LeetCode Multiply Strings 大数相乘

题目要求:

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.
代码：
代码支持 大负数的相乘

class Solution {
public:
string multiply(string num1, string num2)
{
int flag1 = 1, flag2 = 1;
if(num1.empty() || num2.empty())
return " ";
if(num1[0] == '-')
flag1 = -1;
if(num2[0] == '-')
flag2 = -1;
string res;
if(num1[0] == '0' || num2[0] == '0')
return "0";
int len = num1.size();
if(flag1 == -1)
num1 = num1.substr(1);
if(flag2 == -1)
num2 = num2.substr(1);
for(int i = num1.size() - 1; i >= 0; --i)
{
MultiplyByNum(res, num2, num1[i], len - i - 1);
}
if(flag1 != flag2)
res.insert(res.begin(), '-');
return res;
}

void MultiplyByNum(string& res, const string& multi,
char num, int multiply)
{
string tmp;
if(num == '0')
return;
MultiplyByNum(tmp, multi, num);
for(size_t i = 0; i < multiply; ++i)
tmp.push_back('0');
Add(res, tmp);
}
void MultiplyByNum(string& res, const string& multi, char num)
{
int carry = 0;
for(int i = multi.size() - 1; i >= 0; --i)
{
int tmp = (multi[i] - '0') * (num - '0') + carry;
res.push_back((tmp % 10) + '0');
carry = tmp / 10;
}
if(carry != 0)
res.push_back(carry + '0');
reverse(res.begin(), res.end());
}

void Add(string& num1, string& num2)
{
int carry = 0;
string tmp;
int len1 = num1.size(), len2 = num2.size();
int i = len1 - 1, j = len2 - 1;
while(i >= 0 || j >= 0)
{
int sum = 0;
if(i >= 0)
{
sum += num1[i] - '0';
--i;
}
if(j >= 0)
{
sum += num2[j] - '0';
--j;
}
sum += carry;
tmp.push_back((sum % 10) + '0');
carry = sum / 10;
}
if(carry != 0)
tmp.push_back(carry + '0');
reverse(tmp.begin(), tmp.end());
num1 = tmp;
}
};