大数的乘法 Multiply Strings

一、简单版

# include<stdio.h>
# include<string.h>
# include<malloc.h>
 
void multiply(char* a,char* b,char* c)
{
    int i,j,ca,cb,* s;
    ca=strlen(a);
    cb=strlen(b);
    s=(int*)malloc(sizeof(int)*(ca+cb));
    for (i=0;i<ca+cb;i++)
        s[i]=0;
    for (i=0;i<ca;i++)
        for (j=0;j<cb;j++)
            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');
    for (i=ca+cb-1;i>=0;i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }
    i=0;
    while (s[i]==0)
        i++;
       for (j=0;i<ca+cb;i++,j++)
           c[j]=s[i]+'0';
    c[j]='';
    free(s);
}

二、增强版?

加入了正负号的处理

# include<stdio.h>
# include<string.h>
# include<malloc.h>
 
void multiply(char* a,char* c)
{
	int i,*s,cs;
	int a_sign=1,b_sign=1,c_sign;
	int a_left,b_left;

	if(a[0] == '-')
	{
		a_left = 1;
		a_sign = -1;
	}
	else{
		a_left = 0;
	}

	if(b[0] == '-')
	{
		b_left = 1;
		b_sign = -1;
	}
	else{
		b_left = 0;
	}

	ca = strlen(a);
	cb = strlen(b);
	cs = ca+cb-a_left-b_left;
    s=(int*)malloc(sizeof(int)*cs);	
    for(i=0; i<cs; i++)
		s[i]=0;

    for(i=a_left; i<ca; i++)
        for (j=b_left; j<cb; j++)
            s[i+j+1-a_left-b_left] += (a[i]-'0')*(b[j]-'0');

    for(i=ca+cb-1; i>=0; i--)
        if (s[i]>=10)
        {
            s[i-1]+=s[i]/10;
            s[i]%=10;
        }

    i=0;
    while (s[i]==0)
        i++;

	c_sign = a_sign*b_sign;
	j=0;
	if(c_sign == -1)
	{
		c[0] = '-';
		j++;
	}
    for(; i<cs; i++,j++)
    	c[j] = s[i] + '0';
    c[j]='';
    free(s);
}

void main()
{
	char a[] = "-123453";
	char b[] = "987654";
	char c[50];
	multiply(a,b,c);

	printf("%sn",c);

}

三、String版本

class Solution {
public:
    string multiply(string num1,string num2) {
        if(num1.empty() || num2.empty())
            return "";
        if(num1 == "0" || num2 == "0")
            return "0";
        int len1 = num1.size();
        int len2 = num2.size();
        int *mul = new int[len1+len2]();//申请动态数组，并初始化为全0
        
        for(int i=0;i<len1;i++)
            for(int j=0;j<len2;j++)
                mul[i+j+1] += (num1[i] - '0') * (num2[j] - '0');
        
        for(int i=len1+len2-1;i>0;i--)
            if(mul[i] >= 10)
            {
                mul[i-1] += mul[i]/10;
                mul[i] %= 10;
            }
        string result;
        int i = 0;
        if(mul[0] == 0)
            i++;
        for(;i<len1+len2;i++)
            result += (mul[i] + '0');
        delete mul;
        mul = NULL;
        return result;
    }
};