面试题35：大数（字符串）相乘

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.

class Solution {
public:
    string add(string num1, string num2) {
        string res;
        char carry = '0';
        int n1 = num1.length() - 1;
        int n2 = num2.length() - 1;
        char c1, c2;
        while (n1 >= 0 || n2 >= 0 || carry != '0') {
            if (n1 < 0) c1 = '0';
            else c1 = num1[n1];
            if (n2 < 0) c2 = '0';
            else c2 = num2[n2];
            int sum = c1 - '0' + c2 - '0' + carry - '0';
            res.push_back(sum % 10 + '0');
            carry = sum / 10 + '0';
            n1--;
            n2--;
        }
        reverse(res.begin(), res.end());
        return res;
    }
    string multiply(string num1, string num2) {
        int n1 = num1.length() - 1;
        int n2 = num2.length() - 1;

        string res;
        for (int i = n1; i >= 0; i--) {
            string sum;
            for (int j = n2; j >= 0; j--) {
                string a = to_string((num1[i] - '0') * (num2[j] - '0'));
                for (int k = 0; k < n1 + n2 - i - j; k++) {
                    a.push_back('0');
                }
                sum = add(sum, a);
            }
            res = add(res, sum);
        }
        size_t i = 0;
        while(i < res.size() && res[i]=='0'){
            i++;
        }
        if(res.substr(i).empty()) return "0";
        return res.substr(i);
    }
};