罗马数字与整数相互转化

leetcode中的题目

直接上代码：

def int2roman(num):
    ret = ""
    lists = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
    chars = ("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I")
    while num > 0:
        i = 0
        while i <= 12:
            if lists[i] <= num:
                break
            i += 1
        ret += chars[i]
        num -= lists[i]
    return ret

def int2roman2(num):
    M = ("", "M", "MM", "MMM")
    C = ("","C","CC","CCC","CD","D","DC","DCC","DCCC","CM")
    X = ("","X","XX","XXX","XL","L","LX","LXX","LXXX","XC")
    I = ("","I","II","III","IV","V","VI","VII","VIII","IX")
    return M[(num/1000)%10]+C[(num/100)%10]+X[(num/10)%10]+I[num%10]

上面是整数转罗马数字的，第一种方法是把所有可能的字符表示出来，然后1-3999的数字是由其相加组成的，依次从最大的先加上去，懂得原理会比较好理解；

第二中是将每一位的可能表示出来，直接算出每一位再替换上去，也比较简单。

下面是罗马数字转整数：

def roman2int(s):
    mapval = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
    cur = 0
    pre = 1001
    ret = 0
    for i in s:
        ret += mapval[i]
        cur = mapval[i]
        if pre < cur:
            ret -= 2 * pre
        pre = cur
    return ret

前面的数字比后面的数字小是减法，比如IV = 5 - 1

先把所有的数字加起来，如果有前面比后面小的就减去自己的两倍即可。