用Python3实现LeetCode算法题系列——No.013 Roman to Integer [Easy]

版权声明：转载请注明来源！ https://blog.csdn.net/CAU_Ayao/article/details/82262536

目录

---

1. 题目

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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

---

2. 提交的代码

        sum = 0
        s_len = len(s) #字符串长度
        for i in range(0,s_len):
            if s[i] == 'I':
                if (i < s_len-1) and (s[i+1] == 'V' or s[i+1] == 'X'):
                    sum -= 1
                else:
                    sum += 1
            if s[i] == 'V':
                sum += 5
            if s[i] == 'X':
                if (i < s_len-1) and (s[i+1] == 'L' or s[i+1] == 'C'):
                    sum -= 10
                else:
                    sum += 10
            if s[i] == 'L':
                sum += 50
            if s[i] == 'C':
                if (i<s_len-1) and (s[i+1] == 'D' or s[i+1] == 'M'):
                    sum -= 100
                else:
                    sum += 100
            if s[i] == 'D':
                sum += 500
            if s[i] == 'M':
                sum += 1000
        return sum

3. 运行效果

4. 完整代码

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        sLen = len(s)
        res = 0
        for i in range(0, sLen):
            if s[i] == 'M':
                res += 1000
            if s[i] == 'D':
                res += 500
            if s[i] == 'C':
                if (i < sLen - 1) and (s[i + 1] == 'D' or s[i + 1] == 'M'):
                    res -= 100
                else:
                    res += 100
            if s[i] == 'L':
                res += 50
            if s[i] == 'X':
                if (i < sLen - 1) and (s[i + 1] == 'L' or s[i + 1] == 'C'):
                    res -= 10
                else:
                    res += 10
            if s[i] == 'V':
                res += 5
            if s[i] == 'I':
                if (i < sLen - 1) and (s[i + 1] == 'V' or s[i + 1] == 'X'):
                    res -= 1
                else:
                    res += 1
        return res

def stringToString(input):
    return input[1:-1].decode('string_escape')

def main():
    import sys
    def readlines():
        for line in sys.stdin:
            yield line.strip('\n')

    lines = readlines()
    while True:
        try:
            line = next(lines)
            s = stringToString(line);

            ret = Solution().romanToInt(s)

            out = str(ret);
            print(out)
        except StopIteration:
            break

if __name__ == '__main__':
    main()