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 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))