C#LeetCode刷题之#13-罗马数字转整数(Roman to Integer)

C#LeetCode刷题之#13-罗马数字转整数(Roman to Integer)

问题

罗马数字包含以下七种字符: I, V, X, L,C,D 和 M。

字符          数值
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

例如, 罗马数字 2 写做 II ,即为两个并列的 1。12 写做 XII ,即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。

通常情况下,罗马数字中小的数字在大的数字的右边。但也存在特例,例如 4 不写做 IIII,而是 IV。数字 1 在数字 5 的左边,所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地,数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况:

I 可以放在 V (5) 和 X (10) 的左边,来表示 4 和 9。
X 可以放在 L (50) 和 C (100) 的左边,来表示 40 和 90。 
C 可以放在 D (500) 和 M (1000) 的左边,来表示 400 和 900。
给定一个罗马数字,将其转换成整数。输入确保在 1 到 3999 的范围内。

输入: “III”

输出: 3

输入: “IV”

输出: 4

输入: “IX”

输出: 9

输入: “LVIII”

输出: 58

解释: L = 50, V= 5, III = 3.

输入: “MCMXCIV”

输出: 1994

解释: M = 1000, CM = 900, XC = 90, IV = 4.

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.

Input: “III”

Output: 3

Input: “IV”

Output: 4

Input: “IX”

Output: 9

Input: “LVIII”

Output: 58

Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Input: “MCMXCIV”

Output: 1994

Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

示例

public class Program {

    public static void Main(string[] args) {
        var s = "MCMXCIV";

        var res = RomanToInt(s);
        Console.WriteLine(res);

        s = "LVIII";

        res = RomanToInt2(s);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static Dictionary<char, int> _dic = new Dictionary<char, int> {
        {'I',1},
        {'V',5},
        {'X',10},
        {'L',50},
        {'C',100},
        {'D',500},
        {'M',1000},
        {'a',4},
        {'b',9},
        {'c',40},
        {'d',90},
        {'e',400},
        {'f',900},
    };

    private static int RomanToInt(string s) {
        var res = 0;
        s = s.Replace("IV", "a")
             .Replace("IX", "b")
             .Replace("XL", "c")
             .Replace("XC", "d")
             .Replace("CD", "e")
             .Replace("CM", "f");
        for(var i = 0; i < s.Length; i++) {
            res += _dic[s[i]];
        }
        return res;
    }

    private static Dictionary<char, int> _dic2 = new Dictionary<char, int> {
        {'I',1},
        {'V',5},
        {'X',10},
        {'L',50},
        {'C',100},
        {'D',500},
        {'M',1000},
    };

    private static int RomanToInt2(string s) {
        var res = 0;
        for(var i = 0; i < s.Length; i++) {
            if(i == s.Length - 1 || _dic2[s[i + 1]] <= _dic2[s[i]]) {
                res += _dic2[s[i]];
            } else {
                //右边比左边大要减
                res -= _dic2[s[i]];
            }
        }
        return res;
    }

}

以上给出2种算法实现,以下是这个案例的输出结果:

1994
58

分析:

显而易见,以上2种算法的时间复杂度均为: O(n) 。

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