5.Integer to Roman && Roman to Integer

Roman chart: http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm

Integer to Roman

Given an integer, convert it to a roman numeral.

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

解析：只有数字为 9 和 4 时，取一个小数放在左边。（把数字为 9 和 4 时的罗马符号看作一个原子，避免取小数过程。）

class Solution {
public:
    string intToRoman(int num) {
        string Roman[13] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        int val[13] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
        string s;
        for(int i = 0; i < 13; ++i){
            while(num >= val[i]){
                num -= val[i];
                s += Roman[i];
            }
        }
        return s;
    }
};

 Roman to Integer

Given a roman numeral, convert it to an integer.

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

解析：1. 最直观想法，对字符串按罗马值从大到小比对。 (372 ms)

bool inHead(string s1, string& s2)
{
    if(s2 == "") return false;
    int k = 0;
    auto it = s1.begin();
    for(; it != s1.end(); ++it)
    {
        if(*it != s2[k++]) return false;
    }
    if(it == s1.end()){
        s2.erase(0, k);
        return true;
    }
}

class Solution {
public:
    int romanToInt(string s) {
        string sigmal[13] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        int a[13] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
        int num = 0;
        int k = 0;
        for(int i = 0; i < 13; ++i){
            while(inHead(sigmal[i], s)){
                num += a[i];
            }
        }
        return num;
    }
};

2. 利用 hash 函数来做。(每个罗马字符范围都在 'A' - 'Z' 之间，对字符串从左到右扫描，若是出现逆序（按罗马值），则这个逆序对为一个值) (260ms)

class Solution {
public:
    int romanToInt(string s) {
        char c[13] = {'M', 'D', 'C', 'L', 'X', 'V', 'I'};
        int v[13] = {1000, 500, 100, 50, 10, 5, 1 };
        int hash[256] = {0};
        for(int i = 0; i < 7; ++i){
            hash[c[i]] = v[i];
        }
        int num = 0;
        int len = s.length();
        for(int i = 0; i < len; ++i){
            if(i == len-1){
                num += hash[s[i]];
                return num;
            }
            if(hash[s[i]] < hash[s[i+1]]){
                num += hash[s[i+1]] - hash[s[i]];
                ++i;
            }else num += hash[s[i]];
        }
        return num;
    }
};