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.

思路

经典的题目，但是今天写还是犯了错误，易犯错误如下：

当cur1和cur2有一个为null就终止，然后把剩下的续上。这样的问题在于进位，进位有可能进很多位，例如1+99999，正确做法是cur1和cur2都为null才终止，如果其中一边用完了就当成0处理就完事。最后可能会多一个carry为1的要单独加上。例如99+99.

代码

class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: if l1 == None or l2 == None: return None dummy = ListNode(0) cur = dummy cur1 = l1 cur2 = l2 carry = 0 while cur1 or cur2: v1 = cur1.val if cur1 else 0 v2 = cur2.val if cur2 else 0 cur.next = ListNode((v1 + v2 + carry) % 10) carry = (v1 + v2 + carry) // 10 if cur1: cur1 = cur1.next if cur2: cur2 = cur2.next cur = cur.next if carry != 0: cur.next = ListNode(carry) cur = cur.next return dummy.next