leetcode中的题目
直接上代码：
1 def int2roman(num): 2 ret = "" 3 lists = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1) 4 chars = ("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I") 5 while num > 0: 6 i = 0 7 while i <= 12: 8 if lists[i] <= num: 9 break 10 i += 1 11 ret += chars[i] 12 num -= lists[i] 13 return ret 14 15 def int2roman2(num): 16 M = ("", "M", "MM", "MMM") 17 C = ("","C","CC","CCC","CD","D","DC","DCC","DCCC","CM") 18 X = ("","X","XX","XXX","XL","L","LX","LXX","LXXX","XC") 19 I = ("","I","II","III","IV","V","VI","VII","VIII","IX") 20 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：
1 def roman2int(s): 2 mapval = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000} 3 cur = 0 4 pre = 1001 5 ret = 0 6 for i in s: 7 ret += mapval[i] 8 cur = mapval[i] 9 if pre < cur: 10 ret -= 2 * pre 11 pre = cur 12 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。