12. Integer to Roman[M]整数转罗马数字
题目
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
| Symol | 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.
思路
思考罗马数字与整数的变换关系:罗马数字通常按照从大到小从左到右的顺序排,字母累加以表示数字。但是注意,整数4的罗马数字不是IIII,而是IV。像这样的数字还有9、40等。于是重新考虑建立罗马数字与整数的变换表:
| Symol | Value |
|---|---|
| I | 1 |
| IV | 4 |
| V | 5 |
| IX | 9 |
| X | 10 |
| XL | 40 |
| L | 50 |
| XC | 90 |
| C | 100 |
| CD | 400 |
| D | 500 |
| CM | 900 |
| M | 1000 |
列举出所有情况下的符号,在表中查找数字num:
- 去Value表中找出不大于num的数字x
- 根据对应的Symbol输出字符,并将num-x
- 循环执行2操作,直到num为0
C++
class Solution {
public:
string intToRoman(int num) {
string resStr = "";
vector<int> value{1000,900,500,400,100,90,50,40,10,9,5,4,1};
vector<string> symbol{"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
for(int i=0;i<value.size();i++)
while(num>=value[i])
{
resStr+=symbol[i];
num-=value[i];
}
return resStr;
}
};
Python
def intToRoman(self, num):
"""
:type num: int
:rtype: str
"""
intList = [1000,900,500,400,100,90,50,40,10,9,5,4,1]
strList = ['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I']
resStr = ''
i = 0
count = 0
while num > 0:
count = num/intList[i]
num %= intList[i]
while count > 0:
resStr += strList[i]
count -= 1
i += 1
return resStr
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