reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: if not head: return head p = head q = p.next p.next = None while q: r = q.next q.next = p p = q q = r return p发布于 2024-03-21 10:22・北京 Py
Linked Lists:Reverse a doubly linked list ...Reverse Linked List #记录经典问题,方便以后复习。 题目: Reverse a singly linked list. Example: 1、非递归 2、递归 ...leetcode: Linked List Question Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it ...
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]: i = 0 p = head p_l = None...
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseList(self, head: ListNode) -> ListNode:# solution one: Listpos = head newList = []whilepos: newList.insert(0, pos.val) pos = pos.nextpos = h...
leetcode 【 Reverse Linked List II 】 python 实现 题目: Reverse a linked list from positionmton. Do it in-place and in one-pass. For example: Given1->2->3->4->5->NULL,m= 2 andn= 4, return1->4->3->2->5->NULL. Note:...
Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Example: Answer: ...【Leetcode】92. Reverse Linked List II 题目: Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list...
Reverse Linked List II 题目大意 翻转指定位置的链表 解题思路 将中间的执行翻转,再将前后接上 代码 迭代 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution(object):# 迭代 defreverseBetween(self,head,m,n):""":type head:ListNode:type m:int:type n:int:rtype:ListNode""" ...
```python class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next def reverse_linked_list(head): prev = None curr = head while curr: next_node = curr.next curr.next = prev prev = curr curr = next_node return prev # 测试 node1 = ListNode(1)...
Python Copy 最近在練習程式碼本身就可以自解釋的 Coding style,可以嘗試直接閱讀程式碼理解 算法說明 同上slow, fast 作法,但實現 O(1) 空間,利用 LinkedList 的反轉,同向 two pointer。input handling 處理沒有 head 的情況,return False Boundary conditions ...
def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ dummy = ListNode(None) while head: # 终止条件是head=Null nextnode = head.next # nextnode是head后面的节点 head.next = dummy # dummy.next是Null,所以这样head.next就成为了Null ...