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 ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

#
# 
# @param head1 ListNode类 
# @param head2 ListNode类 
# @return ListNode类
#
class Solution:
    def addInList(self , head1 , head2 ):
        h1=self.reverse(head1)
        h2=self.reverse(head2)
        N=ListNode(-1)
        N1=N
        c=0
        while h1 is not None and h2 is not None:
            
            m=(h1.val+h2.val+c)%10
            c=(h1.val+h2.val+c)//10
            p=ListNode(m)
            N1.next=p
            N1=N1.next
            h1=h1.next
            h2=h2.next
        while h1 is not None:
            m=(h1.val+c)%10
            c=(h1.val+c)//10
            p=ListNode(m)
            N1.next=p
            N1=N1.next
            h1=h1.next
        while h2 is not None:
            m=(h2.val+c)%10
            c=(h2.val+c)//10
            p=ListNode(m)
            N1.next=p
            N1=N1.next
            h2=h2.next
        if c==1:
            N1.next=ListNode(1)
            N1=N1.next
        return self.reverse(N.next)
        
        
        
        
        
    def reverse(self,head):
        last=None
        while head is not None:
            temp=last
            last=head
            head=head.next
            last.next=temp
        return last
##### 反转链表，这么做虽然代码啰嗦了点，但是不回超时。