Leetcode 2

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Question

Solution

首先创建一个首位元素为0的空链表，然后创建carry = 0. 如果l1和l2不为空，则以此读取当前位置，然后加到carry上，之后考虑如何将carry添加到开始创建的空链表上，注意carry的位数，以及如果进位了需要保留。

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# 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:
        new = cur = ListNode(0)
        carry = 0
        while l1 or l2 or carry:
            if l1:
                carry += l1.val
                l1 = l1.next
            if l2:
                carry += l2.val
                l2 = l2.next
            cur.next = ListNode(carry % 10)
            cur = cur.next
            carry = carry // 10
        return new.next