def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
stk1, stk2 = [], []
# 放入栈中
while l1 or l2:
if l1:
stk1.append(l1.val)
l1 = l1.next
if l2:
stk2.append(l2.val)
l2 = l2.next
dummy = ListNode(-1)
carry = 0
# 对栈中元素进行相加
while stk1 or stk2:
val = carry
if stk1:
val += stk1.pop()
if stk2:
val += stk2.pop()
carry, val = divmod(val, 10)
# 头插
node = ListNode(val)
node.next = dummy.next
dummy.next = node
# 处理最后一次进位
if carry:
node = ListNode(1)
node.next = dummy.next
dummy.next = node
return dummy.next
def reverseList2(self, head):
dummy = ListNode(-1)
while head:
_next = head.next
head.next = dummy.next
dummy.next = head
head = _next
return dummy.next
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
dummy = ListNode(-1)
dummy.next = head
slow = fast = dummy
for i in range(n + 1):
fast = fast.next
while fast:
slow = slow.next
fast = fast.next
# slow.next即为被删节点，不为None
slow.next = slow.next.next
return dummy.next
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
dummy = ListNode(-1)
dummy.next = head
ptr1 = dummy
ptr2 = head
for i in range(n):
ptr2 = ptr2.next
while ptr2:
ptr1 = ptr1.next
ptr2 = ptr2.next
ptr1.next = ptr1.next.next
return dummy.next
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(-1)
ptr = dummy
carry = 0
while l1 or l2:
val = carry
if l1:
val += l1.val
l1 = l1.next
if l2:
val += l2.val
l2 = l2.next
carry, val = divmod(val, 10)
ptr.next = ListNode(val)
ptr = ptr.next
# 最后一次进位，总是错！！！
if carry:
ptr.next = ListNode(1)
return dummy.next
def reverseList(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
# 先递归，reverse head.next开头的链表，返回了将来的头结点
# head.next在这个过程中，没有被修改
node = self.reverseList(head.next)
# 这里才真正reverse
head.next.next = head
head.next = None
return node
:type n: int
:rtype: ListNode
"""
dummy = ListNode(-1)
dummy.next = head
ptr1 = dummy
ptr2 = head
for i in range(n):
ptr2 = ptr2.next
while ptr2:
ptr1 = ptr1.next
ptr2 = ptr2.next
ptr1.next = ptr1.next.next
return dummy.next
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
node5 = ListNode(5)
node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node5
Solution().removeNthFromEnd(node1, 5)
a = 1
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
from util.List import ListNode
class Solution(object):
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
node.val = node.next.val
node.next = node.next.next
node1 = ListNode(4)
node2 = ListNode(5)
node3 = ListNode(1)
node4 = ListNode(9)
node1.next = node2
node2.next = node3
node3.next = node4
Solution().deleteNode(node2)
a = 1