# 字符串相乘
给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
示例 1:
输入: num1 = "2", num2 = "3"
输出: "6"
示例 2:
输入: num1 = "123", num2 = "456"
输出: "56088"
说明：
num1 和 num2 的长度小于110。num1 和 num2 只包含数字 0-9。num1 和 num2 均不以零开头，除非是数字 0 本身。- 不能使用任何标准库的大数类型（比如 BigInteger）或直接将输入转换为整数来处理。
以下错误的选项是？
## aop
### before
```cpp
#include
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
string num1 = "2", num2 = "3";
string res;
res = sol.multiply(num1, num2);
cout << res;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
string multiply(string num1, string num2)
{
if (num1 == "0" || num2 == "0")
return "0";
int size1 = num1.size(), size2 = num2.size();
string str(size1 + size2, '0');
for (int i = size2 - 1; i >= 0; --i)
{
int mulflag = 0, addflag = 0;
for (int j = size1 - 1; j >= 0; --j)
{
int temp1 = (num2[i] - '0') * (num1[j] - '0') + mulflag;
mulflag = temp1 / 10;
temp1 = temp1 % 10;
int temp2 = str[i + j + 1] - '0' + temp1 + addflag;
str[i + j] = temp2 % 10 + '0';
addflag = temp2 / 10;
}
str[i] += mulflag + addflag;
}
if (str[0] == '0')
str = str.substr(1, str.size());
return str;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
string multiply(string num1, string num2)
{
string res(num1.length() + num2.length(), '0');
for (int i = num2.length() - 1; i >= 0; i--)
{
int j, carry = 0;
for (j = num1.length() - 1; j >= 0; j--)
{
carry += (num1[j] - '0') * (num2[i] - '0') + (res[i + j + 1] - '0');
res[i + j + 1] = carry % 10 + '0';
carry /= 10;
}
res[i + j + 1] = carry + '0';
}
int i;
for (i = 0; i < res.length() - 1; i++)
{
if (res[i] != '0')
{
break;
}
}
return res.substr(i);
}
};
```
### B
```cpp
class Solution
{
public:
string multiply(string num1, string num2)
{
if (num1 == "0" || num2 == "0")
{
return "0";
}
int n1 = num1.size(), n2 = num2.size();
int n = n1 + n2;
vector ves(n, 0);
int k = n - 2;
for (int i = 0; i < n1; i++)
for (int j = 0; j < n2; j++)
{
ves[k - i - j] += (num1[i] - '0') * (num2[j] - '0');
}
int carry = 0;
for (int i = 0; i < n; i++)
{
ves[i] += carry;
carry = ves[i] / 10;
ves[i] %= 10;
}
int t = n - 1;
while (ves[t] == 0)
{
--t;
}
string result;
while (t >= 0)
{
result.append(1, ves[t] + '0');
--t;
}
return result;
}
};
```
### C
```cpp
class Solution
{
public:
string multiply(string num1, string num2)
{
string &upper = num1;
string &down = num2;
deque res, adv;
for (int i = 0; i < down.length(); i++)
{
if (!res.empty())
{
for (int k = 0; k < upper.length() - 1; k++)
{
adv.push_back(res.back());
res.pop_back();
}
}
for (int j = 0; j < upper.length(); j++)
{
if (adv.empty())
{
res.push_back(
(down[i] - '0') * (upper[j] - '0'));
}
else
{
res.push_back(
adv.back() + (down[i] - '0') * (upper[j] - '0'));
adv.pop_back();
}
}
}
deque::iterator it = res.end() - 1;
while (it != res.begin() - 1)
{
if (*it > 9)
{
if (it == res.begin())
{
int tmp = *it;
*it = tmp % 10;
res.push_front(tmp / 10);
}
else
{
int tmp = *it;
*it = tmp % 10;
*(it - 1) += tmp / 10;
}
}
it--;
}
string _r;
for (it = res.begin(); it != res.end(); it++)
{
_r += (*it + '0');
}
return _r;
}
};
```