LeetCode 题解（73）: Roman to Integer
题目：
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
题解：笨办法，switch-case，好办法，hashmap。
C++：
class Solution {
public:
int romanToInt(string s) {
int result = 0;
if(!s.length())
return result;
for(int i = 0; i < s.length(); i++) {
switch(s[i]) {
case 'M':
result += 1000;
break;
case 'D':
result += 500;
break;
case 'C':
if(i + 1 < s.length()) {
switch(s[i+1]) {
case 'D':
result += 400;
i++;
break;
case 'M':
result += 900;
i++;
break;
default:
result += 100;
break;
}
} else {
result += 100;
}
break;
case 'L':
result += 50;
break;
case 'X':
if(i + 1 < s.length()) {
switch(s[i+1]) {
case 'L':
result += 40;
i++;
break;
case 'C':
result += 90;
i++;
break;
default:
result += 10;
break;
}
}
else {
result += 10;
}
break;
case 'V':
result += 5;
break;
case 'I':
if(i + 1 < s.length()) {
switch(s[i+1]){
case 'V':
result += 4;
i++;
break;
case 'X':
result += 9;
i++;
break;
default:
result += 1;
break;
}
}
else {
result += 1;
}
break;
default:
break;
}
}
return result;
}
};
Java:
public class Solution {
public int romanToInt(String s) {
int result = 0;
if(s.length() == 0)
return 0;
for(int i = 0; i < s.length(); i++) {
switch(s.charAt(i)){
case 'M':
result += 1000;
break;
case 'D':
result += 500;
break;
case 'C':
if(i + 1 < s.length()) {
switch(s.charAt(i+1)) {
case 'D':
result += 400;
i++;
break;
case 'M':
result += 900;
i++;
break;
default:
result += 100;
break;
}
} else {
result += 100;
}
break;
case 'L':
result += 50;
break;
case 'X':
if(i + 1 < s.length()) {
switch(s.charAt(i+1)) {
case 'L':
result += 40;
i++;
break;
case 'C':
result += 90;
i++;
break;
default:
result += 10;
break;
}
} else {
result += 10;
}
break;
case 'V':
result += 5;
break;
case 'I':
if(i + 1 < s.length()) {
switch(s.charAt(i+1)) {
case 'V':
result += 4;
i++;
break;
case 'X':
result += 9;
i++;
break;
default:
result += 1;
break;
}
} else {
result += 1;
}
break;
}
}
return result;
}
}
Python:
class Solution:
# @return an integer
def romanToInt(self, s):
if len(s) == 0:
return 0
d = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D': 500, 'M': 1000}
result = 0
while len(s) > 0:
if len(s) == 1 or d[s[0]] >= d[s[1]]:
result += d[s[0]]
s = s[1:]
else:
result = result + d[s[1]] - d[s[0]]
s = s[2:]
return result