Roman To Integer
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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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.
Solution:
public class Solution { | |
public int RomanToInt(string s) { | |
int len = s.Length; | |
int num = 0; | |
int i=0; | |
while(i<len){ | |
if(i-1>=0 && s[i] == 'V' && s[i-1] == 'I'){ | |
num = num + 4; | |
} | |
else if(s[i] == 'V'){ | |
int j = i; | |
int five = 0; | |
while(j<len && s[j] == 'V'){ | |
five++; | |
j++; | |
} | |
num = num + 5*five; | |
i = j; | |
i--; | |
} | |
else if(i-1>=0 && s[i] == 'X' && s[i-1] == 'I'){ | |
num = num + 9; | |
} | |
else if(s[i] == 'X'){ | |
int j = i; | |
int ten = 0; | |
while(j<len && s[j] == 'X'){ | |
ten++; | |
j++; | |
} | |
if(j<len && (s[j] == 'L' || s[j] == 'C')){ | |
ten--; | |
} | |
num = num + 10*ten; | |
i = j; | |
i--; | |
} | |
else if(i-1>=0 && s[i] == 'L' && s[i-1] == 'X'){ | |
num = num + 40; | |
} | |
else if(s[i] == 'L'){ | |
int j = i; | |
int fifty = 0; | |
while(j<len && s[j] == 'L'){ | |
fifty++; | |
j++; | |
} | |
num = num + 50*fifty; | |
i = j; | |
i--; | |
} | |
else if(i-1>=0 && s[i] == 'C' && s[i-1] == 'X'){ | |
num = num + 90; | |
} | |
else if(s[i] == 'C'){ | |
int j = i; | |
int hundred = 0; | |
while(j<len && s[j] == 'C'){ | |
hundred++; | |
j++; | |
} | |
if(j<len && (s[j] == 'D' || s[j] == 'M')){ | |
hundred--; | |
} | |
num = num + 100*hundred; | |
i = j; | |
i--; | |
} | |
else if(i-1>=0 && s[i] == 'D' && s[i-1] == 'C'){ | |
num = num + 400; | |
} | |
else if(s[i] == 'D'){ | |
int j = i; | |
int fivehundred = 0; | |
while(j<len && s[j] == 'D'){ | |
fivehundred++; | |
j++; | |
} | |
num = num + 500*fivehundred; | |
i = j; | |
i--; | |
} | |
else if(i-1>=0 && s[i] == 'M' && s[i-1] == 'C'){ | |
num = num + 900; | |
} | |
else if(s[i] == 'M'){ | |
int j = i; | |
int thousand = 0; | |
while(j<len && s[j] == 'M'){ | |
thousand++; | |
j++; | |
} | |
num = num + 1000*thousand; | |
i = j; | |
i--; | |
} | |
else{ | |
int j = i; | |
int one = 0; | |
while(j<len && s[j] == 'I'){ | |
one++; | |
j++; | |
} | |
if(j<len && (s[j] == 'V' || s[j] == 'X')){ | |
one--; | |
} | |
num = num + one; | |
i = j; | |
i--; | |
} | |
i++; | |
} | |
return num; | |
} | |
} |
Space Complexity: O(1)
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