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2.两数相加.py
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'''
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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-two-numbers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
'''
# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:#60ms,13.6mb
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
def jinwei(v1, v2, ad):
if v1 + v2 + ad >= 10:
return True
else:
return False
p1 = l1
p2 = l2
if p1.val + p2.val < 10:
p3 = ListNode(p1.val + p2.val)
ad = 0
else:
p3 = ListNode((p1.val + p2.val) % 10)
ad = 1
p1 = p1.next
p2 = p2.next
first = p3
while (p1 != None and p2 != None):
if jinwei(p1.val, p2.val, ad):
tmp = ListNode((p1.val + p2.val + ad) % 10)
ad = 1
else:
tmp = ListNode(p1.val + p2.val + ad)
ad = 0
p3.next = tmp
p3 = p3.next
p1 = p1.next
p2 = p2.next
while p1:
if ad:
if jinwei(p1.val, 0, ad):
tmp = ListNode((p1.val + ad) % 10)
ad = 1
else:
tmp = ListNode(p1.val + ad)
ad = 0
else:
tmp = ListNode(p1.val)
p3.next = tmp
p3 = p3.next
p1 = p1.next
while p2:
if ad:
if jinwei(p2.val, 0, ad):
tmp = ListNode((p2.val + ad) % 10)
ad = 1
else:
tmp = ListNode(p2.val + ad)
ad = 0
else:
tmp = ListNode(p2.val)
p3.next = tmp
p3 = p3.next
p2 = p2.next
if ad == 1:
tmp = ListNode(1)
p3.next = tmp
return first