LeetCode43 Multiply Strings 字符串相乘

题目描述：
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:

The length of both num1 and num2 is < 110.
Both num1 and num2 contain only digits 0-9.
Both num1 and num2 do not contain any leading zero, except the number 0 itself.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
题源：here；完整实现：here
思路：
不用想多了，就是小学的乘法规则就可以了：从个位开始乘。当然，每一次的结果都是存在字符串中，也就是说我们还需要完成字符串的加法工作，实现的代码如下：

class Solution {
public:
    string sum2string(string num1, string num2){
        if (num1.size() == 0) return num2;
        string result;
        int carryBit = 0;
        while(num1.size() && num2.size()){
            int S = num1[0] -  '0' + num2[0] - '0' + carryBit;
            result.push_back(S % 10 + '0');
            carryBit = S / 10;
            num1.erase(0, 1); num2.erase(0, 1);
        }

        if (num1.size()){
            while (num1.size()){
                int S = num1[0] - '0' + carryBit;
                result.push_back(S % 10 + '0');
                carryBit = S / 10;
                num1.erase(0, 1);
            }
        }
        else{
            while (num2.size()){
                int S = num2[0] - '0' + carryBit;
                result.push_back(S % 10 + '0');
                carryBit = S / 10;
                num2.erase(0, 1);
            }
        }
        if (carryBit > 0) result.push_back(carryBit + '0');

        return result;
    }

    string multiply(string num1, string num2) {
        if ((num1.size() == 1 && num1[0] == '0') || (num2.size() == 1 && num2[0] == '0')) return "0";
        string result;
        for (int i = num2.size()-1; i >= 0; i--){
            string bitM;
            for (int j = 0; j < num2.size() - i - 1; j++) bitM.push_back('0');
            int carryBit = 0, digit2 = num2[i] - '0';
            for (int j = num1.size() - 1; j >= 0; j--){
                int digit1 = num1[j] - '0';
                int M = digit1*digit2 + carryBit;
                bitM.push_back(M % 10+'0');
                carryBit = M / 10;
            }
            if (carryBit > 0) bitM.push_back(carryBit + '0');

            result = sum2string(result, bitM);
        }

        reverse(result.begin(), result.end());
        return result;
    }
};