# Succeeded on local but failed online.Using input data:"93553535314", "25247452591474"

• Input: "93553535314", "25247452591474"
Output: "236198844760500367431283'`" (why is a '`?)
Expected: "2361988447605003674312836"

``````char* multiply(char* num1, char* num2) {
char r1[100], r2[100], mult[200];			//变量声明和初始化
unsigned long result[200];                  //declare&initialize
long i, j, k;
unsigned long reg;

for (i = 0; i < 200; i++)
{
result[i] = 0;
mult[i]='\0';
}

if (num1[0] == 48 || num2[0] == 48)                     //nil participates the muliplication is no possibile
{
return "0";
}
else
{
j = strlen(num1);										//乘数被乘数倒序，方便进位运算；char转long，避免溢出
for (i = 0; i < j; i++)                                 //reverse the  sequence of the array,in oreder to be imprehensive;
r1[j - i - 1] = num1[i] - 48;                       // the char string was changed into long array, in order to
r1[j] = -1;                                             //reach a higher digit

j = strlen(num2);
for (i = 0; i<j; i++)
r2[j - i - 1] = num2[i] - 48;
r2[j] = -1;

for (i = 0; r1[i] != -1; i++)						//按位相乘并累加进位
{                                                   //every digit of two long data was mutiplied; the result was recorded in
for (j = 0; r2[j] != -1; j++)                   //the result[] one by one; no element would be more than 81;so, if the
{                                               //element is more than 9, the carry digit should be count to the
result[i + j] += r1[i] * r2[j];             //sequencial element.this step will repeat as many times as the digits
if (result[i + j]>9)                        //multiplier(r1) has
{
result[i + j + 1] += result[i + j] / 10;
result[i + j] %= 10;
}
}

}

k = i + j;
if (!result[k - 1])									//去掉最高位的0进位
k--;                                            //to kill the highest 0 digit

for (i = 0; i < k; i++)								//long转char，
mult[k - i - 1] = result[i] + 48;               //long array changes to char string
mult[k] = '\0';

return mult;}
}``````

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