43. Multiply Strings


题目：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.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.

思路：就是用两个字符串进行乘法计算（数据很大），结果也用字符串表示。

（1）先开辟len1+len2长度的整型空间，初始化为0；

（2）对每一个字符串从头扫描至末尾，然后进行存储；

（3）从len1+len2-1到0开始进位

（4）转换成字符串

例子：123*645=79335

0	1	2
1	2	3
6	4	5

a[i+j+1]，加上1是为了方便前面的进位。

a[1]=a[0+0+1]=1*6=6                                                    先6+1=7                                        7%10=7                                         7/10无进位

a[2]=a[1+0+1]=a[0+1+1]=1*4+6*2=16                          先16+3=19                                    19%10=9                                       19/1=1

a[3]=a[1+1+1]=a[0+2+1]=a[2+0+1]=2*4+1*5+6*3=31  先31+2=33                                     33%10=3                                       33/10=3;

a[4]=a[1+2+1]=a[2+1+1]=2*5+3*4=22         先22加上a[5]的进位1，则22+1=23；             23%10=3作为倒数第二位              23/10=2作为进位

a[5]=a[2+2+1]=3*5=15                                                                                                        15 %10=5作为最后一位                 15/10=1作为a[4]的进位；

所以整数数组是：079335；

转换为字符串，如果第一个为0就不用转换为字符串。

代码：

char* multiply(char* num1, char* num2) {
int i=0,j=0;

if(!num1 || !num2 ) return "";

int len1=strlen(num1);
int len2=strlen(num2);

char a[3];
a[0]='0'+79;
a[1]='0'+80;
a[2]='0'+81;

if( (len1==1 && (*num1)=='0' ) || (len2==1 && (*num2)=='0') ) return "0";

int* ret=(int*)malloc(sizeof(int)*(len1+len2));
memset(ret,0,sizeof(int)*(len1+len2));

char* ret1=(char*)malloc(sizeof(char)*(len1+len2+1));
memset(ret1,'0',(len1+len2+1));

for(i=0;i<len1;i++)
{
int pre=num1[i]-'0';

for(j=0;j<len2;j++)
{
int end=num2[j]-'0';
ret[i+j+1]+=pre*end;
}
}

for(i=len1+len2-1;i>=1;i--)
{
int tmp=ret[i];

ret[i]=tmp%10;
int t=tmp/10;

ret[i-1]=ret[i-1]+t;
}

for(i=0;i<=len1+len2-1;i++)
{
ret1[i]=ret[i]+'0';
}

ret1[i]='\0';
//转换成字符串
if(ret1[0] == '0') return ret1+1;
return ret1;

}