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02.add-two-numbers.py
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# https://leetcode-cn.com/problems/add-two-numbers
from typing import Optional
# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution1:
'''
Date: 2022.03.30
Pass/Error/Bug: 1/3/0
执行用时：72 ms, 在所有 Python3 提交中击败了 15.90% 的用户
内存消耗：15 MB, 在所有 Python3 提交中击败了 35.62% 的用户
'''
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
# 使用函数递归，增加进位功能
def addWithCarry(
l1: Optional[ListNode] = None,
l2: Optional[ListNode] = None,
carry: Optional[int] = 0
) -> ListNode:
val1, val2 = 0, 0
next1, next2 = None, None
flag1, flag2 = 0, 0
if l1 is not None:
val1 = l1.val
next1 = l1.next
flag1 = 1
if l2 is not None:
val2 = l2.val
next2 = l2.next
flag2 = 1
sum_val = val1 + val2 + carry
add_val = sum_val // 10
new_val = sum_val % 10
if (next1 is None) and (next2 is None):
if add_val != 0:
return ListNode(val=new_val, next=ListNode(val=add_val))
else:
return ListNode(val=new_val)
elif flag1 + flag2 > 0:
return ListNode(val=new_val, next=addWithCarry(next1, next2, add_val))
else:
return ListNode(val=new_val)
return addWithCarry(l1, l2)
class Solution2:
'''
Date: 2022.03.31
Pass/Error/Bug: 1/0/0
执行用时：68 ms, 在所有 Python3 提交中击败了 25.46% 的用户
内存消耗：15 MB, 在所有 Python3 提交中击败了 12.26% 的用户
'''
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
n1s = []
n2s = []
while l1:
n1s.append(str(l1.val))
l1 = l1.next
while l2:
n2s.append(str(l2.val))
l2 = l2.next
n1s = int(''.join(n1s[::-1]))
n2s = int(''.join(n2s[::-1]))
s = n1s + n2s
prev = None
for i in str(s):
out = ListNode(int(i), prev)
prev = out
return out
class Solution3:
'''
Date: 2022.04.30
Pass/Error/Bug: 3/3/8
执行用时：60.0 ms, 在所有 Python3 提交中击败了 72.57% 的用户
内存消耗：14.9 MB, 在所有 Python3 提交中击败了 82.98% 的用户
'''
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
idx = 0
carry = 0
head = ListNode(0)
cache1 = None
cache2 = None
while True:
if l1:
v1 = l1.val
if l1.next:
l1 = l1.next
else:
l1 = None
else:
v1 = 0
if l2:
v2 = l2.val
if l2.next:
l2 = l2.next
else:
l2 = None
else:
v2 = 0
sum_val = v1 + v2 + carry
add_val = sum_val // 10
new_val = sum_val % 10
if sum_val != 0:
if idx != 0:
cache2 = ListNode(new_val)
cache1.next = cache2
cache1 = cache2
else:
head = ListNode(new_val)
cache1 = head
carry = add_val
idx += 1
elif (l1 or l2):
if idx != 0:
cache2 = ListNode(0)
cache1.next = cache2
cache1 = cache2
else:
head = ListNode(0)
cache1 = head
idx += 1
else:
break
return head