笔试题26. LeetCode OJ (13)

哈哈，有了上个题的经验，我们是不是应该趁热打铁呢？这次是讲一个罗曼的(罗马)数字转化为一个整形数字，所以我们逆向去推理上个题，其实这个题很有规律的，将string数组遍历一遍就可以求解了，废话不多说，见代码：

class Solution {
public:
    int romanToInt(string s) {
        //和上道题刚好相反
        int lastnum = 0;
        int len = s.size();
        if(len < 1)
        {
            return lastnum;
        }
        int i = 0;
        while( i < len)
        {
            switch(s[i])
            {
                case 'M': //1000
                        lastnum+=1000;
                        ++i;
                    break;
                case 'D': //500
                        lastnum+=500;
                        i++;
                    break;
                case 'C': //100
                {
                    if(i+1 <len && s[i+1] == 'M') //900
                    {
                        lastnum+=900;
                        i=i+2;
                    }
                    else if(i+1 <len && s[i+1] == 'D') //400
                    {
                        lastnum+=400;
                        i=i+2;
                    }
                    else //100
                    {
                        lastnum+=100;
                        ++i;
                    }
                    break;
                }
                case 'L':
                        lastnum+=50;
                        ++i;
                    break;
                case 'X':
                {
                    if(i+1 <len && s[i+1] == 'C')
                    {
                        lastnum+=90;
                        i=i+2;
                    }
                    else if(i+1 <len && s[i+1] == 'L')
                    {
                        lastnum+=40;
                        i=i+2;
                    }
                    else
                    {
                        lastnum+=10;
                        ++i;
                    }
                    break;
                }
                case 'V':
                {
                    lastnum+=5;
                    ++i;
                    break;
                }
                case 'I':
                {
                    if(i+1 <len && s[i+1] == 'X')
                    {
                        lastnum+=9;
                        i=i+2;
                    }
                    else if(i+1 <len && s[i+1] == 'V')
                    {
                        lastnum+=4;
                        i=i+2;
                    }
                    else
                    {
                        lastnum+=1;
                        ++i;
                    }
                    break;
                }
                default :
                    break;
            }
        }
        
        return lastnum;
    }
};

最后还是附上这个吧！^_^

笔试题26. LeetCode OJ (13)

原文：http://blog.csdn.net/zr1076311296/article/details/51164836