Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.
Note:Â 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 -
num1andnum2consist of digits only. - Both
num1andnum2Â do not contain any leading zero, except the number0itself.
SOLUTION:
class Solution:
def addStrings(self, num1: str, num2: str) -> str:
m = len(num1)
n = len(num2)
if m < n:
num1 = "0" * (n - m) + num1
else:
num2 = "0" * (m - n) + num2
carry = 0
i = max(m, n) - 1
op = ""
while i >= 0:
val = carry + int(num1[i]) + int(num2[i])
op = str(val % 10) + op
carry = val // 10
i -= 1
if carry > 0:
op = "1" + op
return op
def mulNtimes(self, num, n):
if n == 0:
return "0"
if n == 1:
return num
op = "0"
if n & 1:
a = num
b = self.mulNtimes(num, (n - 1) // 2)
b = self.addStrings(b, b)
return self.addStrings(a, b)
else:
a = self.mulNtimes(num, n // 2)
return self.addStrings(a, a)
def multiply(self, num1: str, num2: str) -> str:
op = "0"
n = len(num2)
for i in range(n):
curr = self.mulNtimes(num1, int(num2[n - i - 1]))
if curr != "0":
curr += "0" * i
op = self.addStrings(op, curr)
return op
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