【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";
}