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:
    #反转链表
    def ReverseList(self, pHead:ListNode): 
        if pHead == None:
            return None
        cur = pHead
        pre = None
        while cur:
            #断开链表，要记录后续一个
            temp = cur.next 
            #当前的next指向前一个
            cur.next = pre 
            #前一个更新为当前
            pre = cur 
            #当前更新为刚刚记录的后一个
            cur = temp 
        return pre
    
    def addInList(self , head1: ListNode, head2: ListNode) -> ListNode:
        #任意一个链表为空，返回另一个
        if head1 == None: 
            return head2
        if head2 == None:
            return head1
        #反转两个链表
        head1 = self.ReverseList(head1) 
        head2 = self.ReverseList(head2)
        #添加表头
        res = ListNode(-1) 
        head = res
        #进位符号
        carry = 0 
        #只要某个链表还有或者进位还有
        while head1 != None or head2 != None or carry != 0: 
            #链表不为空则取其值
            val1 = 0 if head1 == None else head1.val 
            val2 = 0 if head2 == None else head2.val
            #相加
            temp = val1 + val2 + carry 
            #获取进位
            carry = (int)(temp / 10) 
            temp %= 10
            #添加元素
            head.next = ListNode(temp) 
            head = head.next
            #移动下一个
            if head1: 
                head1 = head1.next
            if head2:
                head2 = head2.next
        #结果反转回来
        return self.ReverseList(res.next)