AI代码解释 publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=nu...
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Jupyter 中的实现 class Solution: def reverseList(self, head:list) -> list: head.re...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseBetween(ListNode* head, int m, int n) { //ListNode* pre=NULL; //ListNode* cur=head; ListNode ...
}ListNodep=reverseList(head.next); head.next.next = head; head.next =null;returnp; } } Python 实现 # Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseList(self, head):""" :type head: ListNode :...
问题链接英文网站:92. Reverse Linked List II中文网站:92. 反转链表 II问题描述Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the l…
LeetCode 206题 反转链表(Reverse Linked List) /** Definition for singly-linked list. public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */classSolution{publicListNodereverseList(ListNode head){ListNode pre=null;ListNode next=null;while(head!=null){next=head....
return newHead; } }; Python: class Solution(object): def reverseList(self, head): if(head == None or head.next == None): return head newHead = self.reverseList(head.next) head.next.next = head head.next = None return newHead分享至 投诉或建议评论2 赞与转发21...
class Solution: def reverseList2(self, head: ListNode) -> ListNode: """ 双指针法 :param head: :return: """ # 空表时直接返空 if head is None: return None # 初始化cur为head cur = head while head.next != None: # 当head的next非空时, 完成局部反转 ...
public class Solution { public ListNode reverseList(ListNode head) { ListNode prev = null; ListNode cur = head; while(cur!=null){ ListNode next = cur.next; cur.next = prev; prev = cur; cur = next; } return prev; } } recursive 解法: ...
Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 递归法 复杂度 时间O(N) 空间 O(N) 递归栈空间 思路 基本递归 代码 public class Solution { ...