2. Add Two Numbers

题目要求：

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 contains a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

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

解题思路：

• 巧用头指针，本题目也需要使用额外的头指针
• 遍历链表直至为空：while...
• 考虑进位的情况，设置一个标志位。
• 考虑在中间进位和在最后一位进位两种情况。

代码一：

#Time:  O(n)
#Space: O(1)

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

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        current, carry = dummy, 0

        while l1 or l2:
            val = carry
            if l1:
                val += l1.val
                l1 = l1.next
            if l2:
                val += l2.val
                l2 = l2.next
            carry, val = divmod(val, 10)
            current.next = ListNode(val)
            current = current.next

        if carry == 1:
            current.next = ListNode(1)

        return dummy.next

补充：Python divmod()函数

• Python 内置函数：divmod()函数把除数和余数运算结果结合起来，返回一个包含商和余数的元组(a // b, a % b)。
  [>>>11 % 10
  1
  [>>>12 % 10
  2
  [>>>2 % 10
  2
  [>>>11 // 10
  1
  [>>>11 / 10
  1.1
  [>>> 12 // 10
  1
  [>>>12 / 10
  1.2
  [>>>1 // 10
  0
  [>>> 1 / 3
  0.3333333
  [>>>divmod(12,10)
  1, 2
  [>>>divmod(2,10)
  0, 2