二进制求和（LintCode）

 1 public class Solution {
 2     /**
 3      * @param a a number
 4      * @param b a number
 5      * @return the result
 6      */
 7     public String addBinary(String a, String b) {
 8         int c = 0;
 9         int al = a.length() - 1;
10         int bl = b.length() - 1;
11         
12         String s = "";
13         for(;al >= 0 && bl >= 0;al--,bl--) {
14             int ax = Integer.valueOf(a.substring(al,al + 1)).intValue();
15             int bx = Integer.valueOf(b.substring(bl,bl + 1)).intValue();
16             s = (ax + bx + c) % 2 + s;
17             c = (ax + bx + c) / 2;
18         }
19         
20         if(bl >= 0) {
21             while(bl != -1) {
22                 int x = Integer.valueOf(b.substring(bl,bl + 1)).intValue();
23                 s = (c + x) % 2 + s;
24                 c = (c + x) / 2;
25                 bl--;
26             }
27         }
28         
29         if(al >= 0) {
30             while(al != -1) {
31                 int x = Integer.valueOf(a.substring(al,al + 1)).intValue();
32                 s = (c + x) % 2 + s;
33                 c = (c + x) / 2;
34                 al--;
35             }
36         }
37         
38         if(c != 0) {
39             s = c + s;
40         }
41         
42         return s;
43     }
44 }