[leetcode]Roman to Integer
阿新 • • 發佈:2018-12-17
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
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
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: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
分析:
羅馬數字轉整數主要需要特別注意“IV、IX、XL、XC、CD、CM”這幾種情況,其餘情況直接加上羅馬字母對應的值即可。
class Solution {
public:
int romanToInt(string s) {
int res = 0;
for(int i=0; i<s.size(); i++)
{
if(s[i] == 'V')
res += 5;
else if(s[i] == 'L')
res += 50;
else if(s[i] == 'D')
res += 500;
else if(s[i] == 'M')
res += 1000;
else if(s[i] == 'I')
{
if(s[i+1]=='V')
{
res += 4;
i++;
}
else if(s[i+1]=='X')
{
res += 9;
i++;
}
else
res +=1;
}
else if(s[i] == 'X')
{
if(s[i+1]=='L')
{
res += 40;
i++;
}
else if(s[i+1]=='C')
{
res += 90;
i++;
}
else
res +=10;
}
else if(s[i] == 'C')
{
if(s[i+1]=='D')
{
res += 400;
i++;
}
else if(s[i+1]=='M')
{
res += 900;
i++;
}
else
res +=100;
}
}
return res;
}
};