LC的罗马转数字规则是什么？# JobHunting - 待字闺中
A*e
1 楼
怎么觉得有二义性呢？
MCMXCVI，按结果是M+(C-M)+(C-X)+(V+I) = 1996，为何不是(M+C) +(M+X) + (C+V)+I
= 2216？左减优先？
MCMXCVI，按结果是M+(C-M)+(C-X)+(V+I) = 1996，为何不是(M+C) +(M+X) + (C+V)+I
= 2216？左减优先？
A*e
2 楼
百度百科的规则，没有排除二义性。
http://baike.baidu.com/view/42061.htm
I
【在 A*******e 的大作中提到】
: 怎么觉得有二义性呢？
: MCMXCVI，按结果是M+(C-M)+(C-X)+(V+I) = 1996，为何不是(M+C) +(M+X) + (C+V)+I
: = 2216？左减优先？
http://baike.baidu.com/view/42061.htm
I
【在 A*******e 的大作中提到】
: 怎么觉得有二义性呢？
: MCMXCVI，按结果是M+(C-M)+(C-X)+(V+I) = 1996，为何不是(M+C) +(M+X) + (C+V)+I
: = 2216？左减优先？
t*5
3 楼
应该是只要左边比右边大，就要相减。建议看wiki的规则，非常清楚明了
A*e
4 楼
明白了。中文版说的不清楚，还得看英文维基。
没有右加规则，只有左减，仅限于I,X,C
【在 A*******e 的大作中提到】
: 百度百科的规则，没有排除二义性。
: http://baike.baidu.com/view/42061.htm
:
: I
没有右加规则，只有左减，仅限于I,X,C
【在 A*******e 的大作中提到】
: 百度百科的规则，没有排除二义性。
: http://baike.baidu.com/view/42061.htm
:
: I
A*e
5 楼
说反了，是左小右大。而且这个总结也不准确，是以下具体三条规则：
1. I placed before V or X indicates one less, so four is IV (one less than
five) and nine is IX (one less than ten)
2. X placed before L or C indicates ten less, so forty is XL (ten less than
fifty) and ninety is XC (ten less than a hundred)
3. C placed before D or M indicates a hundred less, so four hundred is CD (a
hundred less than five hundred) and nine hundred is CM (a hundred less than
a thousand)[5]
【在 t********5 的大作中提到】
: 应该是只要左边比右边大，就要相减。建议看wiki的规则，非常清楚明了
1. I placed before V or X indicates one less, so four is IV (one less than
five) and nine is IX (one less than ten)
2. X placed before L or C indicates ten less, so forty is XL (ten less than
fifty) and ninety is XC (ten less than a hundred)
3. C placed before D or M indicates a hundred less, so four hundred is CD (a
hundred less than five hundred) and nine hundred is CM (a hundred less than
a thousand)[5]
【在 t********5 的大作中提到】
: 应该是只要左边比右边大，就要相减。建议看wiki的规则，非常清楚明了
t*5
6 楼
哈哈 翻了一下LC，原来我是hardcode的，应该更简洁一些
其实应该用一个index i的，我比较懒直接修改了原string，面试想修改原string的同
学最好征求面试官的意见
class Solution:
# @param s, a string
# @return an integer
def romanToInt(self, s):
ROMAN_INT = {'M': 1000, 'CM': 900, 'D': 500, 'CD': 400,
'C': 100, 'XC': 90, 'L': 50, 'XL': 40,
'X': 10, 'IX': 9, 'V': 5, 'IV': 4, 'I': 1}
result = 0
while s:
if s[:2] in ROMAN_INT:
result += ROMAN_INT[s[:2]]
s = s[2:]
elif s[:1] in ROMAN_INT:
result += ROMAN_INT[s[:1]]
s = s[1:]
else:
return -1
return result
than
(a
than
【在 A*******e 的大作中提到】
: 说反了，是左小右大。而且这个总结也不准确，是以下具体三条规则：
: 1. I placed before V or X indicates one less, so four is IV (one less than
: five) and nine is IX (one less than ten)
: 2. X placed before L or C indicates ten less, so forty is XL (ten less than
: fifty) and ninety is XC (ten less than a hundred)
: 3. C placed before D or M indicates a hundred less, so four hundred is CD (a
: hundred less than five hundred) and nine hundred is CM (a hundred less than
: a thousand)[5]
其实应该用一个index i的，我比较懒直接修改了原string，面试想修改原string的同
学最好征求面试官的意见
class Solution:
# @param s, a string
# @return an integer
def romanToInt(self, s):
ROMAN_INT = {'M': 1000, 'CM': 900, 'D': 500, 'CD': 400,
'C': 100, 'XC': 90, 'L': 50, 'XL': 40,
'X': 10, 'IX': 9, 'V': 5, 'IV': 4, 'I': 1}
result = 0
while s:
if s[:2] in ROMAN_INT:
result += ROMAN_INT[s[:2]]
s = s[2:]
elif s[:1] in ROMAN_INT:
result += ROMAN_INT[s[:1]]
s = s[1:]
else:
return -1
return result
than
(a
than
【在 A*******e 的大作中提到】
: 说反了，是左小右大。而且这个总结也不准确，是以下具体三条规则：
: 1. I placed before V or X indicates one less, so four is IV (one less than
: five) and nine is IX (one less than ten)
: 2. X placed before L or C indicates ten less, so forty is XL (ten less than
: fifty) and ninety is XC (ten less than a hundred)
: 3. C placed before D or M indicates a hundred less, so four hundred is CD (a
: hundred less than five hundred) and nine hundred is CM (a hundred less than
: a thousand)[5]
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