[LeetCode-13]-Roman to Integer (将罗马数字转换为整数)

 

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题目相关

【题目解读】
将罗马数字转换为整数，其中「I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000」。IV、IX、XL、XC、CD、CM这几种情况需要做减法，要特殊处理。

【原题描述】原题链接
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.

【难度】Easy

Solution

该题比较直观，采用的是逐个遍历并通过switch判断，遇到IV、IX、XL、XC、CD、CM这些情况时再恢复数据。

采用switch代码逻辑清晰，不好的是代码过长。

该程序耗时： 16 ms，运行时间超过了99.46 % 的C++提交代码，内存使用情况超过100%的C++代码。

下面是程序代码：

    int romanToInt(string s)
    {
        int ret = 0;
        for (size_t i = 0; i < s.size(); i++)
        {
            /* code */
            switch (s[i])
            {
                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'I':
                ret += 1;
                break;

                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'V':
                ret += 5;

                if (s[i - 1] == 'I')
                {
                    //IV的这种情况需要恢复
                    ret -= 2;
                }
                break;
                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'X':
                ret += 10;
                if (s[i - 1] == 'I')
                {
                    //IX的这种情况需要恢复
                    ret -= 2;
                }
                break;

                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'L':
                ret += 50;
                if (s[i - 1] == 'X')
                {
                    //XL的这种情况需要恢复
                    ret -= 20;
                }
                break;

                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'C':
                ret += 100;
                if (s[i - 1] == 'X')
                {
                    //XC的这种情况需要恢复
                    ret -= 20;
                }
                break;

                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'D':
                ret += 500;
                if (s[i - 1] == 'C')
                {
                    //CD的这种情况需要恢复
                    ret -= 200;
                }
                break;

                //I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; M = 1000
            case 'M':
                ret += 1000;
                if (s[i - 1] == 'C')
                {
                    //CM的这种情况需要恢复
                    ret -= 200;
                }
                break;

            default:
                return 0;
            }//end switch
        }//end for

        return ret;
    }

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