[LeetCode]Add Binary

题目描述：
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
解题方案：
 1 class Solution {
 2 public:
 3     string addBinary(string a, string b) {
 4         int LocationA = a.size() - 1;
 5         int LocationB = b.size() - 1;
 6 
 7         if (LocationA == -1) {
 8             return b;
 9         }
10         if (LocationB == -1) {
11             return a;
12         }
13         string result;
14         int carry = 0;
15 
16         while((LocationA >= 0) && (LocationB >= 0)) {
17             int temp = ((a[LocationA] - '0') + (b[LocationB] - '0') + carry) % 2;
18             carry    = ((a[LocationA] - '0') + (b[LocationB] - '0') + carry) / 2;
19             result.append(1, temp + '0');
20             --LocationA;
21             --LocationB;
22         }
23         if (LocationA != -1) {
24             while(LocationA >= 0) {
25                 int temp = ((a[LocationA] - '0') + carry) % 2;
26                 carry    = ((a[LocationA] - '0') + carry) / 2;
27                 result.append(1, temp + '0');
28                 --LocationA;
29             }
30         }
31         if (LocationB != -1) {
32             while(LocationB >= 0) {
33                 int temp = ((b[LocationB] - '0') + carry) % 2;
34                 carry    = ((b[LocationB] - '0') + carry) / 2;
35                 result.append(1, temp + '0');
36                 --LocationB;
37             }
38         }
39 
40         if (carry != 0) {
41             result.append(1, carry + '0');
42         }
43 
44         int i = 0;
45         int j = result.size() - 1;
46         //将字符串反转
47         while (i <= j) {
48             char temp = result[i];
49             result[i] = result[j];
50             result[j] = temp;
51             ++i;
52             --j;
53         }
54         return result;
55     }
56 };