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
原因:342 + 465 = 807
# Definition for singly-linked list.
# class ListNode(object):#定义链表
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1,l2):
if(l1 is None and l2 is None):
return None
temp = 0
sum_node = ListNode(None)
head = sum_node
while(l1 is not None or l2 is not None or temp ==1):
sum = 0
if(l1 is not None):
sum += l1.val
if(l2 is not None):
sum+=l2.val
sum+=temp
temp = 0
if(sum<10):
sum_node.val = sum
else:
sum_node.val = sum-10
temp = 1
if(l1 is not None):
l1 = l1.next
if(l2 is not None):
l2 = l2.next
if(l1 is None and l2 is None and temp == 0):
sum_node.next = None
else:
sum_node.next = ListNode(None)
sum_node = sum_node.next
return head
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
该题关键是要保存进位,即将相加后的个位和十位分开存储。
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
re = ListNode(0)
r=re
carry=0
while(l1 or l2):
x= l1.val if l1 else 0
y= l2.val if l2 else 0
s=carry+x+y
carry=s//10
r.next=ListNode(s%10)
r=r.next
if(l1!=None):l1=l1.next
if(l2!=None):l2=l2.next
if(carry>0):
r.next=ListNode(1)
return re.next