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  • [LeetCode] 12. Integer to Roman 整数转为罗马数字

    Roman numerals are represented by seven different symbols: IVXLCD 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: 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;
        }
    };
    

      

    类似题目:

    [LeetCode] 13. Roman to Integer 罗马数字转为整数  

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8493212.html
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