leetcode 第2题(Add Two Numbers)

给定两个非空列表用来表示两个正整数，给定的数以反序存储在链表中，链表中每个节点的值表示一位数，将两个数字求和并以链表形式返回

可以假定除0外，所有给定的数字都不包含前导0

例：

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解法1（理解较容易）：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        list1 = []
        list2 = []
        while l1:
            list1.insert(0, l1.val)
            l1 = l1.next
        while l2:
            list2.insert(0, l2.val)
            l2 = l2.next
        
        result = int(''.join(map(str, list1))) + int(''.join(map(str, list2)))
        
        s = str(result)
        k = None
        
        for i in range(len(s)):
            node = ListNode(int(s[i]))
            node.next = k
            k = node
        return k

解法2（pythonic，性能更高）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        
        head = ListNode(0)
        tail = head
        carry = 0
        while l1 or l2 or carry:
            val1 = l1.val if l1 else 0
            val2 = l2.val if l2 else 0
            
            carry, val = divmod(val1 + val2 + carry, 10)
            
            tail.next = ListNode(val)
            tail = tail.next
            
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return head.next