LeetCode: No.2 Add Two Numbers

No.2 Add Two Numbers

原题：

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

题目大意：给出两个链表，存储非负数，两个链表都是按倒序方式存储数字（个位，十位，百位……）要求将两个链表相加并以链表形式返回。

Example 1:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807

Solution one:

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        head = ListNode(0)
        ptr = head
        carry  = 0
        while True:
            if l1 != None:
                carry += l1.val
                l1 = l1.next
            if l2 != None:
                carry += l2.val
                l2 = l2.next
            ptr.val = int(carry % 10)
            carry = int(carry / 10)
            # 运算未结束新建一个节点用于储存答案，否则退出循环
            if l1 != None or l2 != None or carry != 0:
                ptr.next = ListNode(0)
                ptr = ptr.next
            else: 
                break
        return head         

第一次一直跑不通的原因是上述代码中的carry = int(carry / 10)没有做int强制类型转换，这就导致carry是一个小说，即便是每次除10依旧是小数不是0，导致最后输出结果多出很多0。

Reference solution:

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = dummy = ListNode(-1)
        #进位标志，低位有进位时为1，默认为0
        carry = 0
        #当两个链表的同一位置同时不为空时（即小詹举例的低三位数）
        while l1 and l2:
            #p.next为指针所指 l1.val为对应的数据
            p.next = ListNode(l1.val + l2.val + carry)
            #除法和求模运算获取p.next的十位个位
            carry = int(p.next.val / 10)    #注意必须要做的强制类型转换
            p.next.val = int(p.next.val % 10)
            p = p.next
            l1 = l1.next
            l2 = l2.next
        res = l1 or l2    #或运算即对应小詹举得高位例子（第四位第五位）
        while res:
            p.next = ListNode(res.val + carry)
            carry = p.next.val / 10
            p.next.val %= 10
            p = p.next
            res = res.next
        if carry:
            p.next = ListNode(1)
        return dummy.next    #返回该链表