Multiply Strings

Multiply Strings

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.

 Math String

Hide Similar Problems

 (M) Add Two Numbers (E) Plus One (E) Add Binary

 

class Solution {
public:

    string multiply(string num1, string num2) 
    {
        string addDeci(string a,string b);
        
           string str;//默觉得空串 //存储最后结果
		   string strBit;//存储一位相乘的结果.
		   if (num2.size()>num1.size())
		   {
			   string tmp=num1;
			   num1=num2;
			   num2=tmp;
		   }
		   if (num2=="0")
		   {
			   return "0";
		   }
		   int lengthA=num1.size();
		   int lengthB=num2.size();  //小的
		   int carry =0; //进位标志
		   char bitSum;
		   int MultiplySum;//每一次乘数个位的乘
		   for(int j=0;j<lengthB;j++)//j个相加..1,10,100
		   {
			   strBit="";
			   for(int i=0;i<lengthA;i++)   //每一位和他相加..
			   {
				    MultiplySum=(num2[lengthB-j-1]-'0')*(num1[lengthA-i-1]-'0')+carry;
				   carry=MultiplySum/10;
				  if(carry>0) //有进位
				  {
                     MultiplySum=MultiplySum%10;
					 bitSum='0'+MultiplySum;
                     strBit=bitSum+strBit;  //每一次一个右边最小添加进来..
				  }
				  else
				  {
					  bitSum='0'+MultiplySum;
					  strBit=bitSum+strBit;
				  }
			   }
			   if(carry>0)
			   {
				   MultiplySum=carry;
				   bitSum='0'+MultiplySum; //char型
				   strBit=bitSum+strBit;

			   }
			    carry=0;
			   //加一位，10,20。两个数相加..
			  for (int Cntdigit=0;Cntdigit<j;Cntdigit++)//看移动多少位
			  {
				  strBit=strBit+"0";
			  }
			   str=addDeci(strBit,str);
		   }
		   return str;
 
    }
};

string addDeci(string a,string b)
{

	string str;//默觉得空串
	if (b.size()>a.size())
	{
		string tmp=b;
		b=a;
		a=tmp;
	}
	int lengthA=a.size();
	int lengthB=b.size();
		//最后一位与最后一位相加..
		int i=lengthA-1;
		int j=lengthB-1;
		int carry=0;
		int bit_Sum=0;
		char bitCurrent;
		while (j>=0) //先加最低位
		{
			bit_Sum=(a[i]-'0'+b[j]-'0')+carry; //计算每一位添加的数字
			if (bit_Sum/10>=1)  //1+1//有进位，自己变为2的余数
			{
				//之间记录到a[i]里面
				carry=bit_Sum/10; //进位
				bitCurrent=('0'+bit_Sum%10);
				str=bitCurrent+str;
			}
			else
			{
				carry=0;
				bitCurrent=('0'+bit_Sum); //convert to char
				str=bitCurrent+str;	
			}
			i--;
			j--;
		}
		while (i>=0)
		{
			bit_Sum=(a[i]-'0')+carry;
			if (bit_Sum/10>=1)  //1+1
			{
				//之间记录到a[i]里面
				carry=bit_Sum/10; //进位
				bitCurrent=('0'+bit_Sum%10);  //1+1=2,+1=3
				str=bitCurrent+str;
			}
			else
			{
				carry=0; //没有进位
				bitCurrent=('0'+bit_Sum);
				str=bitCurrent+str;
			}
			i--;
		}
		if (carry>0)
		{
			bitCurrent=('0'+carry);
			str=bitCurrent+str;
			return str;
		}
		return str;
}