来自专栏 · python算法题笔记# Definition for singly-linked list.# class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: ''' # 递归 def reverseList(self, head: Opti
王几行xing:【Python-转码刷题】LeetCode 203 移除链表元素 Remove Linked List Elements 提到翻转,熟悉Python的朋友可能马上就想到了列表的 reverse。 1. 读题 2. Python 中的 reverse 方法 list1 = list("abcde") list1.reverse() list1 [Out: ] ['e', 'd', 'c', 'b', 'a'] 2.1 reverse 在...
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ prev = None while head: curr = head # 1 head =...
Givenm,nsatisfy the following condition: 1≤m≤n≤ length of list. 代码:oj测试通过 Runtime: 65 ms 1#Definition for singly-linked list.2#class ListNode:3#def __init__(self, x):4#self.val = x5#self.next = None67classSolution:8#@param head, a ListNode9#@param m, an integer10#@...
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""" ...
Python3代码 # 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: ...
11-9 LeetCode 92. Reverse Linked List II Reverse Linked List II Description Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Note: Given m, n satisfy the follow...leetcode 92. Reverse Linked List II https://leetcode.com/problems/reverse-li...
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 ...
```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)...
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 class Solution(object): # 迭代 def reverseBetween(self, head, m, n): """ :type head: ListNode :type m: int :type n: int :rtype: ListNode ...