43. Multiply Strings #

Problem #

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

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

Example 1:

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

Example 2:

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

Constraints:

• 1 <= num1.length, num2.length <= 200
• num1 and num2 consist of digits only.
• Both num1 and num2 do not contain any leading zero, except the number 0 itself.

Problem Summary #

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

Solution Approach #

• Use an array to simulate multiplication. Create an array of length len(num1) + len(num2) to store the product. For any 0 ≤ i < len(num1), 0 ≤ j < len(num2), the result of num1[i] * num2[j] is located at tmp[i+j+1]; if tmp[i+j+1]≥10, add the carry part to tmp[i+j]. Finally, convert the array tmp into a string, and discard the highest digit if it is 0.

Code #

package leetcode

func multiply(num1 string, num2 string) string {
	if num1 == "0" || num2 == "0" {
		return "0"
	}
	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
	for i := 0; i < len(b1); i++ {
		for j := 0; j < len(b2); j++ {
			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
		}
	}
	for i := len(tmp) - 1; i > 0; i-- {
		tmp[i-1] += tmp[i] / 10
		tmp[i] = tmp[i] % 10
	}
	if tmp[0] == 0 {
		tmp = tmp[1:]
	}
	res := make([]byte, len(tmp))
	for i := 0; i < len(tmp); i++ {
		res[i] = '0' + byte(tmp[i])
	}
	return string(res)
}