leetcode 13: Roman to Integer

最新推荐文章于 2018-12-18 10:29:51 发布

原创最新推荐文章于 2018-12-18 10:29:51 发布 · 180 阅读

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本文介绍了一种将罗马数字转换为整数的算法，详细解释了罗马数字的构成原则和特殊情况处理，提供了两种Python实现方案，一种速度较快，另一种虽然较慢但易于理解。

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: L = 50, V= 5, III = 3.

Example 5:

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

解答一：

这是第二次的答案faster than 43%, 还凑合

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        d1 = {
            'I':1,
            'V':5,
            'X':10,
            'L':50,
            'C':100,
            'D':500,
            'M':1000
        }
        length = len(s)
        result = 0
        for i in range(length):
            if i == length-1:
                result += d1[s[i]]
                return result
            if d1[s[i]]<d1[s[i+1]]:
                result -= d1[s[i]]
            else:
                result += d1[s[i]]
        return result

解答二：

这是第一次的解答，leetcode数据显示属于贼慢的那种faster than 2%，哭T_T

class Solution:
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        d1 = {
            'I':1,
            'V':5,
            'X':10,
            'L':50,
            'C':100,
            'D':500,
            'M':1000
        }
        length = len(s)
        result = 0
        for i in range(length):
            if i == length-1:
                result += d1[s[i]]
                return result
            if d1[s[i]]<d1[s[i+1]]:
                result -= d1[s[i]]
            else:
                result += d1[s[i]]
        return result