43. Multiply Strings (String)

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.

思路：首先确定结果的位数，难点在于知道结果中的每一位，是乘数中的哪两位相乘贡献的：是num2中的i-len1+1到i中的数分别乘以num1中的i-j相加而得的，并且要注意不能超过num2的界限[0,len2-1]

class Solution {
public:
    string multiply(string num1, string num2) {
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        int carry = 0, sum =0;
        string result="";
        int len1 = num1.length();
        int len2 = num2.length();
        int resLen = len1+len2-1;
        
        for(int i = 0; i < resLen; i++){ //traverse the result digit
            for(int j = max(0,i-len1+1) ; j <= min(i,len2-1); j++){ //traverse the num2 digit
                sum += (num2[j]-'0')*(num1[i-j]-'0');
            }
            sum += carry;
            carry = sum/10;
            result += ((sum%10) + '0');
            sum=0;a
        }
        
        if(carry) result += (carry+'0');
        reverse(result.begin(),result.end());
        if(result[0]=='0') return "0"; //全0的情况
        return result;
    }
};