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.

I was not very familiar with the large integers represented by strings in the multiplication of large integers, and it is not clear how to handle overflow. So I have referred to others' practices, then I redo it with my impressions.

• Direct multiplication overflows, so two single digit pairs must be multiplied each time. The maximum value is 81, and no overflow occurs.
• For example, 385*97 means a single digit = 5*7, ten digits = 8*7 + 5*9, and a hundred digits = 3*7 + 8*9...
  Each digit can be represented by an int, which exists in an int.
• The maximum length of this array is num1.len + num2.len. For example, 99*99, the maximum length is no more than 10000, so four digits are enough.
• This kind of BIT is behind, it is not easy to do (10 to the power of 0, but unfortunately the index of the corresponding bit array is not 0 but N-1 ),
  So the code of string reverse is much clearer first.
• The first 0 in the final result must be cleared.

The methods here are clever and typical and need to be mastered. Also pay attention to the overflow processing. Here, the processing is: if the highest bit produces carry, then ignore this carry directly.

 1 public class Solution { 2     public String multiply(String num1, String num2) { 3         num1 = new StringBuffer(num1).reverse().toString(); 4         num2 = new StringBuffer(num2).reverse().toString(); 5         int[] n1 = new int[num1.length()]; 6         int[] n2 = new int[num2.length()]; 7         int[] temp = new int[num1.length() + num2.length()]; 8         for (int i = 0; i < num1.length(); i++) { 9             n1[i] = (int)(num1.charAt(i) - ‘0‘); 10             for (int j = 0; j < num2.length(); j++) {11                 n2[j] = (int)(num2.charAt(j) - ‘0‘);12                 temp[i + j] += n1[i] * n2[j]; 13             }14         }15         16         StringBuffer sb = new StringBuffer();17         int digit = 0;18         int carry = 0;19         for (int k = 0; k < temp.length; k++) {20             digit = temp[k] % 10;21             carry = temp[k] / 10;22             sb.insert(0, digit);23             if (k < temp.length - 1) {24                 temp[k + 1] += carry;25             }26         }27         28         while (sb.length() > 0 && sb.charAt(0) == ‘0‘) {29             sb.deleteCharAt(0);30         }31         return sb.length() == 0? "0" : sb.toString();32     }33 }

 

Leetcode: multiply strings