你有兩個用鏈表代表的整數(shù)，其中每個節(jié)點(diǎn)包含一個數(shù)字。數(shù)字存儲按照在原來整數(shù)中相反的順序，使得第一個數(shù)字位于鏈表的開頭。寫出一個函數(shù)將兩個整數(shù)相加，用鏈表形式返回和。
您在真實(shí)的面試中是否遇到過這個題？
樣例

給出兩個鏈表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

"""
Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
"""


class Solution:
    """
    @param: l1: the first list
    @param: l2: the second list
    @return: the sum list of l1 and l2 
    """
    def addLists(self, l1, l2):
        # write your code here
        if l1 is None and l2 is None:
            return None
        
        cur1 = l1.next
        cur2 = l2.next
        head = ListNode((l1.val + l2.val)%10)
        cur = head
        
        carry = (l1.val + l2.val) // 10
        #當(dāng)其中一條鏈表為空的時候,為該鏈表添加一個值為0的節(jié)點(diǎn)
        #直到兩條都為None,這種方式會修改原先的列表.
        while cur1 is not None or cur2 is not None:
            if cur1 is None:cur1 = ListNode(0)
            if cur2 is None:cur2 = ListNode(0)
            
            val = cur1.val + cur2.val + carry
            carry = val // 10
            cur.next = ListNode(val%10)
            
            cur = cur.next
            cur1 = cur1.next
            cur2 = cur2.next
            
        if carry != 0:
            cur.next = ListNode(carry)
        
        return head