[LeetCode] 12. Integer to Roman

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Problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

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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:

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Input: 3
Output: "III"

Example 2:

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Input: 4
Output: "IV"

Example 3:

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Input: 9
Output: "IX"

Example 4:

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Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 5:

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Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Explanation 1

Character I V X L C D M
Number 1 5 10 50 100 500 1000

  1. For example, 1437 in roman is MCDXXXVII. We find out the thousand digit, hundred digit, thenth digit, and single digit’s numbers all can be represented by roman character. 1000 is M, 400 is CD, 30 is XXX, 7 is VII. So, we can get every digit by division 1000, 100, 10, 1, and represent them.

  2. We can have 4 groups. 1-3 is one group, 4 is one group, 5-8 is one group, 9 is one group. For example:

  • 100 - C

  • 200 - CC

  • 300 - CCC

  • 400 - CD

  • 500 - D

  • 600 - DC

  • 700 - DCC

  • 800 - DCCC

  • 900 - CM

  1. We will first create a roman character array roman{'M', 'D', 'C', 'L', 'X', 'V', 'I'} and value array value{1000, 500, 100, 50, 10, 5, 1}. We will iterate two steps each time, in other words, 1000, then 100, then 10, then 1, so we can divide it to get the digit.

Solution 1

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class Solution {
    public String intToRoman(int num) {
        StringBuilder res = new StringBuilder();
        char[] roman = {'M', 'D', 'C', 'L', 'X', 'V', 'I'};
        int[] value = {1000, 500, 100, 50, 10, 5, 1};

        for (int i = 0; i < 7; i += 2) {
            int x = num / value[i];
            if (x < 4) {
                for (int j = 0; j < x; j++) {
                    res.append(roman[i]);
                }
            } else if (x == 4) {
                res.append(roman[i]);
                res.append(roman[i-1]);
            } else if (x > 4 && x < 9) {
                res.append(roman[i-1]);
                for (int j = 0; j < x-5; j++) {
                    res.append(roman[i]);
                }
            } else if (x == 9) {
                res.append(roman[i]);
                res.append(roman[i-2]);
            }

            num %= value[i];
        }

        return res.toString();
    }
}

Explanation 2

  1. We can also use greedy approach to solve this problem. From large to small, we create two one dimensional array. int[] val = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1} and String[] str {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}. Loop through the var array, every iteration, while the number num is greater than the current val[i], we subtract var[i], and add the corresponding str[i] to the result.

Solution 2

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

        return res.toString();
    }
}