[LeetCode] 12. Integer to Roman 整数转为罗马数字
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 beforeV
(5) andX
(10) to make 4 and 9. -
X
can be placed beforeL
(50) andC
(100) to make 40 and 90. -
C
can be placed beforeD
(500) andM
(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: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: 1994 Output: "MCMXCIV" Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
将阿拉伯整数转为罗马数字,首先要对罗马数字有了解,找到两种数字转换的规律,然后用一个Hash map来保存这些规律,然后把整数进行相应的转换。输入数字的范围(1 - 3999)。
Java:
public static String intToRoman(int num) { String M[] = {"", "M", "MM", "MMM"}; String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}; String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"}; String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"}; return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10]; }
Java:
public class Solution { public String intToRoman(int number) { int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 }; String[] numerals = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" }; StringBuilder result = new StringBuilder(); for (int i = 0; i < values.length; i++) { while (number >= values[i]) { number -= values[i]; result.append(numerals[i]); } } return result.toString(); } }
Python:
class Solution(object): def intToRoman(self, num): """ :type num: int :rtype: str """ numeral_map = {1: "I", 4: "IV", 5: "V", 9: "IX", 10: "X", 40: "XL", 50: "L", 90: "XC", 100: "C", 400: "CD", 500: "D", 900: "CM", 1000: "M"} keyset, result = sorted(numeral_map.keys()), [] while num > 0: for key in reversed(keyset): while num / key > 0: num -= key result += numeral_map[key] return "".join(result)
Python:
strs = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'] nums = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1] ret = "" for i, j in enumerate(nums): while num >= j: ret += strs[i] num -= j if num == 0: return ret
Python:
def intToRoman1(self, num): values = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ] numerals = [ "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" ] res, i = "", 0 while num: res += (num//values[i]) * numerals[i] num %= values[i] i += 1 return res
Python:
def intToRoman(self, num): values = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ] numerals = [ "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" ] res = "" for i, v in enumerate(values): res += (num//v) * numerals[i] num %= v return res
C++:
class Solution { public: string intToRoman(int num) { const vector<int> nums{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}; const vector<string> romans{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}; string result; int i = 0; while (num > 0) { int times = num / nums[i]; while (times--) { num -= nums[i]; result.append(romans[i]); } ++i; } return result; } };
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