13. Roman to Integer (E)

Roman to Integer (E)

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

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

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

Example 5:

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

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题意

将给定的罗马数字转化为整数。

思路

罗马数字的规律是，除了4、9等这类特殊的数字，所有的大数字都在小数字的左边，当小数排在大数左侧时，说明这个小数是需要被减去的。参照此规律，从右往左遍历字符，如果当前数字比右边的数字(若存在)小，则需要在总和中减去当前数，否则加上当前数。
为了提高效率，先将字符串转化为数组，每一个位置存储各位字符对应的数字。

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代码实现

class Solution {
    public int romanToInt(String s) {
        int ans = 0;
        Map<Character, Integer> map = new HashMap<>();
        map.put('I', 1);
        map.put('V', 5);
        map.put('X', 10);
        map.put('L', 50);
        map.put('C', 100);
        map.put('D', 500);
        map.put('M', 1000);
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (i < s.length() - 1 && map.get(c) < map.get(s.charAt(i + 1))) {
                ans -= map.get(c);
            } else {
                ans += map.get(c);
            }
        }
        return ans;
    }
}

代码实现 - 简化

class Solution {
    public int romanToInt(String s) {
        int ans = 0;
        int[] nums = new int[s.length()];
        for (int i = 0; i < s.length(); i++) {
            switch (s.charAt(i)) {
                case 'I':
                    nums[i] = 1;
                    break;
                case 'V':
                    nums[i] = 5;
                    break;
                case 'X':
                    nums[i] = 10;
                    break;
                case 'L':
                    nums[i] = 50;
                    break;
                case 'C':
                    nums[i] = 100;
                    break;
                case 'D':
                    nums[i] = 500;
                    break;
                case 'M':
                    nums[i] = 1000;
                    break;
            }
        }

        for (int i = nums.length - 1; i >= 0; i--) {
            if (i < nums.length - 1 && nums[i] < nums[i + 1]) {
                ans -= nums[i];
            } else {
                ans += nums[i];
            }
        }
        return ans;
    }
}