题目：
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
题解：
二进制加法都是从最低位（从右加到左）。所以对两个字符串要从最后一位开始加，如果遇见长度不一的情况，就把短的字符串高位补0.
每轮计算要加上进位，最后跳出循环后要坚持进位是否为1，以便更新结果。
代码如下（from discussion）：
1 public String addBinary(String a, String b) {
2 int m = a.length();
3 int n = b.length();
4 int carry = 0;
5 String res = "";
6 // the final length of the result depends on the bigger length between a and b,
7 // (also the value of carry, if carry = 1, add "1" at the head of result, otherwise)
8 int maxLen = Math.max(m, n);
9 for (int i = 0; i < maxLen; i++) {
10 // start from last char of a and b
11 // notice that left side is int and right side is char
12 // so we need to minus the decimal value of '0'
13 int p=0,q=0;
14 if(i<m)
15 p = a.charAt(m-1-i) - '0';
16 else
17 p = 0;
18
19 if(i<n)
20 q = b.charAt(n-1-i)-'0';
21 else
22 q = 0;
23
24 int tmp = p + q + carry;
25 carry = tmp / 2;
26 res += tmp % 2;
27 }
28 return (carry == 0) ? res : "1" + res;
29 }
2 int m = a.length();
3 int n = b.length();
4 int carry = 0;
5 String res = "";
6 // the final length of the result depends on the bigger length between a and b,
7 // (also the value of carry, if carry = 1, add "1" at the head of result, otherwise)
8 int maxLen = Math.max(m, n);
9 for (int i = 0; i < maxLen; i++) {
10 // start from last char of a and b
11 // notice that left side is int and right side is char
12 // so we need to minus the decimal value of '0'
13 int p=0,q=0;
14 if(i<m)
15 p = a.charAt(m-1-i) - '0';
16 else
17 p = 0;
18
19 if(i<n)
20 q = b.charAt(n-1-i)-'0';
21 else
22 q = 0;
23
24 int tmp = p + q + carry;
25 carry = tmp / 2;
26 res += tmp % 2;
27 }
28 return (carry == 0) ? res : "1" + res;
29 }