【leetcode】Multiply Strings（middle）

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：直观思路，就是模拟乘法过程。注意进位。我写的比较繁琐。

string multiply(string num1, string num2) {
        vector<int> v1, v2, vmulti;
        vector<vector<int>> vvtmp;
        int l1 = num1.size();
        int l2 = num2.size();

        if(l1 ==  && num1[] == '') return "";
        if(l2 ==  && num2[] == '') return "";

        int maxl = ;
        //用vector存储两个数字 v[0]是最高位
        for(int i = ; i < l1; i++)
            v1.push_back(num1[i] - '');
        for(int i = ; i < l2; i++)
            v2.push_back(num2[i] - '');

        //乘步骤
        for(int i = l1 - ; i >= ; i--)
        {
            vector<int> vtmp; //v[0]是最低位
            int t = i;
            while(t < l1 - ) //后面补错位的0
            {
                vtmp.push_back();
                t++;
            }
            int c = ; //记录进位
            for(int j = l2 - ; j >= ; j--)
            {
                int cur = v1[i] * v2[j] + c;
                vtmp.push_back(cur % );
                c = cur / ;
            }
            if(c != )
                vtmp.push_back(c);
            vvtmp.push_back(vtmp);
            maxl = (vtmp.size() > maxl) ? vtmp.size() : maxl;
        }

        //加步骤
        int c = ; //记录进位
        for(int i = ; i < maxl; i++)
        {
            int cur = c;
            for(int j = ; j < vvtmp.size(); j++)
            {
                if(i >= vvtmp[j].size())
                    continue;
                cur += vvtmp[j][i];
            }
            vmulti.push_back(cur % );
            c = cur / ;
        }
        if(c != )
            vmulti.push_back(c);

        //转换为string
        string ans;
        for(int i = vmulti.size() - ; i >= ; i--)
        {
            ans += (vmulti[i] + '');
        }

        return ans;
    }

大神总是能把多个步骤一次到位：

string multiply(string num1, string num2) {
    string sum(num1.size() + num2.size(), '');

    for (int i = num1.size() - ;  <= i; --i) {
        int carry = ;
        for (int j = num2.size() - ;  <= j; --j) {
            int tmp = (sum[i + j + ] - '') + (num1[i] - '') * (num2[j] - '') + carry; //把当前乘出来的数字和之前的数字以及进位相加
            sum[i + j + ] = tmp %  + '';
            carry = tmp / ;
        }
        sum[i] += carry; //这个是最高位多出来的进位 其他的都是i+j+1 这里没有+1
    }

    size_t startpos = sum.find_first_not_of("");
    if (string::npos != startpos) {
        return sum.substr(startpos);
    }
    return "";
}