LeetCode: Add Two Numbers

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.

Solutions:

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 # 方法一：
 7 class Solution:
 8     def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
 9         num1 = ''
10         while l1 and l1.val:
11             num1 += str(l1.val)
12             l1 = l1.next
13         num1 = int(num1[::-1])
14         num2 = ''
15         while l2 and l2.val:
16             num2 += str(l2.val)
17             l2 = l2.next
18         num2 = int(num2[::-1])
19         sum = str(num1 + num2)[::-1]
20         head = ListNode(sum[0])
21         tmp = head
22         i = 1
23         while i < len(sum):
24             tmp.next = ListNode(sum[i])
25             tmp = tmp.next
26             i += 1
27         return head
28              
29

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 # 方法二： 递归实现：
 7 
 8 class Solution:
 9     def addTwoNumbers(self, l1: ListNode, l2: ListNode, c = 0) -> ListNode:
10         val = l1.val + l2.val + c
11         c = val // 10
12         ret = ListNode(val % 10 ) 
13         
14         if (l1.next != None or l2.next != None or c != 0):
15             if l1.next == None:
16                 l1.next = ListNode(0)
17             if l2.next == None:
18                 l2.next = ListNode(0)
19             ret.next = self.addTwoNumbers(l1.next,l2.next,c)
20         return ret
21

原文地址：https://www.cnblogs.com/noodleman/p/11830896.html