描写叙述：
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.
思路：
简而言之，要实现的就是BigInteger(a).Multiply(BigInteger(b))的功能，但非常显然，leetcode中不让用BigInteger
代码：
public class Solution {
public String multiply(String num1, String num2)
{
if(num1==null||num2==null)
return new String();
num1=num1.trim();
num2=num2.trim();
if(num1.equals("0")||num2.equals("0"))
return "0";
List<Integer>listNum1=new ArrayList<>();
List<Integer>listNum2=new ArrayList<>();
List<Integer>listResult=new ArrayList<>();
int len1=num1.length();
int len2=num2.length();
int i=0,j=0,lenResult=0,index=0;
int baseNum=0,flowNum=0,tempNum1=0,tempNum2=0;
for( i=len1-1;i>=0;i--)
listNum1.add(num1.charAt(i)-'0');
for( i=len2-1;i>=0;i--)
listNum2.add(num2.charAt(i)-'0');
tempNum2=listNum2.get(0);
for(i=0;i<len1;i++)
{
tempNum1=listNum1.get(i);
tempNum1=tempNum1*tempNum2+flowNum;
baseNum=tempNum1%10;
flowNum=tempNum1/10;
listResult.add(baseNum);
}
if(flowNum!=0)
{
listResult.add(flowNum);
flowNum=0;
}
for(j=1;j<len2;j++)
{
baseNum=0;flowNum=0;
tempNum2=listNum2.get(j);
lenResult=listResult.size();
for(i=0;i<len1;i++)
{
index=i+j;
if(index<lenResult)
{
tempNum1=listNum1.get(i);
tempNum1=tempNum1*tempNum2+flowNum+listResult.get(index);
baseNum=tempNum1%10;
flowNum=tempNum1/10;
listResult.set(index, baseNum);
}
else {
tempNum1=listNum1.get(i);
tempNum1=tempNum1*tempNum2+flowNum;
baseNum=tempNum1%10;
flowNum=tempNum1/10;
listResult.add(baseNum);
}
}
if(flowNum!=0)
{
listResult.add(flowNum);
flowNum=0;
}
}
if(flowNum!=0)
listResult.add(flowNum);
StringBuilder sBuilder=new StringBuilder();
for(int num:listResult)
sBuilder.append(num);
sBuilder.reverse();
return sBuilder.toString();
}
}