问题：
给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
示例：
输入:
[2,4,3]
[5,6,4]
输出:
[7,0,8]
预期结果:
[7,0,8]
# 基于链表直接输出结果（执行速度很快）
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
# 直接基于链表输出结果
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
p,q = l1,l2
curr = ListNode(0)
result = curr
jinzhi = 0
while(p or q):
if p:x = p.val
else:x = 0
if q:y = q.val
else:y = 0
temp = x+y+jinzhi
jinzhi = temp//10
curr.next = ListNode(temp%10)
curr = curr.next
if p:p = p.next
if q:q = q.next
if jinzhi >0:
curr.next = ListNode(jinzhi)
return result.next
# 直接基于链表输出结果（和上面大体思想是一样的，但执行效率很低）
def addTwoNumbers2(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
res = [] # 存储输出结果列表
next1 = l1.next
next2 = l2.next
value1 = l1.val
value2 = l2.val
temp = value1+value2
jinzhi = 0
if temp>9:
jinzhi = 1
temp = temp%10
res.append(temp)
while next1 or next2:
if next1 and next2:
temp = next1.val +next2.val +jinzhi
if temp >9:
jinzhi=1
temp = temp%10
else:
jinzhi = 0
res.append(temp)
next1 = next1.next
next2 = next2.next
elif next1 and next2 is None:
temp = next1.val + jinzhi
if temp >9:
jinzhi=1
temp = temp%10
else:
jinzhi =0
res.append(temp)
elif next2 and next1 is None:
temp = next2.val + jinzhi
if temp >9:
jinzhi=1
temp = temp%10
else:
jinzhi = 0
res.append(temp)
if jinzhi==1:
res.append(jinzhi)
return res
# 直接基于链表输出结果 （鉴于addTwoNumbers()，修改的addTwoNumbers2()，实现效率和addTwoNumbers()一样）
# 直接基于链表输出结果
def addTwoNumbers3(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
res = [] # 存储输出结果列表
next1 = l1
next2 = l2
jinzhi = 0
while next1 or next2:
if next1 and next2:
temp = next1.val +next2.val +jinzhi
next1 = next1.next
next2 = next2.next
elif next1 and next2 is None:
temp = next1.val + jinzhi
next1 = next1.next
elif next2 and next1 is None:
temp = next2.val + jinzhi
next2 = next2.next
jinzhi = temp//10
temp = temp%10
res.append(temp)
if jinzhi==1:
res.append(jinzhi)
return res
# 先得到两个链表的值，得到和后，在取余输出结果（相比addTwoNumbers2()好点，但仍然很高）
def addTwoNumbers4(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
value1 = l1.val
i=1
next1 = l1.next
while next1!=None:
value1 = value1 +next1.val*(10**i)
i = i+1
next1 = next1.next
value2 = l2.val
i=1
next2 = l2.next
while next2!=None:
value2 = value2 +next2.val*(10**i)
i = i+1
next2 = next2.next
result = value1+value2
res = [] # 存储数据结果列表
while result >9:
res.append(result%10)
result = result//10
res.append(result)
return res