两数相加

题目介绍

难度中等

给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的，并且每个节点只能存储一位数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例 1：

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

示例 2：

输入：l1 = [0], l2 = [0]
输出：[0]

示例 3：

输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

提示：

• 每个链表中的节点数在范围 [1, 100] 内
• 0 <= Node.val <= 9
• 题目数据保证列表表示的数字不含前导零

解答

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        p1 = l1
        p2 = l2
        tail = l1
        c = 0
        while p1 is not None and p2 is not None:
            val = p1.val + p2.val + c
            if val > 9:
                p1.val = val % 10
                c = 1
            else:
                p1.val = val
                c = 0
            tail = p1
            p1 = p1.next
            p2 = p2.next
        
        while p1 is not None:
            val = p1.val + c
            if val > 9:
                p1.val = val % 10
                c = 1
            else:
                p1.val = val
                c = 0
            tail = p1
            p1 = p1.next
        
        while p2 is not None:
            val = p2.val + c
            tail.next = ListNode()
            tail = tail.next
            p1 = tail
            if val > 9:
                p1.val = val % 10
                c = 1
            else:
                p1.val = val
                c = 0
            p2 = p2.next

        if c == 1:
            tail.next = ListNode()
            tail = tail.next
            p1 = tail
            p1.val = 1
        return l1