leetcode 2. 两数相加(python)

给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

---

1、正常思路

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

class Solution(object):
    # maybe standard version
    def _addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = dummy = ListNode(-1)
        carry = 0
        while l1 and l2:
            p.next = ListNode(l1.val + l2.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            l1 = l1.next
            l2 = l2.next
        
        res = l1 or l2
        while res:
            p.next = ListNode(res.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            res = res.next
        if carry:
            p.next = ListNode(1)
        return dummy.next

2、大神解法：

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

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = dummy = ListNode(-1)
        carry = 0
        while l1 or l2 or carry:
            val = (l1 and l1.val or 0) + (l2 and l2.val or 0) + carry
            carry = val / 10
            p.next = ListNode(val % 10)
            l1 = l1 and l1.next
            l2 = l2 and l2.next
            p = p.next
        return dummy.next