Roman to Integer

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

 

 

 

Runtime: 59ms

 1 class Solution {
 2 public:
 3     int romanToInt(string s) {
 4         if (s.empty()) return 0;
 5         
 6         int result = 0;
 7         int n = s.size();
 8         unordered_map<char, int> roman;
 9         roman['I'] = 1;
10         roman['V'] = 5;
11         roman['X'] = 10;
12         roman['L'] = 50;
13         roman['C'] = 100;
14         roman['D'] = 500;
15         roman['M'] = 1000;
16         
17         for (int i = 0; i < n; i++) {
18             if (i < n - 1) {
19                 if (s[i] == 'I' && s[i + 1] == 'V') {result += 4; i++;}
20                 else if (s[i] == 'I' && s[i + 1] == 'X') {result += 9; i++;}
21                 else if (s[i] == 'X' && s[i + 1] == 'L') {result += 40; i++;}
22                 else if (s[i] == 'X' && s[i + 1] == 'C') {result += 90; i++;}
23                 else if (s[i] == 'C' && s[i + 1] == 'D') {result += 400; i++;}
24                 else if (s[i] == 'C' && s[i + 1] == 'M') {result += 900; i++;}
25                 else result += roman[s[i]];
26             }
27             else {
28                 result += roman[s[i]];
29             }
30         }
31         return result;
32     }
33 };

 

Runtime: 92ms

 1 class Solution {
 2 public:
 3     int romanToInt(string s) {
 4         if (s.empty()) return 0;
 5         
 6         int result = 0;
 7         int n = s.size();
 8         unordered_map<string, int> roman;
 9         roman["I"] = 1;
10         roman["V"] = 5;
11         roman["X"] = 10;
12         roman["L"] = 50;
13         roman["C"] = 100;
14         roman["D"] = 500;
15         roman["M"] = 1000;
16         roman["IV"] = 4;
17         roman["IX"] = 9;
18         roman["XL"] = 40;
19         roman["XC"] = 90;
20         roman["CD"] = 400;
21         roman["CM"] = 900;
22         
23         for (int i = 0; i < n;) {
24             if (i < n - 1) {
25                 string temp = s.substr(i, 2);
26                 if (roman.find(temp) != roman.end()) {
27                     result += roman[temp];
28                     i += 2;
29                     continue;
30                 }
31             }
32             result += roman[s.substr(i, 1)];
33             i++;
34         }
35         return result;
36     }
37 };

 

posted @ 2016-09-29 05:41  amazingzoe  阅读(88)  评论(0)  编辑  收藏  举报