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...
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: Optional[ListNode]) -> Optional[ListNode]: ...
第一种方法:迭代 代码语言:javascript 代码运行次数:0 classListNode(object):def__init__(self,x):self.val=x self.next=NoneclassSolution(object):defreverseList(self,head):""":type head:ListNode:rtype:ListNode""" pre=cur=Noneifhead:pre=head cur=head.next pre.next=Noneelse:returnNonewhilecur:...
publicclassSolution{publicListNodeReverseList(ListNode head){if(head==null)returnnull;//head为当前节点,如果当前节点为空的话,那就什么也不做,直接返回null;ListNode pre=null;ListNode nextnode=null;while(head!=null){nextnode=head.next;head.next=pre;pre=head;head=nextnode;}returnpre;}} Reverse Lin...
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 two: two pointcur, pre =None, headwhilepre: ...
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: Givenm,nsatisfy the following condition: 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 ...
链接:https://leetcode-cn.com/problems/reverse-linked-list https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/ python # 反转单链表 迭代或递归 class ListNode: def __init__(self, val): ...
```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 class Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ p1 = None p2 = head while(p2 is not None): tmp = p2.next p2.next = p1 p1 = p2 p2 = tmp return p1 Reverse Linked List II ...