新手村100题汇总:王几行xing:【Python-转码刷题】LeetCode 力扣新手村100题,及刷题顺序 2个链表相关题目: 王几行xing:【LeetCode-转码刷题】LeetCode 21「合并链表」 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
代码(Python3) # 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]: # 定义一个哨兵结点...
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...
# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseList(self, head: ListNode) -> ListNode:# solution three: recursionifnotheadornothead.next:returnhead pos = head.nextnewList = self.reverseList(pos) 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:...
Leetcode260反转链表(java/c++/python) JAVA: class Solution { public ListNode reverseList(ListNode head) { if( head == null || head.next == null) return head; ListNode newHead = reverseList(head.next); head.next.next = head; head.next = null; return newHead; } } C++: class Solution...
Reverse Linked List 题目大意 翻转链表 解题思路 必看: http://blog.csdn.net/autumn20080101/article/details/7607148 以下代码若理解不通请务必务必务必务必务必务必务必看上方网页 还可以参考(迭代+递归): https://blog.csdn.net/u011608357/article/details/36933337 ...
```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)...
非递归实现代码4[Python] #Runtime: 63 ms# Definition for singly-linked list.classListNode:def__init__(self, x):self.val = xself.next=NoneclassSolution:# @param {ListNode} head# @return {ListNode}defreverseList(self, head):ifhead ==None:returnhead ...
1 <= n <= 500 -500 <= Node.val <= 500 1 <= left <= right <= n 进阶: 你可以使用一趟扫描完成反转吗? 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/reverse-linked-list-ii python # 0092.反转链表II # https://leetcode-cn.com/problems/reverse-linked-list-ii/solution/...