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This article will explain in detail how to multiply large numbers with golang leetcode skills. The editor thinks it is very practical, so I share it with you as a reference. I hope you can get something after reading this article.

Given two non-negative integers num1 and num2 represented as a string, the product of num1 and num2 is returned, and their product is also expressed as a string.

Example 1:

Enter: num1 = "2", num2 = "3"

Output: "6"

Example 2:

Enter: num1 = "123", num2 =" 456 "

Output: "56088"

Description:

The length of num1 and num2 is less than 110.

Num1 and num2 contain only the numbers 0-9.

Neither num1 nor num2 starts with zero unless it is the number 0 itself.

You cannot use any of the standard library's large number types (such as BigInteger) or directly convert input to integers.

Ideas for solving the problem:

1, the maximum length of multiplying two numbers is the addition of the length of two numbers

2Quinum1 [I] and num2 [j] multiply, the result only affects the isimj bit and isimj + 1 bit.

R [i+j+1] + = (r [I + j] + N1 [I] * N2 [j]) / 10 r [iSj] = (r [I + j] + N1 [I] * N2 [j])

3. If the calculation result is transferred to string, you need to pay attention to the 0 of the header. If all 0, keep a 0.

Code implementation:

Func multiply (num1 string, num2 string) string {n1:=str2int (num1) n2:=str2int (num2) if n1==nil | | n2==nil {return ""}

R:=make ([] int64,len (N1) + len (N2))

For iPUR.

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