leetcode 43 Multiply Strings

1. 

class Solution {
public:
    string multiply(string num1, string num2) {
        string str;
        int len1=num1.size(),len2=num2.size();
        for(int i=len1-1;i>=0;--i) {
            int n1=num1[i]-'0';
            if(!n1) continue;
            for(int j=len2-1;j>=0;--j) {
                int n2=num2[j]-'0';
                if(!n2) continue;
                string sum=to_string(n1*n2);
                int k=len1-1-i+len2-1-j;
                while(k--) sum+='0';
                addstring(str,sum);
            }
        }
        return str.empty()?"0":str;
    }
    void addstring(string& str1,string &str2) {
        int k=0,i=0;
        reverse(str1.begin(),str1.end());
        reverse(str2.begin(),str2.end());
        for(i=0;i<str1.size()&&i<str2.size();++i) {
            int sum=str1[i]-'0'+str2[i]-'0'+k;
            str1[i]=sum%10+'0';
            k=sum/10;
        }
        for(;i<str1.size();++i) {
            if(!k) break;
            int sum=str1[i]-'0'+k;
            str1[i]=sum%10+'0';
            k=sum/10;
        }
        for(;i<str2.size();++i) {
            int sum=str2[i]-'0'+k;
            str1+=(sum%10+'0');
            k=sum/10;
        }
        if(k) str1+='1';
        reverse(str1.begin(),str1.end());
    }
};

 2. https://leetcode.com/problems/multiply-strings/discuss/573860/C%2B%2B-easy-short-100sc-95tc

num1*num2，乘积的最长长度为num1.size()+num2.size()；每两位相乘，乘积的位置在i+j+1，进的一位在i+j；num1的每一位与num2相乘过后的进位，在i处。

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1=="0"||num2=="0") return "0";
        string str(num1.size()+num2.size(),'0');
        for(int i=num1.size()-1;i>=0;--i) {
            int carry=0;
            for(int j=num2.size()-1;j>=0;--j) {
                int sum=str[i+j+1]-'0'+(num1[i]-'0')*(num2[j]-'0')+carry;
                carry=sum/10;
                str[i+j+1]=sum%10+'0';
            }
            str[i]=carry+'0';
        }
        int pos=0;
        while(str[pos]=='0') {++pos;}
        return str.substr(pos);
    }
};