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//! # Description
//!
//! Roman numerals are represented by seven different symbols: `I`, `V`, `X`,
//! `L`, `C`, `D` and `M`.
//!
//! ```plain
//! Symbol Value
//! I 1
//! V 5
//! X 10
//! L 50
//! C 100
//! D 500
//! M 1000
//! ```
//!
//! For example, `2` is written as `II` in Roman numeral, just two ones added
//! together. `12` is written as `XII`, which is simply `X` + `II`. The number
//! `27` 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 a roman numeral, convert it to an integer.
//!
//!
//!
//! Example 1:
//!
//! ```plain
//! Input: s = "III"
//! Output: 3
//! Explanation: III = 3.
//! ```
//!
//! Example 2:
//!
//! ```plain
//! Input: s = "LVIII"
//! Output: 58
//! Explanation: L = 50, V= 5, III = 3.
//! ```
//!
//! Example 3:
//!
//! ```plain
//! Input: s = "MCMXCIV"
//! Output: 1994
//! Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
//! ```
//!
//! Constraints:
//!
//! - `1 $\leqslant$ s.length $\leqslant$ 15`
//! - `s` contains only the characters ('`I`', '`V`', '`X`', '`L`', '`C`',
//! '`D`', '`M`').
//! - It is guaranteed that `s` is a valid roman numeral in the range `[1, 3999]`.
//!
//! Sources: <https://leetcode.com/problems/roman-to-integer/description/>
////////////////////////////////////////////////////////////////////////////////
/// Integer to Roman
///
/// # Arguments
/// * `s` - input string
pub fn roman_to_int(s: String) -> i32 {
let mut val = 0;
let mut last_val = 0;
let mut temp_digit = 0;
for ch in s.as_bytes().iter() {
let this_val = match *ch {
b'I' => 1,
b'V' => 5,
b'X' => 10,
b'L' => 50,
b'C' => 100,
b'D' => 500,
b'M' => 1000,
_ => 0,
};
if last_val == 0 {
temp_digit = this_val;
} else if this_val == last_val {
temp_digit += this_val;
} else if this_val < last_val {
val += temp_digit;
temp_digit = this_val;
} else {
val += this_val - temp_digit;
temp_digit = 0;
}
last_val = this_val;
}
val += temp_digit;
val
}