原題描述：

你有兩個用鏈表代表的整數，其中每個節點包含一個數字。數字存儲按照在原來整數中相反的順序，使得第一個數字位於鏈表的開頭。寫出一個函數將兩個整數相加，用鏈表形式返回和。

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

函數addLists，入參為兩個鏈表，長度無限制，即可能存在長度：list1 > list2; list1 = list2; list1 < list2;

最終鏈表長度依據最長鏈表長度n，返回鏈表長度（n~n+1）

# 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: return l2
        if l2 is None: return l1
        
        head1 = l1
        head2 = l2
        flag = 0
        
        while head1.next is not None or head2.next is not None:
            # 存在某一鏈表next為空時，構造next.val = 0，不影響加法結果
            if head1.next is None:
                head1.next = ListNode(0)
            if head2.next is None:
                head2.next = ListNode(0)
                
            sumNum = head1.val + head2.val
            if sumNum >= 10:
                head1.val = sumNum%10
                flag = 1
                head1.next.val += 1
            else:
                head1.val = sumNum
                flag = 0
            head1 = head1.next
            head2 = head2.next
        else:
            # 鏈表末尾時，單獨處理，其和大於10時，追加節點
            head1.val = head1.val + head2.val
            if head1.val >= 10:
                head1.val = head1.val%10
                head1.next = ListNode(1)
        return l1

LintCode Python 簡單級題目 鏈表求和