This year, I am solving more data structures and algorithms.
P.S.: This post was first seen on Twitter. See the link below.
The Problem is 2. Add Two Numbers from LeetCode. Check it out here.
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
Constraints:
The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.
Here is my solution:
def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
current_l1 = l1
current_l2 = l2
l1_list, l2_list = [], []
# Convert both LinkedLists to Python Lists
while current_l1 is not None or current_l2 is not None:
if current_l1 is not None:
l1_list.append(str(current_l1.val))
current_l1 = current_l1.next
if current_l2 is not None:
l2_list.append(str(current_l2.val))
current_l2 = current_l2.next
l1_list.reverse() # Reverse lists e.g. [2, 4, 3] -> [3, 4, 2]
l2_list.reverse() # Reverse lists e.g. [2, 4, 3] -> [3, 4, 2]
# Convert the lists to integers. e.g. [3, 4, 2] -> 342
l1_value = int(''.join(l1_list))
l2_value = int(''.join(l2_list))
sum_values = l1_value + l2_value
sum_values = str(sum_values)
# Create the head node
head = ListNode(sum_values[-1])
current = head
# Insert the elements to LinkedList in reverse order
for digit in sum_values[-2::-1]:
current.next = ListNode(digit)
current = current.next
return head
Lines 2-3:
I create current_l1
and current_l2
to be nodes pointing to the heads of l1
and l2
, respectively.
Line 4 creates empty lists l1_list
and l2_list
to store each digit of l1
and l2
.
Lines 7-14:
I convert the linked lists l1
and l2
into Python lists (l1_list
and l2_list
) by traversing through them, extracting each digit from the linked lists and appending it to the respective lists.
Lines 16-17:
I reverse l1_list
and l2_list
lists. I do this because the digits are in reverse order in the linked lists
Lines 20-21:
I convert the reversed lists to integers (l1_value
and l2_value
) by using the join method to convert to a string and then to an integer ultimately
Lines 23-24:
I calculate the sum of both integers and then convert the sum back to a string (sum_values
) to easily extract each digit.
Lines 27-34:
I create a head node (last digit of the sum), create a pointer to it, and iterate through the remaining digits of the sum in reverse order, creating new nodes for each digit in the sum string (sum_values). The head is then returned.
Here's the performance of my approach as of today:
Runtime: 54ms, beats 79.12% of users with Python3
Memory: 16.60MB, beats 67.14% of users with Python3
Do you have a better-performing approach? What problems would you want to see me solve?
Let me know in the comments.
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