# [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/) ## 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, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` 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 an integer, convert it to a roman numeral. **Example 1:** ``` Input: num = 3 Output: "III" ``` **Example 2:** ``` Input: num = 4 Output: "IV" ``` **Example 3:** ``` Input: num = 9 Output: "IX" ``` **Example 4:** ``` Input: num = 58 Output: "LVIII" Explanation: L = 50, V = 5, III = 3. ``` **Example 5:** ``` Input: num = 1994 Output: "MCMXCIV" Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. ``` **Constraints:** - `1 <= num <= 3999` ## 题目大意 通常情况下,罗马数字中小的数字在大的数字的右边。但也存在特例,例如 4 不写做 IIII,而是 IV。数字 1 在数字 5 的左边,所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地,数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况: - I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。 - X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。 - C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。 给定一个整数,将其转为罗马数字。输入确保在 1 到 3999 的范围内。 ## 解题思路 - 依照题意,优先选择大的数字,解题思路采用贪心算法。将 1-3999 范围内的罗马数字从大到小放在数组中,从头选择到尾,即可把整数转成罗马数字。 ## 代码 ```go package leetcode func intToRoman(num int) string { values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1} symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"} res, i := "", 0 for num != 0 { for values[i] > num { i++ } num -= values[i] res += symbols[i] } return res } ```