AC代码：

class Solution {
public:
    string multiply(string num1, string num2) {
        string res(num1.size()+num2.size(),'0');
        for(int i = num1.size() - 1;i >= 0;i--)
        {
            int carry = 0;
            for(int j = num2.size() - 1;j >= 0;j--)
            {
                int tmp = res[i+j+1]-'0' + (num1[i] - '0')*(num2[j]-'0') + carry;
                res[i+j+1] = tmp%10 + '0';
                carry = tmp/10;
            }
            res[i] += carry;
        }
        
        size_t start = res.find_first_not_of("0");
        if(start != string::npos)
            return res.substr(start);
        
        return "0";
    }
};

这道题目最关键的一点是要理解，针对一个m位 和n位字符串相乘，结果最多m+n位。
我们以"123" "234"为例，循环刚开始时i为2，j为2，结果位应该保存在res[5]，即res[i+j+1];
最后就是需要去掉前导0，返回无前导0子串。进位问题也需要注意，以"123"和"456"为例添加打印：

000368  i = 2  carry = 1
009488  i = 1  carry = 0
055088  i = 0  carry = 0 

实际上此时可以理解成j==-1,最终更新一次res[i+j+1]，即res[i].
运行效率：

Runtime: 4 ms, faster than 88.31% of C++ online submissions for Multiply Strings.
Memory Usage: 6.2 MB, less than 99.03% of C++ online submissions for Multiply Strings.

更好理解的代码如下：

class Solution {
public:
    string multiply(string num1, string num2) {
        string res(num1.size() + num2.size(),'0');
        int m = num1.size(),n = num2.size();
        for(int i=m-1;i>=0;i--)
        {
            for(int j=n-1;j>=0;j--)
            {
                int sum = res[i+j+1] - '0' + (num1[i] - '0')*(num2[j] - '0');
                res[i+j+1] = sum % 10 + '0';
                res[i+j] += sum /10;
            }
        }
        for(int i = 0;i < m + n;i++)
        {
            if(res[i] != '0')
                return res.substr(i);
        }
        return "0";
    }
};

运行效率：

Runtime: 4 ms, faster than 88.46% of C++ online submissions for Multiply Strings.
Memory Usage: 6.4 MB, less than 93.69% of C++ online submissions for Multiply Strings.