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。