package q43;

/**
 * o(n) 可基于乘数某位与被乘数某位相乘产生结果的位置的规律优化
 */
class Solution {

    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        String res = "0";

        for (int i = num2.length() - 1; i >= 0; i--) {
            int carry = 0;
            StringBuilder temp = new StringBuilder();
            for (int j = 0; j < num2.length() - 1 - i; j++) {
                temp.append(0);
            }
            int n2 = num2.charAt(i) - '0';

            for (int j = num1.length() - 1; j >= 0 || carry != 0; j--) {
                int n1 = j < 0 ? 0 : num1.charAt(j) - '0';
                int product = (n1 * n2 + carry) % 10;
                temp.append(product);
                carry = (n1 * n2 + carry) / 10;
            }
            res = addStrings(res, temp.reverse().toString());
        }
        return res;
    }

    public String addStrings(String num1, String num2) {
        StringBuilder builder = new StringBuilder();
        int carry = 0;
        for (int i = num1.length() - 1, j = num2.length() - 1;
             i >= 0 || j >= 0 || carry != 0;
             i--, j--) {
            int x = i < 0 ? 0 : num1.charAt(i) - '0';
            int y = j < 0 ? 0 : num2.charAt(j) - '0';
            int sum = (x + y + carry) % 10;
            builder.append(sum);
            carry = (x + y + carry) / 10;
        }
        return builder.reverse().toString();
    }
}