大数据相乘

　　下面是我写的一个关于大数据相乘的算法，核心思想就是通过小学竖式乘法进行运算，具体代码C#如下所示：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace BigNumberMultiplication
{
    class Program
    {
        static void Main(string[] args)
        {
            try
            {
                int first = 4916;
                int second = 12345;
                long result = first * second;
                Console.WriteLine(string.Format("{0} * {1} = {2}\n\n", first.ToString(), second.ToString(), result.ToString()));

                string firstStr = "100000000000000000000";
                string secondStr = "100000000000000000000";
                string resultStr = MultipFunction(firstStr, secondStr);
                Console.WriteLine("The result is: {0}", resultStr.TrimStart('0'));
                Console.WriteLine("The length of the result is: {0}", resultStr.TrimStart('0').Length);
                Console.ReadKey();
            }
            catch (Exception ex)
            { }
        }

        //大数据乘法
        private static string MultipFunction(string firstNumStr, string secondNumStr)
        {
            try
            {
                int firstNumLength = firstNumStr.Length;
                int secondNumLength = secondNumStr.Length;
                int resultNumLength = firstNumLength + secondNumLength;
                int[] firstNumValue = new int[firstNumLength];
                int[] secondNumValue = new int[secondNumLength];
                int[] resultNumValue = new int[resultNumLength];
                //遍历字符串，将每一位的字符转换成为int整形插入整形数组中
                for (int i = 0; i < firstNumLength; i++)
                {
                    firstNumValue[i] = firstNumStr[i] - 48;
                }
                for (int i = 0; i < secondNumLength; i++)
                {
                    secondNumValue[i] = secondNumStr[i] - 48;
                }
                //定义的整形数组初始化各个位就是0；所以下面赋0的过程可以省略
                for(int i = 0; i < resultNumLength; i++)
                {
                    resultNumValue[i] = 0;
                }

                //算法的核心（小学笔算乘法的流程），将两个数按位进行相乘-->组合成结果的整形数组
                //然后对此结果整形数组进行遍历，结果的低位取整附加给临近的高位，低位取余赋给本低位（注：这里说的低位恰好是数组的高位，即数组下标大的位）
                for (int i = firstNumLength - 1; i >= 0; i--)
                {
                    for (int j = secondNumLength - 1; j >= 0; j--)
                    {
                        resultNumValue[i + j + 1] += firstNumValue[i] * secondNumValue[j];
                        resultNumValue[i + j] += resultNumValue[i + j +1] /10;
                        resultNumValue[i + j + 1] = resultNumValue[i + j + 1] % 10;
                    }
                }

                //将整形数组转化成字符数组
                char[] temp = new char[resultNumLength];
                for (int i = 0; i < resultNumLength; i++)
                {
                    temp[i] = (char)(resultNumValue[i] + 48);
                }
                //将字符数组转化为字符串
                string resultStr = new string(temp);
                return resultStr;
            }
            catch (Exception ex)
            {
                return string.Empty;
            }
        }
    }
}

　　。。。。。。