feat(0002_add-two-numbers): add py3 solution
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0002_add-two-numbers/README.md
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0002_add-two-numbers/README.md
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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.
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You may assume the two numbers do not contain any leading zero, except the number 0 itself.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg)
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Input: l1 = [2,4,3], l2 = [5,6,4]
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Output: [7,0,8]
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Explanation: 342 + 465 = 807.
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**Example 2:**
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Input: l1 = [0], l2 = [0]
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Output: [0]
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**Example 3:**
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Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
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Output: [8,9,9,9,0,0,0,1]
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**Constraints:**
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* The number of nodes in each linked list is in the range `[1, 100]`.
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* `0 <= Node.val <= 9`
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* It is guaranteed that the list represents a number that does not have leading zeros.
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0002_add-two-numbers/python3/solution.py
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0002_add-two-numbers/python3/solution.py
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# Time: O(max(m, n))
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# Space: O(max(m, n) + 1)
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
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sum_head = None
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l1ptr, l2ptr, curr = l1, l2, None
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carry = 0
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# Iterate until l1 and l2 are exhausted
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while l1ptr != None or l2ptr != None:
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l1val = 0 if l1ptr is None else l1ptr.val
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l2val = 0 if l2ptr is None else l2ptr.val
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# Basic sum with carry on the digits
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raw_sum = l1val + l2val + carry
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sum_wo_carry = raw_sum % 10
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carry = 1 if raw_sum >= 10 else 0
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if sum_head is None:
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curr = ListNode(sum_wo_carry)
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sum_head = curr
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else:
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curr.next = ListNode(sum_wo_carry)
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curr = curr.next
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l1ptr = None if l1ptr is None else l1ptr.next
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l2ptr = None if l2ptr is None else l2ptr.next
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if carry > 0:
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curr.next = ListNode(carry)
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return sum_head
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