2 两数相加

2. 两数相加 - 力扣（LeetCode）

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1 题目

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]

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2 解题

# 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]:if l1 is None :return l2if l2 is None:return l1prehead = ListNode(0) #创建一个val为0的单节点head = prehead #指针开始的地方carry = 0while l1 and l2: #同时retain = (l1.val + l2.val + carry)%10carry = (l1.val + l2.val + carry )//10head.next = ListNode(retain)l1 = l1.nextl2 = l2.nexthead = head.nextwhile l1: # l1 too longretain = (l1.val +carry)%10carry = (l1.val +carry)//10head.next = ListNode(retain)l1 = l1.nexthead = head.nextwhile l2 :retain = (l2.val + carry)%10carry = (l2.val + carry)//10head.next = ListNode(retain)l2 = l2.nexthead = head.nextif carry == 1:head.next  = ListNode(carry)return prehead.next