Add Two Numbers

LeetCode - The World's Leading Online Programming Learning Platform You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
您将获得两个非空链表，代表两个非负整数。数字以相反的顺序存储，其每个节点都包含一个数字。将两个数字相加，并将总和作为链表返回。

You may assume the two numbers do not contain any leading zero, except the number 0 itself.
您可以假设这两个数字不包含任何前导零，除了数字 0 本身。

Example 1: 示例 1：

Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.

Example 2: 示例 2：

Input: l1 = [0], l2 = [0] Output: [0]

Example 3: 例3：

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]

Constraints: 约束：

• The number of nodes in each linked list is in the range [1, 100].
  每个链表中的节点数在 [1, 100] 范围内。
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.
  保证列表表示没有前导零的数字。

# Definition for singly-linked list.  
class ListNode(object):  
    def __init__(self, val=0, next=None):  
        self.val = val  
        self.next = next  
  
class Solution(object):  
    def addTwoNumbers(self, l1, l2):  
        """  
        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode  
        """        head = ListNode(0)  # Dummy head node  
        current = head  # Pointer to the current node  
        carry = 0  # Variable to store the carry value  
  
        while l1 or l2 or carry:  
            # Get the values of the current nodes, or 0 if one of them is None  
            val1 = l1.val if l1 else 0  
            val2 = l2.val if l2 else 0  
  
            # Calculate the sum of the current digits and the carry value  
            total = val1 + val2 + carry  
  
            # Calculate the carry for the next iteration  
            carry = total // 10  
  
            # Create a new node with the units digit of the sum  
            current.next = ListNode(total % 10)  
  
            # Move the pointers of the linked lists to the next nodes  
            l1 = l1.next if l1 else None  
            l2 = l2.next if l2 else None  
  
            # Move the current pointer to the next node  
            current = current.next  
  
        return head.next

1. Create a dummy head node and initialize a pointer current to it.
   创建一个虚拟头节点并初始化指向它的指针 current 。
2. Initialize a variable carry to 0 to store the carry value.
   将变量 carry 初始化为 0 以存储进位值。
3. Iterate through both linked lists simultaneously using a while loop while either of the linked lists or the carry value exists.
   当链表或进位值存在时，使用 while 循环同时循环访问两个链表。
4. Get the values of the current nodes val1 and val2, or 0 if one of them is None.
   获取当前节点的值， val2 如果其中一个节点 val1 为 None，则为 0。
5. Calculate the sum of the current digits and the carry value, and store it in total.
   计算当前数字和账位价值的总和，并将其存储在 total .
6. Calculate the carry for the next iteration by dividing total by 10.
   除以 total 10 计算下一次迭代的进位。
7. Create a new node with the units digit of total and set the next pointer of the current node to it.
   创建一个单位数字为 的新 total 节点，并将节点的下一个指针设置为该 current 节点。
8. Move the pointers of the linked lists to the next nodes if they exist.
   将链表的指针移动到下一个节点（如果存在）。
9. Move the current pointer to the next node.
   将 current 指针移动到下一个节点。
10. Once the while loop finishes, return the next pointer of the dummy head node.
    while 循环完成后，返回虚拟头节点的下一个指针。