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
# 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...
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 without using extra space? Solution......
pos = pos.nextpos = headfor_innewList: pos.val = _ pos = pos.nextreturnhead 解法二 双指针 用cur 和 pre 双指针实现反转链表。 可以动手在纸上画画,就能理解过程了。 时间复杂度:O(n),n 为链表长度。 空间复杂度:O(1) Python3代码 # Definition for singly-linked list.# class ListNode:# de...
C++ program to reverse a single linked list #include<bits/stdc++.h>usingnamespacestd;classnode{public:intdata;// data fieldnode*next;};node*reverse(node*head){node*next=NULL,*cur=head,*prev=NULL;//initialize the pointerswhile(cur!=NULL){//loop till the end of linked listnext=cur->nex...
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 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)...
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...
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 ...