python-3.x - 两个数字相加问题（链表） - Python - Leetcode - AttributeError

Question

我正在尝试解决这个问题 - 添加两个在Leetcode上的数字

我尝试将两个链表都转换为数组，然后执行添加操作。现在，我正在努力将它们转换回链接列表，这是问题的所需输出。

谁能检查我哪里出错了？我也收到一个属性错误：

AttributeError：“NoneType”对象没有属性“val”

这是我写的代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
  def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    a = l1 #pointers
    b = l2 #pointers
    arr1 = []
    arr2 = []

    while a.next is not None:
      arr1.append(a.val)
      a = a.next
    arr1.append(a.val) #storing the values of linked lists in arrays/lists

    while b.next is not None:
      arr2.append(b.val)
      b = b.next
    arr2.append(b.val) #storing the values of linked lists in arrays/lists

    rev1 = reversed(arr1) #reversed list
    rev2 = reversed(arr2) #reversed list

    inta = "".join(str(rev1)) #converting list to strings
    intb = "".join(str(rev2))

    c = str(inta + intb) #performing addition - the answer we wanted
    revc = reversed(c) #answer in string form - reversed (output in string at present)

    #trying to convert into linked list and return it
    q = l1
    for i in revc:
      q.val = i
      q = q.next
    return l1

标签： python-3.xlistdata-structureslinked-listattributeerror

Answer 1

class Solution:

    
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
    carry = 0
    head = curr = ListNode()
        
    while l1 and l2:
        total = l1.val + l2.val + carry
        curr.next = ListNode(total% 10)
        carry = total // 10
        l1,l2,curr = l1.next, l2.next,curr.next
            
    while l1:
        total = l1.val + carry
        curr.next = ListNode(total%10)
        carry = total // 10
        l1, curr = l1.next, curr.next
            
    while l2:
        total = l2.val + carry
        curr.next = ListNode(total%10)
        carry = total//10
        l2, curr = l2.next, curr.next
    if carry > 0:
        curr.next  = ListNode(carry)
                
    return head.next