LeetCode 刷题记录 12. Integer to Roman

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本文详细介绍如何将整数转换为罗马数字，包括贪心法和打表法两种算法实现，适用于1到3999范围内的整数转换。

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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.

方法1：贪心法
c++：

class Solution {
public:
    string intToRoman(int num) {
        int radix[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        string symbol[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        string roman;
        int n = sizeof(radix) / sizeof(radix[0]);
        for(int i = 0; i < n; i++){
            while(num >= radix[i]){
                num -= radix[i];
                roman += symbol[i];
            }
        }
        return roman;
        
    }
};

java：
字符串需要初始化为空而c++则不需要

class Solution {

    public String intToRoman(int num) {
        int[] radix= {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        String[] symbol = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        String roman = "";
        int n = radix.length;
        for(int i = 0; i < n; i++){
            while(num >= radix[i]){
                num -= radix[i];
                roman += symbol[i];
            }
        }
        return roman;
        
    }
}

python:

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        radix= [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
        symbol = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
        roman = ""
        n = len(radix)
        for i in range(n):
            while num >= radix[i]:
                num -= radix[i]
                roman += symbol[i]
            
        
        return roman

方法2：
打表法：将所有情况列出直接取即可
c++：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
        vector<string> v1{"", "M", "MM", "MMM"};
        vector<string> v2{"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
        vector<string> v3{"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
        vector<string> v4{"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10];
    }
};

java:

class Solution {

    public String intToRoman(int num) {
         String res = "";
         String[] v1 = {"", "M", "MM", "MMM"};
         String[] v2 = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
         String[] v3 = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
         String[] v4 = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
        return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10];
        
    }
}

python:

class Solution(object):
    def intToRoman(self, num):
        """
        :type num: int
        :rtype: str
        """
        v1 = ["", "M", "MM", "MMM"]
        v2 = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"]
        v3 = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]
        v4 = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"]
        return v1[num / 1000] + v2[(num % 1000) / 100] + v3[(num % 100) / 10] + v4[num % 10]