Topic link
https://leetcode-cn.com/problems/roman-to-integer
Title description
Roman numerals contain the following seven characters: I, V, X, L, C, D and M.
Character Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, the Roman numeral 2 is written as II, which means two parallel ones. 12 is written as XII, which means X + II. 27 is written as XXVII, which is XX + V + II.
Normally, the small numbers in Roman numerals are to the right of the large numbers. But there are special cases. For example, 4 is not written as IIII, but as IV. The number 1 is to the left of the number 5, and the number represented is equal to the number 4 obtained by subtracting the number 1 from the large number 5. Similarly, the number 9 is represented as IX. This special rule only applies to the following six situations:
I can be placed to the left of V (5) and X (10) to represent 4 and 9.
X can be placed to the left of L (50) and C (100) to represent 40 and 90.
C can be placed to the left of D (500) and M (1000) to represent 400 and 900.
Given a Roman numeral, convert it to an integer. Ensure that the input is in the range of 1 to 3999.
input Output
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, IV = 4.
Problem solving ideas
The value is simply to judge the next digit of the next character. If it can form a number with the next digit, then the group will be grouped. If not, the value will be calculated directly.
Problem-solving code
#include <iostream>
#include <string>
using namespace std;
class Solution {
public:
int getTwoNum(char num1, char num2) {
// I可以放在V(5) 和 X (10) 的左边，来表示 4 和 9。
// X可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。
// C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
if (num1=='I' && (num2 =='V' || num2=='X')){
if (num2 =='V'){
return 4;
} else{
return 9;
}
}
if (num1=='X' && (num2 =='L' || num2=='C')){
if (num2 =='L'){
return 40;
} else{
return 90;
}
}
if (num1=='C' && (num2 =='D' || num2=='M')){
if (num2 =='D'){
return 400;
} else{
return 900;
}
}
else{
return -1;
}
}
int romanToInt(string s) {
int index = 0;
int sum = 0;
while (index < s.length()) {
char romeNum1 = s[index];
if (index == s.length()) {
sum += getNum(romeNum1);
} else {
char romeNum2 = s[index + 1];
int twosum = getTwoNum(romeNum1, romeNum2);
if (twosum > 0) {
sum += twosum;
index++;
} else {
sum += getNum(romeNum1);
}
}
index++;
}
return sum;
}
private:
int getNum(char roma) {
switch (roma) {
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:
return 0;
}
}
};
//int main() {
// Solution solution;
// int k = solution.romanToInt("L");
// cout<<k;
// return 0;
//}