LeetCode002__Add Two Numbers

照常题目先放上来：

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

没啥想法，强力破解呗~

代码如下：

# Definition for singly-linked list.
# class ListNode(object):
#     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
"""

solution = ListNode(0)
p=solution
flag=0
while l2 and l1:
p.next=ListNode((l2.val+l1.val+flag)%10)
if l2.val+l1.val+flag>=10:
flag=1
p.next.var=1
else:
flag=0
l2=l2.next
l1=l1.next
p=p.next
return solution.next

结果，恩不错~然后提交……

问题来了：



WTF！！！不能忍，没说一位数要变成两位呀，坑！

改吧……

分情况了：有进位，无进位，两个数位数不相同……

讨论多种情况再来：

# Definition for singly-linked list.
class ListNode(object):
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
"""

solution = ListNode(0)
p=solution
flag=0
while l2 and l1:
p.next=ListNode((l2.val+l1.val+flag)%10)
if l2.val+l1.val+flag>=10:
flag=1
p.next.next=ListNode(1)
print flag,p.next.val
else:
flag=0
#p.next.next=ListNode(0)
l2=l2.next
l1=l1.next
p=p.next

while l1:
p.next=ListNode((flag+l1.val)%10)
if l1.val+flag>=10:
flag=1
p.next.next=ListNode(1)
else:
flag=0
l1=l1.next
p=p.next

while l2:
p.next=ListNode((flag+l2.val)%10)
if l2.val+flag>=10:
flag=1
p.next.next=ListNode(1)
else:
flag=0
l2=l2.next
p=p.next

return solution.next

提交，搞定收工！

结果图：



为什么java和c++可以那么快？等我刷完python要不要用java重写一把呢？