> For the complete documentation index, see [llms.txt](https://liuyang89116.gitbook.io/my-leetcode-book/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuyang89116.gitbook.io/my-leetcode-book/chapter_2_string/problem_12_integer_to_roman.md).

# Problem 12: Integer to Roman

> <https://leetcode.com/problems/integer-to-roman/>

## 思路

像做除法运算一样，提前把 千位数，百位数，个位数都存好，然后一位一位地求出来

```java
public class Solution {
    public String intToRoman(int num) {
        String M[] = {"", "M", "MM", "MMM"}; //1000, 2000, 3000
        String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};  //100,200,300,...,900
        String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}; //10,20,...,90
        String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"}; //1,2,3,...,9
        return M[num / 1000] + C[num % 1000 / 100] + X[num % 100 / 10] + I[num % 10];
    }
}
```

## 易错点

1. 每个String数组开始都有一个“”空字符串，这个位置是给0预备的
2.
3. ```java
   M[num / 1000] + C[num % 1000 / 100] + X[num % 100 / 10] + I[num % 10]
   ```

   直接想的时候有时候会出现一些小的错误，如果能直接举例子，就会好很多了。eg: 3560，然后自己手动拿手除一下