【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 == 1 && num1[0] == '0') return "0";

        if(l2 == 1 && num2[0] == '0') return "0";



        int maxl = 0;

        //用vector存储两个数字 v[0]是最高位

        for(int i = 0; i < l1; i++)

            v1.push_back(num1[i] - '0');

        for(int i = 0; i < l2; i++)

            v2.push_back(num2[i] - '0');



        //乘步骤

        for(int i = l1 - 1; i >= 0; i--)

        {

            vector<int> vtmp; //v[0]是最低位

            int t = i;

            while(t < l1 - 1) //后面补错位的0

            {

                vtmp.push_back(0);

                t++;

            }

            int c = 0; //记录进位

            for(int j = l2 - 1; j >= 0; j--)

            {

                int cur = v1[i] * v2[j] + c;

                vtmp.push_back(cur % 10);

                c = cur / 10;

            }

            if(c != 0)

                vtmp.push_back(c);

            vvtmp.push_back(vtmp);

            maxl = (vtmp.size() > maxl) ? vtmp.size() : maxl;

        }



        //加步骤

        int c = 0; //记录进位

        for(int i = 0; i < maxl; i++)

        {

            int cur = c;

            for(int j = 0; j < vvtmp.size(); j++)

            {

                if(i >= vvtmp[j].size())

                    continue;

                cur += vvtmp[j][i];

            }

            vmulti.push_back(cur % 10);

            c = cur / 10;

        }

        if(c != 0)

            vmulti.push_back(c);



        //转换为string

        string ans;

        for(int i = vmulti.size() - 1; i >= 0; i--)

        {

            ans += (vmulti[i] + '0');

        }



        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(), '0');



    for (int i = num1.size() - 1; 0 <= i; --i) {

        int carry = 0;

        for (int j = num2.size() - 1; 0 <= j; --j) {

            int tmp = (sum[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0') + 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 + 1] = tmp % 10 + '0';

            carry = tmp / 10;

        }

        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("0");

    if (string::npos != startpos) {

        return sum.substr(startpos);

    }

    return "0";

}