leetcode 【 Add Two Numbers 】 python 实现

题目：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

代码：oj测试通过 Runtime: 171 ms

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

 class Solution:
     # @return a ListNode
     def addTwoNumbers(self, l1, l2):
         if l1 is None:
             return l2
         if l2 is None:
             return l1

         dummyhead = ListNode(0)
         p = ListNode(0)
         dummyhead.next = p

         jinwei = 0
         while l1 is not None and l2 is not None:
             curr_total = l1.val + l2.val + jinwei
             l1 = l1.next
             l2 = l2.next
             curr_digit = curr_total % 10
             jinwei = curr_total / 10
             curr_node = ListNode(curr_digit)
             p.next = curr_node
             p = p.next

         if l1 is not None:
             while l1 is not None:
                 curr_total = l1.val + jinwei
                 l1 = l1.next
                 curr_digit = curr_total % 10
                 jinwei = curr_total / 10
                 curr_node = ListNode(curr_digit)
                 p.next = curr_node
                 p = p.next
         if l2 is not None:
             while l2 is not None:
                 curr_total = l2.val + jinwei
                 l2 = l2.next
                 curr_digit = curr_total % 10
                 jinwei = curr_total / 10
                 curr_node = ListNode(curr_digit)
                 p.next = curr_node
                 p = p.next

         if jinwei == 1:
             curr_node = ListNode(1)
             p.next = curr_node

         return dummyhead.next.next

思路：

就是加法运算 注意两条链表上所有值计算过后 是否有进位；如果有进位 需要再处理一下。