Roman numerals are represented by seven different symbols: I
, V
, X
, L
, C
, D
and M
.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, two is written as II
in Roman numeral, just two one's added together. Twelve is written as, XII
, which is simply X
+ II
. The number twenty seven is written as XXVII
, which is XX
+ V
+ II
.
Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII
. Instead, the number four is written as IV
. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX
. There are six instances where subtraction is used:
-
I
can be placed beforeV
(5) andX
(10) to make 4 and 9. -
X
can be placed beforeL
(50) andC
(100) to make 40 and 90. -
C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
Example 1:
Input: "III"
Output: 3
Example 2:
Input: "IV"
Output: 4
Example 3:
Input: "IX"
Output: 9
Example 4:
Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 5:
Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
罗马数转化成数字问题,我们需要对于罗马数字很熟悉才能完成转换。以下截自百度百科:
I - 1
V - 5
X - 10
L - 50
C - 100
D - 500
M - 1000
class Solution {
public:
int romanToInt(string s) {
int res = ;
unordered_map<char, int> m{{'I', }, {'V', }, {'X', }, {'L', }, {'C', }, {'D', }, {'M', }};
for (int i = ; i < s.size(); ++i) {
int val = m[s[i]];
if (i == s.size() - || m[s[i+]] <= m[s[i]]) res += val;
else res -= val;
}
return res;
}
};
我们也可以每次跟前面的数字比较,如果小于等于前面的数字,先加上当前的数字,比如 "VI",第二个字母 'I' 小于第一个字母 'V',所以要加1。如果大于的前面的数字,加上当前的数字减去二倍前面的数字,这样可以把在上一个循环多加数减掉,比如 "IX",我们在 i=0 时,加上了第一个字母 'I' 的值,此时结果 res 为1。当 i=1 时,字母 'X' 大于前一个字母 'I',这说明前面的1是要减去的,而由于前一步不但没减,还多加了个1,所以此时要减去2倍的1,就是减2,所以才能得到9,整个过程是 res = 1 + 10 - 2 = 9,参见代码如下:
解法二:
class Solution {
public:
int romanToInt(string s) {
int res = ;
unordered_map<char, int> m{{'I', }, {'V', }, {'X', }, {'L', }, {'C', }, {'D', }, {'M', }};
for (int i = ; i < s.size(); ++i) {
if (i == || m[s[i]] <= m[s[i - ]]) res += m[s[i]];
else res += m[s[i]] - * m[s[i - ]];
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/13
类似题目:
参考资料:
https://leetcode.com/problems/roman-to-integer/
https://leetcode.com/problems/roman-to-integer/discuss/6547/Clean-O(n)-c%2B%2B-solution