Multiply Strings ---leetcode

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.

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这是来自于leetcode上的一道算法题目。

算法思路：用char存储大数的每一位，然后控制每一位之间的相乘和进位即可。

我的代码如下：

class Solution {
public:
    string multiply(string num1,string num2) {
	const char *ch1 = num1.c_str();
	const char *ch2 = num2.c_str();
	const unsigned int l1=num1.length();
	const unsigned int l2=num2.length();
	const unsigned int lc = l1 +l2 +1;
	//cout<<str1.length()+str2.length()+1<<endl;
	char *ch3 = (char *)malloc((lc+1)*sizeof(char));
	for (int i = 0; i < lc; i++)
	{
		ch3[i]='0';
	}
	for (int i = l1 - 1; i >= 0 ; i--)
	{
		unsigned int start = (l1-i);
		int re = 0;
		for (int j = l2 - 1; j >= 0 ; j--)
		{
			int temp = (ch1[i]-48)*(ch2[j]-48) + re;
			int result = ch3[lc - (l1-i)-(l2-j)]+(temp%10) -48;
			re =temp/10;
			if (result>9)
			{
				re+=result/10;
				result=result%10;
			}
			ch3[lc - (l1-i)-(l2-j)]= result+48;
			if (j==0)
			{
				ch3[lc - (l1-i)-(l2-j)-1]=re+48;
			}
		}
	}
	string a(ch3);
	free(ch3);
	int i=0;
	for(;i<lc-2;i++){
	    if(a.substr(i,1)=="0")continue;
	    else break;
	}
    return a.substr(i,lc-i-1);
    }
};

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