LeetCode(43) Multiply Strings

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本文详细介绍了如何使用字符串表示的任意大小的非负整数进行相乘运算，并通过模拟传统乘法过程来实现算法。包括乘法原理、代码实现及性能分析。

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题目如下：

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.

分析如下：

以 1234 * 567 = 69978为例。假设

num1 = 1234

num2 = 567

那么可以模拟乘法，用num2 的每一位去乘以num1，得到一个结果, num2 的低位开始，到高位结束，乘积依次是

1234 * 7 = 8638 ----> 8638

1234 * 6 = 7404 ----> 74040

1234 * 5 = 6170 ----> 617000

然后，每次得到的结果要根据情况乘以扩大10的N倍，最后把这个结果加起来。

我的代码：

//209ms
/*
    1234
   X 567
------------
    8638
   7404
  6170
------------
  699678
*/
class Solution {
public:
    string add_string(string & s1, string & s2) {
        if (s1 == "" ) return s2;
        int len1 = s1.length();
        int len2 = s2.length();
        int i = len1 - 1;
        int j = len2 - 1;
        int rounding = 0;
        string res = "";
        while (i >=0 && j >= 0) {
            res = to_string((rounding + s1[i] - '0' + s2[j] - '0') % 10) + res;
            rounding = (rounding + s1[i] - '0' + s2[j] - '0') / 10;
            --i;
            --j;
        }
        while (i >= 0) {
            res = to_string((rounding + s1[i] - '0' ) % 10) + res;
            rounding = (rounding + s1[i] - '0' ) / 10;
            --i;
        }
        while (j >= 0) {
            res = to_string((rounding + s2[j] - '0') % 10) + res;
            rounding = (rounding + s2[j] - '0') / 10;
            --j;
        }
        if (rounding > 0)
            res = to_string(rounding) + res;
        return res;
    }
    
    string multiply(string num1, string num2) {
        if (num1.size() == 0 || num1 == "0" ) return "0";
        if (num2.size() == 0 || num2 == "0" ) return "0";
        string each_res = "";
        string final_res = "";
        string add_zero = "";
        int rounding = 0;
        for (int i = num2.size() - 1; i >= 0; --i) {
            each_res = "";
            rounding = 0;
            for (int j = num1.size() - 1; j >= 0; --j) {
                each_res = to_string((rounding + (num2[i] - '0') * (num1[j] - '0')) % 10) + each_res;
                rounding = (rounding + (num2[i] - '0') * (num1[j] - '0')) / 10;
            }
            if (rounding > 0)
                each_res = to_string(rounding) + each_res;
            each_res = each_res + add_zero;
            add_zero = add_zero + "0";
            final_res = add_string(final_res, each_res);
            //std::cout<<"each_res = "<<each_res<<", final_res = "<<final_res<<std::endl;
        }
        return final_res;
    }
};