43. Multiply Strings

43. Multiply Strings

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.

2. Both num1 and num2 contain only digits 0-9.

3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.

4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

My Solutions:

因为相乘的结果数字可能很大，所以不能按照乘法的正常方法来做（数字相乘再相加）。

class Solution {
    public String multiply(String num1, String num2) {
        if (num1 == null || num2 == null || num1.length() == 0 || num2.length() == 0) return "0";
        if (num1.equals("0") || num2.equals("0")) return "0";

        int[] res = new int[num1.length() + num2.length()];
        
        for (int i = num1.length() - 1; i >= 0; i--) {

            for (int j = num2.length() - 1; j >= 0; j--) {
                int product = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
                // p1是靠左的index，p2是靠右的
                int p1 = i + j, p2 = p1 + 1;
                
                int sum = res[p2] + product; //sum需要加上当下位置的数
                res[p1] += sum / 10; //p1是左边的位置
                res[p2] = sum % 10; //p2是当下位置
            } 
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i : res) {
            if (!(i == 0 && sb.length() == 0)) { // e.g.[0, 0, 1, 0, 3] 跳过leading zero
                sb.append(i);
            }
        }
        return sb.toString();
    }
}

Time: O(n * m); Space: (n + m)