Leetcode_roman-to-integer

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地址：https://oj.leetcode.com/problems/roman-to-integer/

Given a roman numeral, convert it to an integer.

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

思路：罗马数字的维基百科 http://zh.wikipedia.org/zh/%E7%BD%97%E9%A9%AC%E6%95%B0%E5%AD%97

左减数字值限于I, X, C，且只能左减一次, 比如8 VIII 不能是IIX

右加最多连续三个，如9是 IX不能是VIIII

参考代码：

class Solution {public:    inline int c2i(char ch)    {        switch(ch)        {            case 'I' : return 1;            case 'V' : return 5;            case 'X' : return 10;            case 'L' : return 50;            case 'C' : return 100;            case 'D' : return 500;            case 'M' : return 1000;        }    }    int romanToInt(string s) {        int len = s.length(), ans = c2i(s[0]), pre = 0, cur = 0;        if(len==1)            return ans;        for(int i = 1; i<len; ++i)        {            pre = c2i(s[i-1]);            cur = c2i(s[i]);            if(pre >= cur)                ans += cur;            else                ans += (cur - 2*pre);        }        return ans;    }};

SECOND TRIAL

class Solution {public:    inline int c2n(char ch)    {        switch(ch)        {            case 'I': return 1;            case 'V': return 5;            case 'X': return 10;            case 'L': return 50;            case 'C': return 100;            case 'D': return 500;            case 'M': return 1000;            default: return 0;        }    }    int romanToInt(string s) {        int ans = 0, last = 0, cur = 0;        for(int i = 0; i<s.length(); ++i)        {            cur = c2n(s[i]);            ans += cur;            if(i && cur>last)                ans -= 2*last;            last = cur;        }        return ans;    }};

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