# 13 Roman to Integer (有圖)

想法最重要，再來再想可以用那些 js method

LeetCode

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

Example 1:

Input: "III"
Output: 3
Example 2:

Input: "IV"
Output: 4
Example 3:

Input: "IX"
Output: 9
Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:

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

*/
// 先分類個位、十位、百位、千位 一個一個算
// 再全部加起來

/**
 * @param {string} s
 * @return {number}
 */
var romanToInt = function(s) {}

怎麼解

• 先分類個位、十位、百位、千位 一個一個算

• 從最右邊開始算(個位 → 十位 → 百位 → 千位)

  • 若碰到比自己大的代表進位了 i - 1 > i , ex VI 1 + 5 = 6

  • 若碰到比自己小的 i < i - 1，代表要減 ex. IV = 5 - 4

  • 其他: 代表再這個單位裡相加 ex III = 1 + 1 + 1 = 3

• 再慢慢加起來

var romanToInt = function(s) {
    let roman = {
        "I": 1,
        "V": 5,
        "X": 10,
        "L": 50,
        "C": 100,
        "D": 500,
        "M": 1000
    }
    // "MCMXCIV"
    let arr = s.split(''); // ["M", "C", "M", "X", "C", "I", "V"]
    let len = s.length - 1;
    let result = roman[arr[len]]
    for(let i = len; i>0; i--){
        if(roman[arr[i - 1]] > roman[arr[i]]){
            // 下一個位數
            result += roman[arr[i - 1]];
        }else{
            if(roman[arr[i - 1]] < roman[arr[i]]){
                // subtraction
                result = result - roman[arr[i - 1]]
            }else{
                result += roman[arr[i]];
            }
             
        }
    }
    return result;
};
// Runtime: 132 ms, faster than 91.68% of JavaScript online submissions

學道什麼?

這題完全就是想法，想法對了其實答案就出來了