找出字符串是否为罗马数字的最佳方法是什么？

Question

几乎所有的例子都是关于将罗马数字转换为整数。 将所有罗马数字一一转换然后检查该数字是否大于 0 是否有意义？

Answer 1

不确定你是否能从盒子里找到任何东西。 创建自己的函数：

private static Dictionary<char, int> _romanMap = new Dictionary<char, int>
{
   {'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}
};


public static int IsRoman(string text) {
  foreach (var c in text) 
      if (!_romanMap.ContainsKey(char))
          return 0;
   return 1 
 }

Answer 2

你可以做这样的事情......不漂亮但是......

public class Solution {
    public int RomanToInt(string s) {

        var total = 0;
        var currentNumber = 0;
        var lastNumber = 0;
        var lastRoman = '/';

        // Take results from 's' and iterate through chars
        foreach (var roman in s)
        {
            if (roman.Equals('I'))
            {
                currentNumber = 1;
            }
            else if (roman.Equals('V'))
            {
                currentNumber = 5;   
            }
            else if (roman.Equals('X'))
            {
                currentNumber = 10;       
            }
            else if (roman.Equals('L'))
            {
                currentNumber = 50;                 
            }
            else if (roman.Equals('C'))
            {
                currentNumber = 100;          
            }
            else if (roman.Equals('D'))
            {
                currentNumber = 500;  
            }
            else if (roman.Equals('M'))
            {
                currentNumber = 1000;
            }
            // Account for edge cases
            if (roman.Equals('V') && lastRoman.Equals('I') || roman.Equals('X') && lastRoman.Equals('I') || // I can be placed before V (5) and X (10) to make 4 and 9. 
                roman.Equals('L') && lastRoman.Equals('X') || roman.Equals('C') && lastRoman.Equals('X') || // X can be placed before L (50) and C (100) to make 40 and 90. 
                roman.Equals('D') && lastRoman.Equals('C') || roman.Equals('M') && lastRoman.Equals('C') )  // C can be placed before D (500) and M (1000) to make 400 and 900.
            {
                var subtract = currentNumber - lastNumber;                 
                total = total - lastNumber + subtract;                        
            }
            else
            {
                total = currentNumber + total;     
            }

            lastRoman = roman;
            lastNumber = currentNumber;
        }

        return total;
    }
}

我还没有对此进行广泛的测试，但是在我编写它的时候它通过了 leetcode 检查。