[leetcode]43. Multiply Strings高精度乘法

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.

题意：

高精度乘法。

Solution1： Math

Do the simulation like how computer will do multiplication operation

1一个char对应1个digit，

digit相乘后，注意其乘积结果可能需要进位

code

 /*
 Time: O(n^2). We use nested 2 for loop
 Space: O(n). We use int[] to save intermedia infor
 */

 class Solution {
     public String multiply(String num1, String num2) {
         if(num1.length()==0 || num2.length()==0) return "0";
         int len1 = num1.length();
         int len2 = num2.length();
         int [] result = new int [len1+len2];

         for(int i = len1-1; i>=0; i--){
             for(int j = len2-1; j>=0;j--){
                 int mul = (num1.charAt(i)-'0')*(num2.charAt(j)-'0');
                 int idx = i+j+1;
                 // 可能会进位
                 int carryIdx = i+j;
                 mul = mul + result[idx];
                 result[idx] = mul % 10;
                 result[carryIdx] = result[carryIdx] + mul/10;
             }
         }
         StringBuilder sb = new StringBuilder();
         for(int res: result){
             if(sb.length()!=0 || res!=0) sb.append(res);
         }
         return (sb.length() == 0)? "0" : sb.toString();
     }
 }