LeetCode 13. Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

罗马数字：可以查看维基百科的定义

.

最大为MMMCMXCIX,我们需要注意的是9的写法是IX，4的写法是IV

这样我们遍历整个字符串后就可以得到结果，也可以将求值重新封装

成一个新的函数，因为好多部分都是重复的，下面是代码

class Solution {
public:
    int romanToInt(string s) {
        int sum = 0;
        int cnt = 0;
        int i = 0;
            if (i + cnt<s.size() && s[i] == 'M')
            {
                ++cnt;
                for (;s[i + cnt] == 'M';++cnt);
                sum += cnt * 1000;
                i += cnt;
                cnt = 0;
            }
            if (i + cnt < s.size() && s[i] == 'D')
            {
                ++cnt;
                for (;s[i + cnt] == 'C';++cnt);
                sum += (4 + cnt) * 100;
                i += cnt;
                cnt = 0;    
            }
            if (i + cnt < s.size()&&s[i] == 'C')
            {
                ++cnt;
                for (;s[i + cnt] == 'C';++cnt);
                if (1 == cnt)
                {
                    if (s[i + cnt] != 'M'&&s[i + cnt] != 'D')
                    {
                        sum += 100;
                        i += cnt;
                        cnt = 0;
                    }
                    if (s[i + cnt] == 'M')
                    {
                        ++cnt;
                        sum += 900;
                        i += cnt;
                        cnt = 0;
                    }
                    if (s[i + cnt] == 'D')
                    {
                        ++cnt;
                        sum += 400;
                        i += cnt;
                        cnt = 0;
                    }
                }
                if (2 == cnt)
                {
                    sum += 200;
                    i += cnt;
                    cnt = 0;
                }
                if (3 == cnt)
                {
                    sum += 300;
                    i += cnt;
                    cnt = 0;
                }    
            }
            if (i + cnt < s.size() && s[i] == 'L')
            {
                ++cnt;
                for (;s[i + cnt] == 'X';++cnt);
                sum += (4 + cnt) * 10;
                i += cnt;
                cnt = 0;
            }
            if (i + cnt < s.size() && s[i] == 'X')
            {
                ++cnt;
                for (;s[i + cnt] == 'X';++cnt);
                if (1 == cnt)
                {
                    if (s[i + cnt] != 'L'&&s[i + cnt] != 'C')
                    {
                        sum += 10;
                        i += cnt;
                        cnt = 0;
                    }
                    if (s[i + cnt] == 'C')
                    {
                        ++cnt;
                        sum += 90;
                        i += cnt;
                        cnt = 0;
                    }
                    if (s[i + cnt] == 'L')
                    {
                        ++cnt;
                        sum += 40;
                        i += cnt;
                        cnt = 0;
                    }
                }
                if (2 == cnt)
                {
                    sum += 20;
                    i += cnt;
                    cnt = 0;
                }
                if (3 == cnt)
                {
                    sum += 30;
                    i += cnt;
                    cnt = 0;
                }
            }
                if (i + cnt < s.size() && s[i] == 'V')
                {
                    ++cnt;
                    for (;s[i + cnt] == 'I';++cnt);
                    sum += (4 + cnt) * 1;
                    i += cnt;
                    cnt = 0;
                }
                if (i + cnt < s.size() && s[i] == 'I')
                {
                    ++cnt;
                    for (;s[i + cnt] == 'I';++cnt);
                    if (1 == cnt)
                    {
                        if (s[i + cnt] != 'V'&&s[i + cnt] != 'X')
                        {
                            sum += 1;
                            i += cnt;
                            cnt = 0;
                        }
                        if (s[i + cnt] == 'X')
                        {
                            ++cnt;
                            sum += 9;
                            i += cnt;
                            cnt = 0;
                        }
                        if (s[i + cnt] == 'V')
                        {
                            ++cnt;
                            sum += 4;
                            i += cnt;
                            cnt = 0;
                        }
                    }
                    if (2 == cnt)
                    {
                        sum += 2;
                        i += cnt;
                        cnt = 0;
                    }
                    if (3 == cnt)
                    {
                        sum += 3;
                        i += cnt;
                        cnt = 0;
                    }
                }
            return sum;
    }
};