[LeetCode][Python3] 2. Add Two Numbers

Problem :

https://leetcode.com/problems/add-two-numbers/description/

My Solution :

# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution:

def addTwoNumbers(self, l1, l2):

"""

:type l1: ListNode

:type l2: ListNode

:rtype: ListNode

"""

dummy_head = ListNode(0)

current = dummy_head

carry = 0

while l1 and l2:

val = (l1.val + l2.val + carry) % 10

carry = (l1.val + l2.val + carry) // 10

current.next = ListNode(val)

current = current.next

l1 = l1.next

l2 = l2.next

remain = l1 or l2

while remain:

val = (remain.val + carry) % 10

carry = (remain.val + carry) // 10

current.next = ListNode(val)

current = current.next

remain = remain.next

if carry:

current.next = ListNode(1)

return dummy_head.next

Comment :

위 코드는 내가 처음으로 시도했던 방법이고, while문을 중복으로 사용하는게 비 효율적이라 판단되어 좀 더 개선된 아래 코드를 만들어 보았다.

My Solution2 :

# Definition for singly-linked list.

# class ListNode:

# def __init__(self, x):

# self.val = x

# self.next = None

class Solution:

def addTwoNumbers(self, l1, l2):

"""

:type l1: ListNode

:type l2: ListNode

:rtype: ListNode

"""

dummy_head = ListNode(0)

current = dummy_head

carry = 0

while l1 or l2 or carry:

s = carry

if l1:

s += l1.val

l1 = l1.next

if l2:

s += l2.val

l2 = l2.next

current.next = ListNode(s % 10)

current = current.next

carry = s // 10

return dummy_head.next