我的写法是这样

输入: l1=[2,4,3], l2=[5,6,4]
↓
反转: [3,4,2], [4,6,5]
↓
转数字: 342, 465
↓
相加: 807
↓
转链表: [8,0,7]
↓
再反转: [7,0,8]

class Solution:
    def addTwoNumbers(self, l1, l2):
        def reverse_list(head):
            prev = None
            current = head
            while current:
                t = current.next
                current.next = prev
                prev = current
                current = t
            return prev
        def link_to_number(head):
            num = 0
            current = head
            while current:
                num = num * 10 + current.val  # 123 = 1*100 + 2*10 + 3
                current = current.next
            return num
        def number_to_link(num):
            if num == 0:
                return ListNode(0)
            head = None
            while num > 0:
                g = num % 10
                node = ListNode(g)
                node.next = head
                head = node
                num //= 10
            return head
        fanzhuan1 = reverse_list(l1) 
        fanzhuan2 = reverse_list(l2)
        bianli1 = link_to_number(fanzhuan1)  
        bianli2 = link_to_number(fanzhuan2)
        he = bianli1 + bianli2
        result = number_to_link(he)
        final_result = reverse_list(result)
        return final_result  
标准答案不用第一次反转直接逆位

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        标准解法：逐位相加，处理进位
        时间复杂度: O(max(m, n))，m 和 n 分别是两个链表的长度
        空间复杂度: O(max(m, n))，新链表的长度最多为 max(m, n) + 1
        """
        # 创建虚拟头节点，简化代码
        dummy = ListNode(0)
        current = dummy
        carry = 0  # 进位
        
        # 同时遍历两个链表，直到都为空且没有进位
        while l1 or l2 or carry:
            # 获取当前位的值，如果链表为空则为 0
            val1 = l1.val if l1 else 0
            val2 = l2.val if l2 else 0
            
            # 计算当前位的和（包括进位）
            total = val1 + val2 + carry
            
            # 计算当前位的值和新的进位
            carry = total // 10
            digit = total % 10
            
            # 创建新节点并连接到结果链表
            current.next = ListNode(digit)
            current = current.next
            
            # 移动到下一个节点
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        
        # 返回结果链表的头节点（跳过虚拟头节点）
        return dummy.next