[Leetcode 44] 12 Integer To Roman

Problem:

Given an integer, convert it to a roman numeral.

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

 

Analysis:

In roman number system, there are 3 special cases

1. 900  or 400

2. 90 or 40

3. 9 or 4

 

Process the integer number digit by digit, if it's 9 or 4, process it separately. Then if it's 5~8 or it's 0~3, need diferent process.

 

Code:

 1 class Solution {
 2 public:
 3     string intToRoman(int num) {
 4         // Start typing your C/C++ solution below
 5         // DO NOT write int main() function    
 6         
 7         // Roman digit: I:1, V:5, X:10, L:50, C:100, D:500, M:1000
 8         // Special Case: 4:IV, 9:IX, 40:XL, 90:XC, 400:CD, 900:CM
 9         
10         int rom[7] = {1000, 500, 100, 50, 10, 5, 1};
11         char romc[7] = {'M', 'D', 'C', 'L', 'X', 'V', 'I'};
12         char tmp[20];
13         int idx = 0;
14         
15         for (int t = 1000, pt=-1; t > 0; t /= 10, pt+=2) {
16             int digit = num / t;
17             
18             switch(digit) {
19                 case 9: 
20                     tmp[idx++] = romc[pt+1];
21                     tmp[idx++] = romc[pt-1];
22                     break;
23                 case 8: case 7: case 6: case 5: 
24                     tmp[idx++] = romc[pt]; //5
25                     for (int i=0; i<digit-5; i++) {
26                         tmp[idx++] = romc[pt+1];
27                     }
28                     break;
29                 case 4: 
30                     tmp[idx++] = romc[pt+1];
31                     tmp[idx++] = romc[pt];
32                     break;
33                 case 3: case 2: case 1: 
34                     for (int i=0; i<digit; i++) {
35                         tmp[idx++] = romc[pt+1];   
36                     }
37                     break;
38                 case 0: default: break;
39             }
40             num %= t;
41         }
42         
43         tmp[idx] = '\0';
44         string res(tmp);
45         return res;
46     }
47 };
View Code

 

 

Attention:

posted on 2013-05-25 09:40  freeneng  阅读(117)  评论(0)  编辑  收藏  举报

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