给定两个二进制字符串，返回他们的和（用二进制表示）。

输入为非空字符串且只包含数字 1 和 0。

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

 1 #include "_000库函数.h"
 2 
 3 //从末尾一位一位的相加
 4 class Solution {
 5 public:
 6     string addBinary(string a, string b) {
 7         if (a.empty())return b;
 8         if (b.empty())return a;
 9         int p1 = a.size() - 1,p2=b.size()-1;
10         string res = p1 > p2 ? a : b;//选用较长的二进制做基
11         int p = p1 > p2 ? p1 : p2;
12         int s = 0;//进位
13         while (p>=0) {
14             if (min(p1, p2) >= 0)//两数相加
15                 s = a[p1--] - '0' + b[p2--] - '0' + s;
16             else
17                 s = res[p] - '0' + s;//进位想加
18             
19             res[p--] = s % 2 + '0';
20             s = s / 2;
21         }
22         if (s) {//进位
23             a = "1";
24             res = a + res;
25         }
26         return res;
27     }
28 };
29 
30 //下面这种写法又巧妙又简洁，用了两个指针分别指向a和b的末尾，
31 //然后每次取出一个字符，转为数字，若无法取出字符则按0处理，
32 //然后定义进位carry，初始化为0，将三者加起来，对2取余即为当前位的数字，
33 //对2取商即为当前进位的值，记得最后还要判断下carry，如果为1的话，要
34 //在结果最前面加上一个1，参见代码如下：
35 class Solution {
36 public:
37     string addBinary(string a, string b) {
38         string res = "";
39         int m = a.size() - 1, n = b.size() - 1, carry = 0;
40         while (m >= 0 || n >= 0) {
41             int p = m >= 0 ? a[m--] - '0' : 0;
42             int q = n >= 0 ? b[n--] - '0' : 0;
43             int sum = p + q + carry;
44             res = to_string(sum % 2) + res;
45             carry = sum / 2;
46         }
47         return carry == 1 ? "1" + res : res;
48     }
49 };
50 void T067() {
51     Solution s;
52     string a, b;
53     a = "11";
54     b = "1";
55     cout << s.addBinary(a, b) << endl;
56     a = "1010";
57     b = "1011";
58     cout << s.addBinary(a, b) << endl;
59     a = "11010";
60     b = "";
61     cout << s.addBinary(a, b) << endl;
62 
63 
64 }