LeetCode #13:Roman to Integer(罗马数字转整数)

题目描述:

思路:

多个switch-case嵌套即可,注意题目中的六种特例情况。

class Solution {
public:
    int romanToInt(string s) {
        int sum = 0;
        for(int i = 0; i < s.length(); i++)
        {
            switch (s[i])
            {
                case 'I':
                    if (s[i+1] != 'V' && s[i+1] != 'X')
                        sum += 1;
                    else
                    {
                        switch (s[i+1])
                        {
                            case 'V':
                                sum += 4;
                                break;
                            case 'X':
                                sum += 9;
                                break;
                        }
                        i++;
                    }
                    break;
                case 'V':
                    sum += 5;
                    break;
                case 'X':
                    if (s[i+1] != 'L' && s[i+1] != 'C')
                        sum += 10;
                    else
                    {
                        switch (s[i+1])
                        {
                            case 'L':
                                sum += 40;
                                break;
                            case 'C':
                                sum += 90;
                                break;
                        }
                        i++;
                    }
                    break;
                case 'L':
                    sum += 50;
                    break;
                case 'C':
                    if (s[i+1] != 'D' && s[i+1] != 'M')
                        sum += 100;
                    else
                    {
                        switch (s[i+1])
                        {
                            case 'D':
                                sum += 400;
                                break;
                            case 'M':
                                sum += 900;
                                break;
                        }
                        i++;
                    }
                    break;
                case 'D':
                    sum += 500;
                    break;
                case 'M':
                    sum += 1000;
                    break;
            }
        }
        return sum;
    }
};

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