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.

题目解析：

方案一：

模拟数据的乘法

class Solution {
public:
    string multiply(string num1, string num2) {
        string s(num1.size() + num2.size(), '0');
        //将数据翻转，方便求解
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        //num1的每一位与num2相乘
        for(int i = 0; i < num1.size(); i++)
        {
            int flag = 0;
            for(int j = 0; j < num2.size(); j++)
            {
                int digit = s[i+j] - '0';
                int num = (num1[i] - '0') * (num2[j] - '0');
                int res = digit + num + flag;
                s[i+j] = (res % 10) + '0';
                flag = res / 10;
            }
            int index = i + num2.size();
            while(flag) //如果有进位，则修改高位的值
            {
                int digit = s[index] - '0';
                int res = digit + flag;
                s[index] = (res % 10) + '0';
                flag = res / 10;
            }
        }
        //去除高位为0的多余字符
        while(true)
        {
            if (s[s.size()-1] == '0' && s.size() > 1)
                s.erase(s.size() - 1, 1);
            else
                break;
        }
        //翻转
        reverse(s.begin(), s.end());

        return s;
    }
};

方案二：

先乘，然后再处理进位。由于我们的数据是反着的，所有程序中相乘的表达式有res[i+j+1] += n1[i]*n2[j];

按照正常的话：比如45 * 35

流程是：先让35的每一位与45的位分别相乘，得到结果不进位，最后再处理进位的问题。

当然35和45的位从右向左标记的话，就是res[i+j] += n1[i] * n2[j];

class Solution {
public:
    string multiply(string num1, string num2) {
		if(num1=="0" || num2=="0")
            return "0";
		int l1 = num1.length(), l2 = num2.length();
		int* n1 = new int[l1];
		int* n2 = new int[l2];
		int* res = new int[l1+l2];
		memset(res,0,sizeof(int)*(l1+l2));
		for(int i=0; i<l1; ++i)
			n1[i] = num1[i] - '0';
		for(int i=0; i<l2; ++i)
			n2[i] = num2[i] - '0';
		//好好看例子
		for(int i=0; i<l1; ++i)
			for (int j=0; j<l2; ++j)
				res[i+j+1] += n1[i]*n2[j];

		string ss = "";
		for (int k=l1+l2-1; k>=0; --k){
			if(k>0)
                res[k-1] += res[k]/10;
			res[k] %= 10;
			ss = char(res[k]+'0')+ss;
		}
		ss = ss[0]=='0'? ss.substr(1):ss;
		return ss;
    }
};