445.Two numbersII

topic：

445.Two numbers

Thought：

I have been with yesterday’s question，Find by the way
What is deducted is a very strange oneprecompiled.listnode.ListNode，But mine is'__main__.ListNode'。
Because：Repeatedly defined$ListNode$
不反转Linkedof方法可能就是昨天of那种，转化为math做。
The code is still0x3fof
and昨天ofanswer法一样

Code：

class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return head
        new_head = self.reverseList(head.next)
        head.next.next = head  # Direct the next node to yourself
        head.next = None  # 断开指向下一个节点of连接，ensure最终Linkedof末尾节点of next Is an empty node
        return new_head

    # l1 and l2 为当前遍历of节点，carry Be in place
    def addTwo(self, l1: Optional[ListNode], l2: Optional[ListNode], carry=0) -> Optional[ListNode]:
        if l1 is None and l2 is None:  # recursion边界：l1 and l2 都Is an empty node
            return ListNode(carry) if carry else None  # If you are in place，Just create an additional node
        if l1 is None:  # if l1 是空of，Then then l2 一定不Is an empty node
            l1, l2 = l2, l1  # exchange l1 and l2，ensure l1 non empty，从而简化Code
        carry += l1.val + (l2.val if l2 else 0)  # 节点值andcarry加在一起
        l1.val = carry % 10  # Each node saves a digital position
        l1.next = self.addTwo(l1.next, l2.next if l2 else None, carry // 10)  # carry
        return l1

    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        l1 = self.reverseList(l1)
        l2 = self.reverseList(l2)  # l1 and l2 Reversal，Just become【2. Two numbers】Over
        l3 = self.addTwo(l1, l2)
        return self.reverseList(l3)