p = q = head
while q and q.next:
pre = p
p = p.next
q = q.next.next
if q:
pre = p
p = p.next
pre.next = None
def reverse(p):
newHead = ListNode('')
while p:
q = p
p = p.next
q.next = newHead.next
newHead.next = q
return newHead.next
p = reverse(p)
q = head
while q.next:
if q.val != p.val:
return False
q = q.next
p = p.next
return True
head = ListNode.create_list([1,2, 1,2])
print(Solution().isPalindrome(head))
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL
'''
class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
if not head:
return head
p=head
i=1
while i<=k:
if not p.next:
k=k%i
i=1
p=head
else:
p=p.next
i+=1
q=head
if k == 0:
return head
while p.next:
p=p.next
q=q.next
newHead=q.next
q.next=None
p.next=head
return newHead
head=ListNode.create_list([1])
print(Solution().rotateRight(head,1))
return c
def sortedListToBST(self, head: ListNode) -> TreeNode:
'''
根据中序遍历的顺序去构造二叉树，正好中序遍历的顺序就是链表的顺序，中序遍历是递归的
'''
size = self.findSize(head)
def convert(l, r):
nonlocal head
if l > r:
return None
mid = (l + r) // 2
# 中序遍历左子树
left = convert(l, mid - 1)
# 构造左子树之后构造当前节点
node = TreeNode(head.val)
node.left = left
head = head.next
node.right = convert(mid + 1, r)
return node
return convert(0, size - 1)
head = ListNode.create_list([-10, -3, 0, 5, 9])
Solution().sortedListToBST(head).printTree()