Problem

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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

Solution

accumulate the value of each letter.

If the value of current letter is greater than the previous one, deduct twice of the previous value.

e.g. IX, 1 + 10 – 2 * 1 = 9 instead of 1 + 10 = 11

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int romanToInt(string s) {

map<char, int> m{{'I', 1}, {'V', 5},

{'X', 10}, {'L', 50},

{'C', 100}, {'D', 500},

{'M', 1000}};

int ans = 0;

for (int i = 0; i < s.length(); i++) {

ans += m[s[i]];

if (i > 0 && m[s[i]] > m[s[i - 1]])

ans -= 2 * m[s[i - 1]];

}

return ans;

}

};

Java

// Author: Huahua

class Solution {

private int v(char R) {

switch (R) {

case 'I' : return 1;

case 'V' : return 5;

case 'X' : return 10;

case 'L' : return 50;

case 'C' : return 100;

case 'D' : return 500;

case 'M' : return 1000;

default: break;

}

return 0;

}

public int romanToInt(String s) {

int ans = 0;

for (int i = 0; i < s.length(); i++) {

ans += v(s.charAt(i));

if ( i > 0 && v(s.charAt(i)) > v(s.charAt(i - 1)))

ans -= 2 * v(s.charAt(i - 1));

}

return ans;

}

Python3

class Solution:

def romanToInt(self, s):

m = {'I' : 1, 'V': 5, 'X': 10, 'L' : 50,

'C' : 100, 'D': 500, 'M': 1000}

ans = m[s[0]]

for c, p in zip(s[1:], s[0:-1]):

ans += m[c]

if m[c] > m[p]: ans -= 2 * m[p]

return ans

Problem

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

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

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

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

Solution: HashTable + Simulation

Map integer 1,4,5,9,10,40,50,90, …, 1000 to Romain

Start from the largest number y,

if x >= y:
ans += Roman[y]
x -= y

Time complexity: O(x)

Space complexity: O(x)

C++

// Author: Huahua

class Solution {

public:

string intToRoman(int num) {

vector<pair<int,string>> v{{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"},

{ 100, "C"}, { 90, "XC"}, { 50, "L"}, { 40, "XL"},

{ 10, "X"}, { 9, "IX"}, { 5, "V"}, { 4, "IV"},

{ 1, "I"}};

string ans;

auto it = cbegin(v);

while (num) {

if (num >= it->first) {

num -= it->first;

ans += it->second;

} else {

++it;

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def intToRoman(self, num):

m = [[1000, "M"], [900, "CM"], [500, "D"], [400, "CD"],

[ 100, "C"], [ 90, "XC"], [ 50, "L"], [ 40, "XL"],

[ 10, "X"], [ 9, "IX"], [ 5, "V"], [ 4, "IV"],

[ 1, "I"]]

index = 0

ans = ''

while num > 0:

if num >= m[index][0]:

ans += m[index][1]

num -= m[index][0]

else:

index += 1

return ans

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