> For the complete documentation index, see [llms.txt](https://txfs19260817.gitbook.io/leetcode-go-notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://txfs19260817.gitbook.io/leetcode-go-notes/solutions/13.-roman-to-integer.md).

# 13. Roman to Integer

## LeetCode [13. Roman to Integer](https://leetcode-cn.com/problems/roman-to-integer/)

### 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.

### Tags

Math, String

### Solution

遍历字符串，先检查连续的两个字符是否存在映射，如果存在则将相应的数字累加在结果上，如果不存在或者已经遍历到最后一个字符，则将当前这一个字符映射的数字累加在结果上，最后返回结果。

Before start, collate all mapping from symbol to value. Traverse the string and check if there is a mapping for the current 2 consecutive characters first. If it exists, the corresponding number is added to the result. Otherwise the single-character symbol corresponded value is accumulated on the result.

### Complexity

* Time complexity: $$O(1)$$
* Space complexity: $$O(1)$$

### Code

```go
func romanToInt(s string) int {
	var ans int
	for i := 0; i < len(s); i++ {
		var singleFlag bool
		if i+1 < len(s) {
			switch s[i : i+2] {
			case "IV":
				ans += 4
				i++
			case "IX":
				ans += 9
				i++
			case "XL":
				ans += 40
				i++
			case "XC":
				ans += 90
				i++
			case "CD":
				ans += 400
				i++
			case "CM":
				ans += 900
				i++
			default:
				singleFlag = true
			}
		} else {
			singleFlag = true
		}
		if singleFlag {
			switch s[i] {
			case 'I':
				ans += 1
			case 'V':
				ans += 5
			case 'X':
				ans += 10
			case 'L':
				ans += 50
			case 'C':
				ans += 100
			case 'D':
				ans += 500
			case 'M':
				ans += 1000
			}
		}
	}
	return ans
}
```