LeetCode:2. Add Two Numbers

https://leetcode.com/problems/add-two-numbers/description/

内容描述

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 contain a single digit. Add the two numbers and return it as a linked list.You may assume the two numbers do not contain any leading zero, except the number 0 itself.Example:Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

这个题上来的直观想法肯定是链表翻转过来然后求和，之后求出来的数再变成链表再反转回去。但是有没有更好的思路呢？因为要遍历链表，时间复杂度必然是o(N），但是空间复杂度可不可以降低呢，一次出结果，不要中间存储过程。答案的确是可以的，使用递归操作。

这里的思路就是使用加法的办法。对应元素相加，如果相加的和超过1，就要考虑进位。因为是递归操作，所以实际上是链表最后的位置倒着往前加，完美的满足了题目的要求。这里需要特别注意的是关于10的判断和进位操作。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:#因为链表到最后就是None了，所以这里判断一下，如果都是None就返回if not l1 and not l2:return elif not (l1 and l2): #如果L1或者L2有一个是Nonereturn l1 or l2else:if l1.val + l2.val < 10:  # 如果相加不满足10，不执行进位l3 = ListNode(l1.val + l2.val)l3.next = self.addTwoNumbers(l1.next, l2.next)else: l3 = ListNode(l1.val + l2.val - 10)  # 相加满足10的话要有一个进位l3.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next, ListNode(1)))return l3

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