Leetcode第13题：Roman to Integer

题目地址：Roman to Integer

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题目简介：将特定的罗马字符串转换为整数，这里还有一些特定的规则。

罗马数字代码表：

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

罗马数字除了以下六种情况均为直接相加即可：

• 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

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.

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题目解析：理解罗马表，暴力。

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C++版：

class Solution {
public:
    int romanToInt(string s) {
        int ans = 0;
        s += " ";
        for (int i = 0; i < s.length();i++)
        {
            if (s[i] == 'I' && s[i + 1] == 'V')
            {
                ans += 4;
                i++;
            }
                
            else if (s[i] == 'I' && s[i + 1] == 'X')
            {
                ans += 9;
                i++;
            }
                
            else if (s[i] == 'X' && s[i + 1] == 'L')
            {
                ans += 40;
                i++;
            }
                
            else if (s[i] == 'X' && s[i + 1] == 'C')
            {
                ans += 90;
                i++;
                
            }
            else if (s[i] == 'C' && s[i + 1] == 'D')
            {
                ans += 400;
                i++;
            }
                
            else if (s[i] == 'C' && s[i + 1] == 'M')
            {
                ans += 900;
                i++;
            }
                
            else if (s[i] == 'I')
                ans += 1;
            else if (s[i] == 'V')
                ans += 5;
            else if (s[i] == 'X')
                ans += 10;
            else if (s[i] == 'L')
                ans += 50;
            else if (s[i] == 'C')
                ans += 100;
            else if (s[i] == 'D')
                ans += 500;
            else if (s[i] == 'M')
                ans += 1000;
        }
        return ans;
    }
};

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Python版本1：

class Solution:
    def romanToInt(self, s: str) -> int:
        ans = 0
        s += " "
        flag = 0
        for i in range(len(s)):
            if flag == 1:
                flag = 0
            elif (s[i] == 'I' and s[i + 1] == 'V'):
                ans += 4
                flag = 1
            elif (s[i] == 'I' and s[i + 1] == 'X'):
                ans += 9
                flag = 1
            elif (s[i] == 'X' and s[i + 1] == 'L'):
                ans += 40
                flag = 1
            elif (s[i] == 'X' and s[i + 1] == 'C'):
                ans += 90
                flag = 1
            elif (s[i] == 'C' and s[i + 1] == 'D'):
                ans += 400
                flag = 1
            elif (s[i] == 'C' and s[i + 1] == 'M'):
                ans += 900
                flag = 1
            elif (s[i] == 'I'):
                ans += 1
            elif (s[i] == 'V'):
                ans += 5
            elif (s[i] == 'X'):
                ans += 10
            elif (s[i] == 'L'):
                ans += 50
            elif (s[i] == 'C'):
                ans += 100
            elif (s[i] == 'D'):
                ans += 500
            elif (s[i] == 'M'):
                ans += 1000;
        return (ans)

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Python版本2：利用Python的特性：

def romanToInt(self, s):
    value, prevValue, result = {'M': 1000, 'D': 500, 'C': 100, 'L': 50, 'X': 10, 'V': 5, 'I': 1}, None, 0
    for ch in s:
        currValue = value[ch]
        result += currValue
        if prevValue and currValue > prevValue: result -= 2 * prevValue
        prevValue = currValue
    return result