[Leetcode 17] 13 Roman to Integer

Problem:

Given a roman numeral, convert it to an integer.

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

 

Analysis:

Some Roman Number examples can be seen here: http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm. 

Use a one pass scan to decide whether add or subtract a number from current sum.

If a number is greater or equal to its successor or it's the last number, add it.

If a numver is less than its successor, subtract it.

 

Code:

 1 public class Solution {
 2     public int romanToInt(String s) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int sum = 0;
 6         
 7         for (int i=0; i<s.length(); ++i) {
 8             int numi = rtoi(s.charAt(i));
 9             
10             if (i==(s.length()-1)) {
11                 sum += numi;
12             } else if ( numi >= rtoi(s.charAt(i+1))) {
13                 sum += numi;
14             } else {
15                 sum -= numi;
16             }
17         }
18     
19         return sum;
20     }
21     
22     private int rtoi(char x) {
23         int val=0; 
24         
25         switch (x) {
26             case 'I': val = 1; break;
27             case 'V': val = 5; break;
28             case 'X': val = 10; break;
29             case 'L': val = 50; break;
30             case 'C': val = 100; break;
31             case 'D': val = 500; break;
32             case 'M': val = 1000; break;
33         }
34         
35         return val;
36     }
37 }
View Code

 

An efficient version that call helper function only once in each iteration

 1 public class Solution {
 2     public int romanToInt(String s) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int sum = 0, numi=0, numiSucc=rtoi(s.charAt(0));
 6         
 7         for (int i=0; i<s.length(); ++i) {
 8             numi = numiSucc;
 9             
10             if (i==(s.length()-1)) {
11                 sum += numi;
12                 break;
13             }
14             
15             numiSucc = rtoi(s.charAt(i+1));
16             if ( numi >= numiSucc) {
17                 sum += numi;
18             } else {
19                 sum -= numi;
20             }
21         }
22     
23         return sum;
24     }
25     
26     private int rtoi(char x) {
27         int val=0; 
28         
29         switch (x) {
30             case 'I': val = 1; break;
31             case 'V': val = 5; break;
32             case 'X': val = 10; break;
33             case 'L': val = 50; break;
34             case 'C': val = 100; break;
35             case 'D': val = 500; break;
36             case 'M': val = 1000; break;
37         }
38         
39         return val;
40     }
41 }
View Code

 

Attention:

Know what is Roman Number before coding.

posted on 2013-05-19 02:11  freeneng  阅读(573)  评论(0)  编辑  收藏  举报

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