字符串相乘

给定两个代表数字的字符串，返回它们的乘积（字符串形式）
Note:
The numbers can be arbitrarily large and are non-negative.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.
Multiply Strings
遍历两个字符串，依次进位即可
先将vector中无效位的0去除然后逆序得到的string为所求

string multiply(string num1, string num2) {    const int m = num1.size();    const int n = num2.size();    vector<int> a(m + n, 0);    for (int i = 0; i < m; ++i)    {        int dig = 0;        for (int j = 0; j < n; ++j)        {            a[i + j] += (num1[m - 1 - i] - '0')*(num2[n - 1 - j] - '0') + dig;            dig = a[i + j] / 10;            a[i + j] %= 10;        }        if (dig)            a[n + i] += dig;    }    while (a.size() > 1 && a.back() == 0)        a.pop_back();    string result;    for (auto it = a.rbegin(); it != a.rend(); ++it)        result += *it + '0';    return result;}