# [每日LeetCode] 13. Roman to Integer

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Description:

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: L = 50, V= 5, III = 3.
``````

Example 5:

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

``````一个罗马数字重复几次，就表示这个数的几倍。最多只能连续出现三次。
``````

``````在较大的罗马数字的右边记上较小的罗马数字，表示大数字加上小数字。

``````

``````在罗马数字的上方加一条横线或者加上下标M，表示将这个数乘以1000，加两条横线表示乘以1000的平方。
``````

C++代码

``````class Solution {
public:
int intval(char c){
switch(c){
case 'I': return 1;
break;
case 'V': return 5;
break;
case 'X': return 10;
break;
case 'L': return 50;
break;
case 'C': return 100;
break;
case 'D': return 500;
break;
case 'M': return 1000;
break;
}
return 0;
}
int romanToInt(string s) {
int sum = 0;
for(int i = 0;i<s.length(); i++){
if(intval(s[i]) <intval(s[i+1])){
sum -= intval(s[i]);
}
else
sum+=intval(s[i]);
}
return sum;
}
};
``````