Daily LeetCode 2. Add Two Numbers

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

Medium

问题描述：

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.

题目分析及代码：

LeetCode上的第二条题目，一道与链表操作相关的题目，两个链表，将对应位置上的数相加，超过十则进位，输出最终得到的链表。

思路1:

由于链表保存的顺序就是从低位到高位，因此我们直接遍历两个列表，按题目描述那样，对应位置相加，只需要注意最后一位相加需要进位的情况，例如(5) + (5)，在遍历完成后需要对这种情况做额外处理，使得最终输出为0->1而不是0。

代码：

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        res = point = ListNode(0)
        extra = 0
        
        while l1 != None or l2 != None:
            l1_ele = l1.val if l1 != None else 0
            l2_ele = l2.val if l2 != None else 0
            sum = l1_ele + l2_ele + extra
            extra = 1 if sum >= 10 else 0
            point.next = ListNode(int(sum%10))
            point = point.next
            if l1 != None: l1 = l1.next
            if l2 != None: l2 = l2.next
        
        if extra == 1:
            point.next = ListNode(1)
        
        return res.next

思路2:

可以通过递归来减少空间复杂度

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 != None and l2 != None:
            return 
        elif (l1 and l2) == None:
            return l1 or l2
        else:
            if l1.val + l2.val < 10:
                res = ListNode(l1.val + l2.val)
                res.next = self.addTwoNumbers(l1.next, l2.next)
            else:
                res = ListNode(l1.val + l2.val - 10)
                res.next = self.addTwoNumbers(l1.next, self.addTwoNumbers(l2.next, ListNode(1)))
        return res