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廢話不多說開始今天的解題Day~
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

For example, 2 is written as II in Roman numeral, just two one's 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.
Input: s = "III"
Output: 3
Input: s = "IV"
Output: 4
Input: s = "IX"
Output: 9
Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').s is a valid roman numeral in the range [1, 3999].首先先簡單的翻譯一下題目
給你羅馬數字要轉回成十進位的數字表示,然後I可以放在V跟X的前面,X可以放在L跟C的前面,而C可以放在D跟M的前面。
像是:IV,V代表的是5,I代表的是1,IV則代表的是4 (5-1),以此類推。
作法大致上是這樣
sum。class Solution:
    def romanToInt(self, s: str) -> int:
        Symbol_dict = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        
        sum = 0
        list_s = list(s)
        
        for index in range(len(list_s)-1, -1, -1):
            if index == (len(list_s)-1):
                sum += Symbol_dict[list_s[index]]
            else:
                if Symbol_dict[list_s[index+1]] > Symbol_dict[list_s[index]]:
                    sum -= Symbol_dict[list_s[index]]
                else:
                    sum += Symbol_dict[list_s[index]]
        return sum
int romanToInt(char * s){
    int Symbol_dict[] = {1, 5, 10, 50, 100, 500, 1000};
    int sum = 0;
    int Symbol_dict_index1 = -1, Symbol_dict_index2 = -1;
    
    for(int index=strlen(s)-1; index>=0 ; index--) {
        printf("%c\n", s[index]);
        
        if (s[index] == 'I'){
            Symbol_dict_index1 = 0;
        } else if (s[index] == 'V'){
            Symbol_dict_index1 = 1;
        } else if (s[index] == 'X'){
            Symbol_dict_index1 = 2;
        } else if (s[index] == 'L'){
            Symbol_dict_index1 = 3;
        } else if (s[index] == 'C'){
            Symbol_dict_index1 = 4;
        } else if (s[index] == 'D'){
            Symbol_dict_index1 = 5;
        } else if (s[index] == 'M'){
            Symbol_dict_index1 = 6;
        }
        if (index == (strlen(s)-1)){
            sum += Symbol_dict[Symbol_dict_index1];
        } else {
            if (s[index+1] == 'I'){
                Symbol_dict_index2 = 0;
            } else if (s[index+1] == 'V'){
                Symbol_dict_index2 = 1;
            } else if (s[index+1] == 'X'){
                Symbol_dict_index2 = 2;
            } else if (s[index+1] == 'L'){
                Symbol_dict_index2 = 3;
            } else if (s[index+1] == 'C'){
                Symbol_dict_index2 = 4;
            } else if (s[index+1] == 'D'){
                Symbol_dict_index2 = 5;
            } else if (s[index+1] == 'M'){
                Symbol_dict_index2 = 6;
            }
            
            if (Symbol_dict[Symbol_dict_index2] > Symbol_dict[Symbol_dict_index1]){
                sum -= Symbol_dict[Symbol_dict_index1];
            } else {
                sum += Symbol_dict[Symbol_dict_index1];
            }
        }
   
    }
    return sum;
}
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大家明天見
感謝大大分享,我的做法是把羅馬數字6個減法實例直接寫成dictionary參照,效能跑分還可以,也分享給邦友另一種作法
https://www.youtube.com/watch?v=qC2tAhe133Y