solution.md

# 二进制求和

<p>给你两个二进制字符串，返回它们的和（用二进制表示）。</p><p>输入为 <strong>非空 </strong>字符串且只包含数字&nbsp;<code>1</code>&nbsp;和&nbsp;<code>0</code>。</p><p>&nbsp;</p><p><strong>示例&nbsp;1:</strong></p><pre><strong>输入:</strong> a = &quot;11&quot;, b = &quot;1&quot;<strong><br />输出:</strong> &quot;100&quot;</pre><p><strong>示例&nbsp;2:</strong></p><pre><strong>输入:</strong> a = &quot;1010&quot;, b = &quot;1011&quot;<strong><br />输出:</strong> &quot;10101&quot;</pre><p>&nbsp;</p><p><strong>提示：</strong></p><ul>	<li>每个字符串仅由字符 <code>&#39;0&#39;</code> 或 <code>&#39;1&#39;</code> 组成。</li>	<li><code>1 &lt;= a.length, b.length &lt;= 10^4</code></li>	<li>字符串如果不是 <code>&quot;0&quot;</code> ，就都不含前导零。</li></ul>
<p>以下<font color="red">错误</font>的选项是？</p>

## aop

### before

```cpp
#include <bits/stdc++.h>
using namespace std;
```
### after

```cpp
int main()
{
    Solution sol;
    string a = "1010";
    string b = "1011";
    string res;
    res = sol.addBinary(a, b);
    cout << res;
    return 0;
}
```

## 答案

```cpp
class Solution
{
public:
    string addBinary(string a, string b)
    {
        string res;
        int carry = 0;
        int m = a.size() - 1, n = b.size() - 1;
        while (m >= 0 || n >= 0)
        {
            int q = (m >= 0) ? a[m] - '0' : 0;
            int p = (n >= 0) ? b[n] - '0' : 0;
            int k = q + p + carry;
            res = to_string(k % 2) + res;
            carry = k / 2;
        }
        if (1 == carry)
            res = '1' + res;
        return res;
    }
};
```
## 选项


### A

```cpp
class Solution
{
public:
    string addBinary(string a, string b)
    {
        long aa = 0, bb = 0;
        long sum = 0;
        string res;
        int p = 0;
        stringstream outa(a);
        stringstream outb(b);
        outa >> aa;
        outb >> bb;
        sum = aa + bb;
        bool flag = false;
        while (sum != 0)
        {
            p = sum % 10;
            if (flag == true)
                p++;
            flag = false;
            sum /= 10;
            if (p == 0)
                res = res + "0";
            else if (p % 2 == 0)
            {
                flag = true;
                res = res + "0";
            }
            else
                res = res + "1";
        }
        if (res[res.size() - 1] == '0')
            res = res + "1";
        reverse(res.begin(), res.end());
        return res;
    }
};
```

### B

```cpp
class Solution
{
public:
    string addBinary(string a, string b)
    {
        if (b.size() > a.size())
        {
            string temp = b;
            b = a;
            a = temp;
        }
        int i = a.size() - 1;
        int j = b.size() - 1;
        if (i != j)
        {
            for (int k = 0; k < i - j; k++)
                b = "0" + b;
        }
        int count = 0;
        for (int k = i; k >= 0; k--)
        {
            if (a[k] - '0' + b[k] - '0' + count == 0)
            {
                a[k] = '0';
                count = 0;
            }
            else if (a[k] - '0' + b[k] - '0' + count == 1)
            {
                a[k] = '1';
                count = 0;
            }
            else if (a[k] - '0' + b[k] - '0' + count == 3)
            {
                a[k] = '1';
                count = 1;
            }
            else
            {
                a[k] = '0';
                count = 1;
            }
        }
        if (count == 1)
            a = '1' + a;
        return a;
    }
};
```

### C

```cpp
class Solution
{
public:
    string addBinary(string a, string b)
    {
        string result = "", rr = "";
        char aa, bb;
        int l1 = a.length(), l2 = b.length(), i = l1 - 1, j = l2 - 1, carry = 0, sum = 0;
        while (true)
        {
            if (i < 0)
                aa = '0';
            else
                aa = a[i];
            if (j < 0)
                bb = '0';
            else
                bb = b[j];
            sum = (aa - '0') + (bb - '0') + carry;
            result += ((sum % 2) + '0');
            carry = sum / 2;
            j--;
            i--;
            if (i < 0 && j < 0)
            {
                if (carry == 1)
                    result += "1";
                break;
            }
        }
        int l3 = result.length();
        for (int i = l3 - 1; i >= 0; i--)
            rr += result[i];
        return rr;
    }
};
```