计蒜课_罗马数字转换为整数

思路总结：这道题重点在于对罗马数字规则的把握，熟悉了罗马数字的规则以后，实现主要思路是

1、输入的罗马数字字符串str；

2、一个字符数组C，在代码中给出c；

在给定的字符数组C中，不断地判断C的当前元素是否为str的前缀，即分别确定str所代表数字的千位，百位，十位，个位：

实现代码如下：

int getLength(const char*str) {//获取字符串的长度；
    int length_of_str;
    for (length_of_str = 0;str[length_of_str] != '';length_of_str++);
    return length_of_str;
}
bool is_prefix(const char*str1,const char*str2) {//判断str1是否为str2的前缀；
    int length_of_str1 = getLength(str1);
    int length_of_str2 = getLength(str2);
    if (length_of_str1 > length_of_str2)
        return false;
    for (int i = 0;i < length_of_str1;i++) {
        if (str1[i] != str2[i]) {
            return false;
        }
    }
    return true;
}
int RomanToInt(const char*str) {//主要实现函数；
    char*c[4][10] = {
        {"","I","II","III","IV","V","VI","VII","VIII","IX"},//1~9；
        {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},//10~90；
        {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},//100~900；
        {"","M","MM","MMM"},//1000~3000;
    };
    int result = 0;//存放结果；
    int length_of_str = getLength(str);//获取待转换为罗马数字的字符串的长度；
    int currentrow=3;//初始位置的列数；
    int currentcolumn=3;//初始位置的行数；
    int current_length;
    while((currentrow>=0)&&(str[0] != '')){
        current_length = getLength(c[currentrow][currentcolumn]);
        if (is_prefix(c[currentrow][currentcolumn],str)) {
            result = result*10+currentcolumn;//更新result;
            currentrow--;
            currentcolumn = 9;
            str += current_length;//若对字符串的前缀判断成功，则将字符串str的指针右移相应长度；
        }
        else {
            if (--currentcolumn == 0) {
                currentrow--;
                currentcolumn = 9;
            }
        }

    }
    return result;
}