[leetcode] 43.Multiply Strings

题目：
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.
题意：
给定两个字符串形式的整数，每个整数都是非负的，并且是任意大的。求出两者乘积的字符串表示。
思路：
我们不妨思考当初小学学习数学时手动乘法的方法。比如23*14，那么先让4与23相乘，然后让1与23相乘，不过此时要将这一位的乘积往左移动一位，就相当于在这一位的乘积末尾加了一个0。然后让两个位上的乘积相加。
所以我们需要做的就是计算乘数的每一位与被乘数的乘积，并且加上先前位数上的结果。实现一个字符串的加法函数作为辅助函数。
以上。
代码如下：

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.length();
        int len2 = num2.length();
        if (len1 == 0 || len2 == 0)return "";
        if (len2 > len1) {
            swap(num1, num2);
            swap(len1, len2);
        }
        string result = "";
        int addZero = 0;
        for (int i = len2 - 1; i >= 0; i--) {
            int carry = 0;
            string temp(len1, ' ');
            for (int j = len1 - 1; j >= 0; j--) {
                int multiRes = (num2[i] - '0') * (num1[j] - '0') + carry;
                carry = multiRes / 10;
                multiRes %= 10;
                temp[j] = multiRes + '0';
            }
            if (carry != 0)
                temp = char(carry + '0') + temp;
            string zeros(addZero, '0');
            temp += zeros;
            result = add(result, temp);
            addZero++;
        }
        int i = 0;
        while (i < result.size() && result[i] == '0')i++;
        return (i == result.size())?"0":result.substr(i);
    }

    string add(string num1, string num2) {
        int len1 = num1.length();
        int len2 = num2.length();
        if (len2 > len1) {
            swap(len1, len2);
            swap(num1, num2);
        }
        num2 = string(len1 - len2, '0') + num2;
        string result(len1, ' ');
        int carry = 0;
        for (int i = len1 - 1; i >= 0; i--) {
            int sum = (num1[i] - '0') + (num2[i] - '0') + carry;
            carry = sum / 10;
            sum %= 10;
            result[i] = sum + '0';
        }
        if (carry)result = "1" + result;
        return result;
    }
};