43. Multiply Strings

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

my code:

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.length();
        int len2 = num2.length();
        if (len1 == 1 && num1[0] == '0' ||
            len2 == 1 && num2[0] == '0')
            return "0";
        vector<string> v;
        int incidental;
        int multiply;
        for (int i = len2 - 1; i >= 0; --i) {
            incidental = 0;
            int n2 = num2[i] - '0';
            string tem = "";
            for (int k = (len2 - 1) - i; k > 0; --k) {
                tem += '0';
            }
            for (int j = len1 - 1; j >= 0; --j) {
                int n1 = num1[j] - '0';
                multiply = n1 * n2 + incidental;
                incidental = multiply / 10;
                tem += to_string(multiply%10);
            }
            if (incidental != 0) 
                tem += to_string(incidental);
            v.push_back(tem);
        }
        int max_length = v.back().size();
        string ans = "";
        int flag;
        incidental = 0;
        for (int i = 0; i < max_length; ++i) {
            flag = 0;
            for (int j = 0; j < v.size(); ++j) {
                if (i < v[j].size()) {
                    int tage = v[j][i] - '0';
                    flag += tage;
                }
            }
            flag += incidental;
            
            incidental = flag / 10;
            ans = to_string(flag % 10) + ans;
        }
        if (incidental != 0) 
            ans = to_string(incidental) + ans;
        return ans;
    }
};

Runtime: 24 ms, faster than 14.44% of C++ online submissions for Multiply Strings.

efficient code:

class Solution {
public:
    string multiply(string num1, string num2) {
        string sum(num1.size() + num2.size(), '0');

        for (int i = num1.size() - 1; 0 <= i; --i) {
            int carry = 0;
            for (int j = num2.size() - 1; 0 <= j; --j) {
                int tmp = (sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + carry;
                sum[i + j + 1] = tmp % 10 + '0';
                carry = tmp / 10;
            }
            sum[i] += carry;
        }

        size_t startpos = sum.find_first_not_of("0");
        if (string::npos != startpos) {
            return sum.substr(startpos);
        }
        return "0";
    }
};

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Multiply Strings.