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102 lines
2.8 KiB
Markdown
102 lines
2.8 KiB
Markdown
# [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/)
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## 题目
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Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
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```
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Symbol Value
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I 1
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V 5
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X 10
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L 50
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C 100
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D 500
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M 1000
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```
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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`.
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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:
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- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9.
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- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90.
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- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
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Given an integer, convert it to a roman numeral.
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**Example 1:**
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```
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Input: num = 3
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Output: "III"
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```
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**Example 2:**
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```
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Input: num = 4
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Output: "IV"
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```
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**Example 3:**
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```
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Input: num = 9
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Output: "IX"
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```
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**Example 4:**
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```
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Input: num = 58
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Output: "LVIII"
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Explanation: L = 50, V = 5, III = 3.
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```
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**Example 5:**
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```
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Input: num = 1994
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Output: "MCMXCIV"
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Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
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```
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**Constraints:**
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- `1 <= num <= 3999`
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## 题目大意
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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：
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- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。
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- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。
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- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。
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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。
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## 解题思路
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- 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。
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## 代码
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```go
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package leetcode
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func intToRoman(num int) string {
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values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
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symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
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res, i := "", 0
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for num != 0 {
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for values[i] > num {
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i++
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}
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num -= values[i]
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res += symbols[i]
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}
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return res
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}
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``` |