def reverse(self, head: ListNode):
pre = None
while head:
nxt = head.next
head.next = pre
pre = head
head = nxt
return pre
# 还是错了！
# return head
# head1 = init_ln([1, 2, 2, 1])
# check(Solution().isPalindrome, [head1], True)
head1 = init_ln([1, 2, 5, 2, 1])
check(Solution().isPalindrome, [head1], True)
head1 = init_ln([1, 2, 5, 6, 2, 1])
check(Solution().isPalindrome, [head1], False)
"""
O(1)的空间，不考虑链表复原，快慢指针在寻找中间节点的过程中直接反转链表前半部分，找到中间节点之后直接从中间向两边开始比较
"""
"""
其实，借助二叉树后序遍历的思路，不需要显式反转原始链表也可以倒序遍历链表，下面来具体聊聊。
链表兼具递归结构，树结构不过是链表的衍生。那么，链表其实也可以有前序遍历和后序遍历
"""
next = curr.next
curr.next = prev
prev = curr
curr = next
# 返回的是prev
return prev
"""
递归
返回末尾，倒数第二个（当前）指向末尾，所以当前（倒数第二个）的下一个（末尾）的下一个指向当前，然后将当前置空，返回末尾。
"""
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head or head.next is None:
return head
new_head = self.reverseList(head.next)
head.next.next = head
head.next = None
return new_head
# 返回的是prev
head = init_ln([1, 2, 3, 4, 5, 6, 7, 8, 9])
print_ln(Solution().reverseList(head))
"""
pre上一个节点，head当前节点、next下一个节点
当当前节点仍然有下一个节点的时候。将当前节点的方向反转, 移动到下一个节点继续操作。
"""
tmp = head
# 当前节点的前一个节点
# 保留反转后的下一个节点
pre = nxt = None
# 保留下一个节点
while head and k + 1 > 0:
# 当前节点反转到前一个节点
nxt = head.next
head.next = pre
# 顺序下去
pre = head
head = nxt
k -= 1
# 记录之前的头，作为非反转的头
tmp.next = nxt
return pre
head1 = init_ln([1, 2, 3, 4, 5])
print_ln(Solution().reverseTopK2(head1, 2))
# print_ln(Solution().reverseBetween(head1, 2, 4))
# print_ln(Solution().reverseBetween(init_ln([5]), 1, 1))
# print_ln(Solution().reverseBetween(init_ln([3, 5]), 1, 2))
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
cur = head
while cur and cur.next:
next = cur.next
while next and next.val == cur.val:
next = next.next
cur.next = next
cur = next
return head
# head = init_ln([1, 2, 3, 4, 4, 9])
# head = init_ln([1, 2, 3, 3, 4, 4, 5])
head = init_ln([1, 2, 3, 3, 4, 4])
print_ln(Solution().deleteDuplicates(head))
"""
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/solution/shan-chu-pai-xu-lian-biao-zhong-de-zhong-49v5/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
"""
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
if not head:
return head
while l1 and l2:
if l1.val < l2.val:
dump.next = l1
l1 = l1.next
else:
dump.next = l2
l2 = l2.next
dump = dump.next
if l1:
dump.next = l1
else:
dump.next = l2
return tmp.next
head1 = init_ln([1, 3, 5, 7, 9])
head2 = init_ln([1, 2, 3, 4, 5, 6, 7, 8, 9])
print_ln(Solution().mergeTwoLists(head1, head2))
"""
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/he-bing-liang-ge-you-xu-lian-biao-by-leetcode-solu/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
"""
# 迭代
class Solution:
while cur.next and cur.next.val == cur.val:
cur = cur.next
dup = True
if dup:
pre.next = cur.next
else:
# 因为pre为空节点，没有对下一个节点进行关联
# 省下这一步的话，会导致tmp找不到下一个节点
pre.next = cur
pre = cur
cur = cur.next
return tmp.next
# head = init_ln([1, 2, 3, 4, 4, 9])
head = init_ln([1, 2, 3, 3, 4, 4, 5])
print_ln(Solution().deleteDuplicates(head))
"""
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii/solution/shan-chu-pai-xu-lian-biao-zhong-de-zhong-oayn/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
"""
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
if not head:
return head