def list_node_generator(nums):
if len(nums) == 0:
return None
t = ListNode(nums[0])
head = t
for i in range(1, len(nums)):
t.next = ListNode(nums[i])
t = t.next
return head
def add_two_numbers1(l1, l2):
l3 = ListNode(0)
res = l3
temp = 0
while l1 and l2:
l3.val = (l1.val + l2.val + temp) % 10
temp = (l1.val + l2.val + temp) // 10
l1 = l1.next
l2 = l2.next
if l1 or l2:
l3.next = ListNode(0)
l3 = l3.next
while l1:
l3.val = (l1.val + temp) % 10
temp = (l1.val + temp) // 10
l1 = l1.next
if l1:
l3.next = ListNode(0)
l3 = l3.next
while l2:
l3.val = (l2.val + temp) % 10
temp = (l2.val + temp) // 10
l2 = l2.next
if l2:
l3.next = ListNode(0)
l3 = l3.next
if temp > 0:
l3.next = ListNode(0)
l3.next.val = temp
return res
def add_two_numbers2(l1, l2):
l3 = ListNode(0)
res = l3
carry = 0
while l1 or l2:
l3.next = ListNode(0)
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
l3.next.val = carry % 10
carry = carry // 10
l3 = l3.next
if carry > 0:
l3.next = ListNode(0)
l3.next.val = carry
return res.next
def add_two_numbers3(l1, l2):
l3 = ListNode(0)
res = l3
carry = 0
while l1 or l2 or carry:
l3.next = ListNode(0)
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
# 获得商和余数
carry, val = divmod(carry, 10)
l3.next.val = val
l3 = l3.next
return res.next
def insert_at_beginning(self, data):
new_node = ListNode(data)
new_node.next = self.head
self.head = new_node
def test_add_two_numbers():
'''
思路：
时间复杂度：O(n)
'''
def add_two_numbers1(l1, l2):
l3 = ListNode(0)
res = l3
temp = 0
while l1 and l2:
l3.val = (l1.val + l2.val + temp) % 10
temp = (l1.val + l2.val + temp) // 10
l1 = l1.next
l2 = l2.next
if l1 or l2:
l3.next = ListNode(0)
l3 = l3.next
while l1:
l3.val = (l1.val + temp) % 10
temp = (l1.val + temp) // 10
l1 = l1.next
if l1:
l3.next = ListNode(0)
l3 = l3.next
while l2:
l3.val = (l2.val + temp) % 10
temp = (l2.val + temp) // 10
l2 = l2.next
if l2:
l3.next = ListNode(0)
l3 = l3.next
if temp > 0:
l3.next = ListNode(0)
l3.next.val = temp
return res
# 方法一优化
def add_two_numbers2(l1, l2):
l3 = ListNode(0)
res = l3
carry = 0
while l1 or l2:
l3.next = ListNode(0)
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
l3.next.val = carry % 10
carry = carry // 10
l3 = l3.next
if carry > 0:
l3.next = ListNode(0)
l3.next.val = carry
return res.next
# 方法2 优化，pythonic
def add_two_numbers3(l1, l2):
l3 = ListNode(0)
res = l3
carry = 0
while l1 or l2 or carry:
l3.next = ListNode(0)
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
# 获得商和余数
carry, val = divmod(carry, 10)
l3.next.val = val
l3 = l3.next
return res.next
print()
t1 = ListNode(0)
t2 = ListNode(0)
test1 = t1
test2 = t2
for i in range(1):
t1.val = i + 1
t1.next = ListNode(2)
t1 = t1.next
for i in range(1):
t2.val = 9
t2.next = ListNode(9)
t2 = t2.next
print(test1)
print(test2)
print(add_two_numbers3(test1, test2))
sum += carry if carry != 0 else 0
sum += curr1.val if curr1 != None else 0
sum += curr2.val if curr2 != None else 0
if sum >= 10:
sum -= 10
carry = 1
else:
carry = 0
if total == None:
total = ListNode(sum)
current = total
else:
current.next = ListNode(sum)
current = current.next
curr1 = curr1.next if curr1 != None else None
curr2 = curr2.next if curr2 != None else None
if carry > 0:
current.next = ListNode(carry)
return total
l1 = ListNode(1)
l2 = ListNode(9)
l2.next = ListNode(9)
s = Solution()
total = s.addTwoNumbers(l1, l2)