Leetcode_13 Roman numeral to Integer]

Article Directory:

topic
A script and notes
- 　　Uncommented script
A script logic

topic:

Roman numeral characters comprising the following seven: I, V, X, L, C, D and M.

Numerical character
the I. 1
V. 5
X-10
L 50
C 100
D 500
M 1000
, for example, 2 written as Roman numerals II, namely 1.12 written as two parallel XII, namely X + II. 27 written as XXVII, namely XX + V + II.

Typically, small numbers of Roman numerals to the right in large numbers. But there are exceptions, for example, do not write 4 IIII, but IV. In the left number 5 number 1, the number of large numbers equal to the value represented by a reduced number 5 obtained 4. Likewise, the number 9 is represented as IX. This special rule applies only to the following six cases:

I may be on the left V (5) and X (10), and to represent 4 and 9.
X L can be placed on the left (50) and C (100), and 40 and 90 are represented.
C may be placed on D (500) and M (1000) to the left, to 400 and 900 represent.
Given a Roman numeral, to convert it to an integer. To ensure that the input is in the range of 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
explains: L = 50, V = 5 , III = 3.
Example 5:

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

A script and notes:

class Solution:
     DEF romanToInt (Self, S: STR) -> int: 
        SUM1 = s.count ( ' the I ' ) + s.count ( ' V ' ) * + s.count. 5 ( ' X- ' ) * 10 + S .count ( ' L ' ) * 50 + s.count ( ' C ' ) * 100 + s.count ( ' D ' ) * 500 + s.count ( ' M ' ) * 1000 
# Roman numerals of acquiring literally values and 
        jianshu = s.count ( ' IV ' ) * 2 + s.count ( 'IX ') * 2 + s.count ( ' XL ' ) * + 20 is s.count ( ' XC ' ) * + 20 is s.count ( ' the CD ' ) * + 200 is s.count ( ' the CM ' ) * 200 is 
# get input Roman numeral value of the sum of the special representative of the string
         return (SUM1 - jianshu) 
# the literal sum minus the sum of special characters

class Solution:
    def romanToInt(self, s: str) -> int:
        sum1 = s.count('I') + s.count('V') * 5 + s.count('X') * 10 + s.count('L') * 50 + s.count('C') * 100 + s.count('D') * 500 + s.count('M') * 1000
        jianshu = s.count('IV') * 2 + s.count('IX') * 2 + s.count('XL') * 20 + s.count('XC') * 20 + s.count('CD') * 200 + s.count('CM') * 200
        return(sum1 - jianshu)

Uncommented Code

A script logic:

Get its first representative value based on each character Roman
Due to the special string Roman characters, similar to the "IX" literal value of 10 + 1, it is actually representative of 9; Roman character string should be calculated so that the input string contains special number, then the literal minus the sum of the values represented by the sum of the special characters
Note should subtract the sum of the two representatives of special characters

Leetcode_13 Roman numeral to Integer]

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