LeetCode Multiply Strings

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.

class Solution {

public:

    string multiply(string num1, string num2) {

        int len1 = remove_leading_zero(num1);

        int len2 = remove_leading_zero(num2);



        string muls[10];

        muls[0] = "0";

        for (int i=1; i<10; i++) {

            muls[i] = add(muls[i-1], num1);

        }

        

        string ending_zero(""), res("");

        int pos2 = len2 - 1;

        while (pos2 >= 0) {

            int c = num2[pos2--] - '0';

            res = add(res, muls[c] + ending_zero);

            ending_zero.insert(ending_zero.end(), '0');

        }

        return res;

    }

    

    string add(string num1, string num2) {

        string res("");

        int pos1 = remove_leading_zero(num1) - 1;

        int pos2 = remove_leading_zero(num2) - 1;



        int carry = 0;

        

        while (pos1 >=0 && pos2 >= 0) {

            int c1 = num1[pos1--] - '0';

            int c2 = num2[pos2--] - '0';

            int sum= c1 + c2 + carry;

            carry = sum / 10;

            sum = sum % 10;

            res.insert(res.begin(), sum + '0');

        }

        while (pos1 >=0 ) {

            int c = num1[pos1--] - '0';

            int s = c + carry;

            carry = s / 10;

            s = s % 10;

            res.insert(res.begin(), s + '0');

        }

        while (pos2 >=0 ) {

            int c = num2[pos2--] - '0';

            int s = c + carry;

            carry = s / 10;

            s = s % 10;

            res.insert(res.begin(), s + '0');

        }

        if (carry > 0 ) res.insert(res.begin(), carry + '0');

        return res;

    }

    

    int remove_leading_zero(string& str) {

        int len = str.length();

        int pos = 0;

        

        while(pos < (len - 1) && str[pos] == '0') pos++;

        len = len - pos;

        str = str.substr(pos, len);

        return len;

    }

};

200ms，用纯C的话应该可以再快一点,

方法不对，可以直接在各个数字位相乘时使用乘法：

class Solution {

public:

    string multiply(string num1, string num2) {

        int alen = num1.size();

        int blen = num2.size();

        reverse(num1.begin(), num1.end());

        reverse(num2.begin(), num2.end());

        

        vector<int> res(alen + blen, 0);

        

        for (int i=0; i<alen; i++) {

            for (int j=0; j<blen; j++) {

                res[i + j] += (num1[i] - '0') * (num2[j] - '0');

            }

        }

        int pos = alen + blen - 1;

        while (pos >= 0 && res[pos] == 0) {

            pos--;

        }

        string str;

        int carry = 0;

        for (int i=0; i<=pos; i++) {

            res[i]= res[i] + carry;

            carry = res[i] / 10;

            str.push_back(res[i] % 10 + '0');

        }

        if (carry != 0) {

            str.push_back(carry + '0');

        }

        reverse(str.begin(), str.end());

        if (str.size() == 0) {

            str = "0";

        }

        return str;

    }

};