13.roman-to-interger.go

/*
 * [13] Roman to Integer
 *
 * https://leetcode.com/problems/roman-to-integer/description/
 *
 * algorithms
 * Easy (48.88%)
 * Total Accepted:    257.4K
 * Total Submissions: 526.6K
 * Testcase Example:  '"III"'
 *
 * 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: C = 100, L = 50, XXX = 30 and III = 3.
 *
 *
 * Example 5:
 *
 *
 * Input: "MCMXCIV"
 * Output: 1994
 * Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
 *
 */
func romanToInt(s string) int {
	pos := 0
	result := 0

	symbol := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
	value := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
	length := len(symbol)

	for i := 0; i < length; i++ {
		for pos+len(symbol[i]) <= len(s) && s[pos:pos+len(symbol[i])] == symbol[i] {
			pos += len(symbol[i])
			result += value[i]
		}
	}
	return result

}