[LeetCode] Integer to Roman 整數轉化成羅馬數字

Grandyang發表於2014-11-26
LeetCode

 

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.

 

之前那篇文章寫的是羅馬數字轉化成整數 Roman to Integer， 這次變成了整數轉化成羅馬數字，基本演算法還是一樣。由於題目中限定了輸入數字的範圍(1 - 3999), 使得題目變得簡單了不少。

基本字元
I
V
X
L
C
D
M
相應的阿拉伯數字表示為
1
5
10
50
100
500
1000

 

 

例如整數 1437 的羅馬數字為 MCDXXXVII， 我們不難發現，千位，百位，十位和個位上的數分別用羅馬數字表示了。 1000 - M, 400 - CD, 30 - XXX, 7 - VII。所以我們要做的就是用取商法分別提取各個位上的數字，然後分別表示出來：

100 - C

200 - CC

300 - CCC

400 - CD

500 - D

600 - DC

700 - DCC

800 - DCCC

900 - CM

我們可以分為四類，100到300一類，400一類，500到800一類，900最後一類。每一位上的情況都是類似的，程式碼如下：

 

解法一：

class Solution {
public:
    string intToRoman(int num) {
        string res = "";
        vector<char> roman{'M', 'D', 'C', 'L', 'X', 'V', 'I'};
        vector<int> value{1000, 500, 100, 50, 10, 5, 1};
        for (int n = 0; n < 7; n += 2) {
            int x = num / value[n];
            if (x < 4) {
                for (int i = 1; i <= x; ++i) res += roman[n];
            } else if (x == 4) {
                res = res + roman[n] + roman[n - 1]; 
            } else if (x > 4 && x < 9) {
                res += roman[n - 1];
                for (int i = 6; i <= x; ++i) res += roman[n];
            } else if (x == 9) {
                res = res + roman[n] + roman[n - 2];
            }
            num %= value[n];            
        }
        return res;
    }
};

 

本題由於限制了輸入數字範圍這一特殊性，故而還有一種利用貪婪演算法的解法，建立一個數表，每次通過查表找出當前最大的數，減去再繼續查表。參見程式碼如下：

 

解法二：

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

 

下面這種方法個人感覺屬於比較投機取巧的方法，把所有的情況都列了出來，然後直接按位查表，O(1)的時間複雜度啊，參見程式碼如下：

 

解法三：

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];
    }
};

 

類似題目：

Roman to Integer

Integer to English Words

 

參考資料：

https://leetcode.com/problems/integer-to-roman/

https://leetcode.com/problems/integer-to-roman/discuss/6274/simple-solution

https://leetcode.com/problems/integer-to-roman/discuss/6310/my-java-solution-easy-to-understand

 

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