
temp -= ret1*10;
}

if temp == 9 {
ret.push('I');
ret.push('X');
temp = 0;
}

if temp >= 5 {
ret.push('V');
temp -= 5;
}

if temp == 4 {
ret.push('I');
ret.push('V');
temp = 0;
}

if temp > 0 {
for i in 0..temp {
ret.push('I');
}
}

return ret;
}
}``````

2周前 评论

2周前 评论

169

36

219

81