[LeetCode] 13. Roman to Integer

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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 a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

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Input: "III"
Output: 3

Example 2:

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Input: "IV"
Output: 4

Example 3:

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Input: "IX"
Output: 9

Example 4:

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

Example 5:

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Input: "MCMXCIV"
Output: 1994
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. First, we can create a HashMap to map the roman character with its corresponding integer.

  2. Loop through each character, if the current character is the last character or its next character’s corresponding integer is less than the current character’s corresponding integer, then we add the current character’s corresponding integer to the result.

  3. Else, we subtract the current character’s corresponding integer.

Solution 1

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class Solution {
    public int romanToInt(String s) {
        int res = 0;
        Map<Character, Integer> hm = new HashMap<>();
        hm.put('I', 1);
        hm.put('V', 5);
        hm.put('X', 10);
        hm.put('L', 50);
        hm.put('C', 100);
        hm.put('D', 500);
        hm.put('M', 1000);

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (i == s.length() - 1 || hm.get(c) >= hm.get(s.charAt(i+1))) {
                res += hm.get(c);
            } else {
                res -= hm.get(c);
            }
        }

        return res;
    }
}

Explanation 2

  1. Create a HashMap to map the roman character with its corresponding integer.

  2. We can also loop through the roman string. If the current character’s corresponding integer is less than or equal to its previous character’s corresponding integer, then we add the current integer. For example VI, we add 5, add 1. In other words, (5+1=6).

  3. Else if the current character’s corresponding integer is greater than its previous character’s corresponding integer, then we add the current integer and subtract the previous character’s corresponding integer multiple by 2. For example IV, we add 1, add 5, subtract 21. In other words, (1+5-21=4).

Solution 2

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class Solution {
    public int romanToInt(String s) {
        int res = 0;
        Map<Character, Integer> hm = new HashMap<>();
        hm.put('I', 1);
        hm.put('V', 5);
        hm.put('X', 10);
        hm.put('L', 50);
        hm.put('C', 100);
        hm.put('D', 500);
        hm.put('M', 1000);

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (i == 0 || hm.get(c) <= hm.get(s.charAt(i-1))) {
                res += hm.get(c);
            } else {
                res += hm.get(c) - 2*hm.get(s.charAt(i-1));
            }
        }

        return res;
    }
}