[Leetcode] 13. Roman to Integer

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本文介绍了一种将罗马数字转换为整数的方法。通过解析罗马数字的构成规则，设计了一个迭代算法来逐步读取每个字符并累加相应的数值。该算法能够正确处理包括特殊组合在内的各种常见罗马数字。

Problem:

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

思路：
方法与LeetCode 12题同理。

Solution：

class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        returnint = 0
        slen = len(s)
        i = 0
        while i < slen:
            if s[i] == 'M':
                returnint += 1000
                i += 1
            elif s[i] == 'C' and i + 1 != slen and s[i+1] == 'M':
                returnint += 900
                i += 2
            elif s[i] == 'D':
                returnint += 500
                i += 1
            elif s[i] == 'C' and i + 1 != slen and s[i+1] == 'D':
                returnint += 400
                i += 2
            elif s[i] == 'C':
                returnint += 100
                i += 1
            elif s[i] == 'X' and i + 1 != slen and s[i+1] == 'L':
                returnint += 40
                i += 2
            elif s[i] == 'X' and i + 1 != slen and s[i+1] == 'C':
                returnint += 90
                i += 2
            elif s[i] == 'X':
                returnint += 10
                i += 1
            elif s[i] == 'L':
                returnint += 50
                i += 1
            elif s[i] == 'I' and i + 1 != slen and s[i+1] == 'X':
                returnint += 9
                i += 2
            elif s[i] == 'I' and i + 1 != slen and s[i+1] == 'V':
                returnint += 4
                i += 2
            elif s[i] == 'V':
                returnint += 5
                i += 1
            elif s[i] == 'I':
                returnint += 1
                i += 1
        return returnint