Roman to Integer

LeetCode笔记 阅读(60)

编号 : 12      
难度 : <font color="blue">Easy</font>

# 题目 :  
>
>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.  

# 解 :  
```Cpp
class Solution {
public:
    int romanToInt(string s) {
        static map<char, int> roman2int = {
            {'I', 1},
            {'V', 5},
            {'X', 10},
            {'L', 50},
            {'C', 100},
            {'D', 500},
            {'M', 1000}
        };
        
        int ret = 0;
        size_t i = 0;
        for(; i + 1 < s.size(); i++)
        {
            int before = roman2int[s[i]];
            int after  = roman2int[s[i+1]];
            if(before < after)
            {
                ret -= before;
            }
            else
            {
                ret += before;
            }
        }
        
        ret += roman2int[s[i]];
        return ret;
    }
};
```

# 分析 :  
这个题很简单,对于90(XC)这样的数字,是把10(X)放在100(C)前面,表示100 - 10。所以一个位置上的数比他后面一个数小的时候,就减去它的值,除此之外的情况只要加起来就行了。

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