[LeetCode][Python3] 2. Add Two Numbers

snoopybox 2018. 9. 24. 01:27

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