LeetCode 43: multiply strings(大數相乘）

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

思路一：

把輸入數字翻轉，然後注意到 result[i+j] = a[i] * b[j];

最後把結果再翻轉過來：

int mystrlen(char *str)
{
	char *tmp = str;
	while(*tmp++ != '\0');
	
	return tmp - str - 1;
}

void swap(char *str)
{
	int len  = mystrlen(str);
	int i;
	for(i = 0; i < len/2; ++i)
	{
		char tmp = str[i];
		str[i] = str[len-i-1];
		str[len-i-1] = tmp;
	}
}

char * multiply(char * number1, char *number2 )
{
	swap(number1);
	swap(number2);
	
	int tmp[1000] ={0,};
	
	int len1 = mystrlen(number1);
	int len2 = mystrlen(number2);

	char *result = (char *)malloc(len1 + len2 + 1);
	
	int i, j;
	for(i = 0; i < len1; ++i)
	{
		for(j = 0; j < len2; ++j)
		{
			tmp[i +j] += (number1[i] - '0') * (number2[j] - '0');
		}
	}

	int add = 0;
	for(i = 0; i < len1 + len2 - 1; ++i)
	{
		tmp[i] += add;
		if(tmp[i] > 9)
		{
			add = tmp[i] / 10;
			tmp[i] = tmp[i] % 10;	
		}
		else
		{
			add = 0;
		}
	}

	if(add > 0)
	{
		tmp[i] = add;
	}

	while(i >=0 && tmp[i] == 0)
	{
		--i;
	}

	j = 0;
	while( i >= 0)
	{
		result[j] = tmp[i] + '0';
		--i;
		++j;	
	}
	if(j == 0)
	{
		result[j++] = '0';
	}	
	result[j] = '\0';

	return result;
}

思路２：

參考下面的圖　（轉自：https://blog.csdn.net/kangkanglou/article/details/79894208)：

 

 

int mystrlen(char *str)
{
	char *tmp = str;
	while(*tmp++ != '\0');
	
	return tmp - str - 1;
}

char * multiply(char * number1, char *number2 )
{
	int len1 = mystrlen(number1);
	int len2 = mystrlen(number2);

	int *tmp = (int *)malloc(sizeof(int) *(len1+len2));
	
	int i, j;
	for(i = 0; i < len1+len2; ++i)
	{
		tmp[i] = 0;
	}

	for(i = len1 -1; i>= 0; --i)
	{
		for(j = len2 -1; j >=0; --j)
		{
			int multi = (number1[i]-'0')*(number2[j]-'0');					
			tmp[i+j] += (multi+tmp[i+j+1])/10;
			tmp[i+j+1] = (multi+tmp[i+j+1])%10;
		}
	}


	i = 0;
	while( i < len1+len2 && tmp[i] == 0 )
	{
		++i;
	}

	char *result = (char *)malloc(sizeof(char) * (len1+len2+1));
	
	j = 0;
	while(i < len1+len2)
	{
		result[j] = tmp[i] + '0';
		++i;
		++j;
	}

	if(j == 0)
	{
		result[j++] = '0';
	}

	result[j] = '\0';

	return result;
}

 

參考：

https://blog.csdn.net/kangkanglou/article/details/79894208