leetcode_Multiply Strings

来源：互联网发布：华为因数据丢失事件编辑：程序博客网时间：2024/05/21 09:19

描述：

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，但很显然，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();   }}

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