# 12. Integer to Roman

**<font color=red>难度:Medium<font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/rotate-list/

> 内容描述

```
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 an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"
Example 2:

Input: 4
Output: "IV"
Example 3:

Input: 9
Output: "IX"
Example 4:

Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

```



> 思路1

******- 时间复杂度: O(n)******- 空间复杂度: O(n)******

题目不难，直接用一种暴力的写法

```cpp
class Solution {
public:
    string intToRoman(int num) {
        string ans;
        vector<string> table;
        int a = 0;
        int count = 0;
        while(num){
            a = num%10;
            num /= 10;
            ++count;
            if(count==1){
                if(a==1) table.push_back("I");
                else if(a==2) table.push_back("II");
                else if(a==3) table.push_back("III");
                else if(a==4) table.push_back("IV");
                else if(a==5) table.push_back("V");
                else if(a==6) table.push_back("VI");
                else if(a==7) table.push_back("VII");
                else if(a==8) table.push_back("VIII");
                else if(a==9) table.push_back("IX");
            }
            else if(count==2){
                if(a==1) table.push_back("X");
                else if(a==2) table.push_back("XX");
                else if(a==3) table.push_back("XXX");
                else if(a==4) table.push_back("XL");
                else if(a==5) table.push_back("L");
                else if(a==6) table.push_back("LX");
                else if(a==7) table.push_back("LXX");
                else if(a==8) table.push_back("LXXX");
                else if(a==9) table.push_back("XC");
            }
            else if(count==3){
                if(a==1) table.push_back("C");
                else if(a==2) table.push_back("CC");
                else if(a==3) table.push_back("CCC");
                else if(a==4) table.push_back("CD");
                else if(a==5) table.push_back("D");
                else if(a==6) table.push_back("DC");
                else if(a==7) table.push_back("DCC");
                else if(a==8) table.push_back("DCCC");
                else if(a==9) table.push_back("CM");
            }
            else if(count==4){
                if(a==1) table.push_back("M");
                else if(a==2) table.push_back("MM");
                else if(a==3) table.push_back("MMM");
            }
        }
        for(int i = table.size()-1; i >= 0; --i){
            ans += table[i];
        }
        return ans;
    }
};
```

> 思路2

******- 时间复杂度: O(n)******- 空间复杂度: O(1)******

太多的if else看的不好看。我们可以换一种更优雅的写法，让代码更好看。

```cpp
class Solution {
public:
    int arr[7] = {'I','V','X','L','C','D','M'};
    int arr1[13] = {1,4,5,9,10,40,50,90,100,400,500,900,1000};
    void f(int& num,int i,string& ans)
    {
        if(i % 2)
        {
            ans.push_back(arr[i / 4 * 2]);
            ans.push_back(arr[i / 2 + 1]);
        }
        else
            ans.push_back(arr[i / 2]);
        num -= arr1[i];
    }
    string intToRoman(int num) {
        string ans;
        while(num)
        {
            int i;
            for(i = 0;i < 13;++i)
                if(num < arr1[i])
                {
                    f(num,i - 1,ans);
                    break;
                }
            if(i == 13)
                f(num,i - 1,ans);
        }
        return ans;
    }
};
```