leetcode/0002_add-two-numbers/python3/solution.py

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# Time: O(max(m, n))
# Space: O(max(m, n) + 1)
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
sum_head = None
l1ptr, l2ptr, curr = l1, l2, None
carry = 0
# Iterate until l1 and l2 are exhausted
while l1ptr != None or l2ptr != None:
l1val = 0 if l1ptr is None else l1ptr.val
l2val = 0 if l2ptr is None else l2ptr.val
# Basic sum with carry on the digits
raw_sum = l1val + l2val + carry
sum_wo_carry = raw_sum % 10
carry = 1 if raw_sum >= 10 else 0
if sum_head is None:
curr = ListNode(sum_wo_carry)
sum_head = curr
else:
curr.next = ListNode(sum_wo_carry)
curr = curr.next
l1ptr = None if l1ptr is None else l1ptr.next
l2ptr = None if l2ptr is None else l2ptr.next
if carry > 0:
curr.next = ListNode(carry)
return sum_head