题解 | #链表相加(二)#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head1 ListNode类 
# @param head2 ListNode类 
# @return ListNode类
#
class Solution:
    def list_reverse(self, head: ListNode) -> ListNode:
        if head is None:
            return head

        pre = None
        cur = head
        while cur is not None:
            post = cur.next
            cur.next = pre
            pre = cur
            cur = post

        return pre


    def addInList(self , head1: ListNode, head2: ListNode) -> ListNode:
        # write code here
        if head1 is None and head2 is None:
            return None
        if head1 is None:
            return head2
        if head2 is None:
            return head1

        head1 = self.list_reverse(head1)
        head2 = self.list_reverse(head2)

        carry = 0
        head3, p = None, None
        while head1 is not None or head2 is not None:
            sum = 0
            sum += carry
            if head1 is not None:
                sum += head1.val
                head1 = head1.next

            if head2 is not None:
                sum += head2.val
                head2 = head2.next

            carry = sum // 10
            cur = sum % 10
            if head3 is None:
                head3 = ListNode(cur)
                p = head3
            else:
                p.next = ListNode(cur)
                p = p.next

        if carry > 0:
            p.next = ListNode(carry)

        head3 = self.list_reverse(head3)
        return head3

链表翻转，再从低位开始运算，运算结果再翻转