> For the complete documentation index, see [llms.txt](https://r24zeng.gitbook.io/leetcode-notebook/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://r24zeng.gitbook.io/leetcode-notebook/zhong-yu-shua-dao-100-dao-le-wo-shi-fen-shui-ling/bu-chong-120-dao/13.-roman-to-integer.md).

# 13. Roman to Integer

{% hint style="info" %}
Two pointers problem. One is current, the other one is next. Only two situations:

1. roman{cur} < roman{next}, such as 90 = 100 - 10, num{90} = 'XC'
2. roman{cur} >= roman{next}, eg1, 110 = 100 + 10, num{110} = 'CX', eg2, num{80} = 'LXXX'

其实不容易想到。
{% endhint %}

### Solution:

1. if `roman{cur} < roman{next}`, two letters must be computed together, `cur` and `next` pointers are moved **2 steps**.
2. if `roman{cur} >= roman{next}`, then only collect roman{cur}, cur and next pointers are moved **1 step**.
3. Don't forget to consider one last edge situation, if `cur` points to last element, add it to `num` directly, if `cur` points out of range, ignore it.

{% tabs %}
{% tab title="Python" %}

```python
class Solution(object):
    def romanToInt(self, s):
        """
        :type s: str
        :rtype: int
        """
        roman = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C':100, 'D': 500, 'M': 1000}
        # edge case
        if len(s) == 1:
            return roman[s[0]]
        
        # regular case
        cur = 0
        nex = 1
        num = 0
        while nex < len(s):
            if roman[s[cur]] < roman[s[nex]]:   # bind two letters (such as 90, 'XC')
                num += roman[s[nex]] - roman[s[cur]]
                cur += 2
                nex += 2
            else:   # only one single letter (such as 100~'C', 80~'LXXX')
                num += roman[s[cur]]
                cur += 1
                nex += 1
        
        if cur == len(s) - 1:
            num += roman[s[cur]]
            
        return num
        
```

{% endtab %}
{% endtabs %}


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