2. Python 中的 reverse 方法 2.1 reverse 在Jupyter 中的实现 2.2 试试在 LeetCode 中提交 解法一:逐个遍历,双指针 逻辑分析 复杂度分析 解法二:递归解法 关键步骤解释: 复杂度分析: 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:【...
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...
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...
publicListNodereverseList(ListNode head){ListNodeprev=null;ListNodecurr=head;while(curr !=null) {ListNodenextTemp=curr.next; curr.next = prev; prev = curr; curr = nextTemp; }returnprev; } 二刷,python。 # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x...
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 three: recursionifnotheadornothead.next:returnhead ...
lass Solution(object): # 迭代 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): ...
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:newList.insert(0,pos.val)pos=pos.nextpos=headfor_in...
LeetCode 206. 反转链表(Reverse Linked List) 示例: 输入:1->2->3->4->5->NULL输出:5->4->3->2->1->NULL 切题 一、Clarification 只需注意为空链表的情况 二、Possible Solution 1、迭代 2、递归 可利用哨兵简化实现难度 Python3 实现
Problem: 复杂度 时间复杂度: O(n) 空间复杂度: O(n) Code Python3 Java C++ 数学 2+ 1 270 0LSY ・ 2024.12.31 150. 逆波兰表达式求值 Problem: 思路 当遇到数字时直接进栈,不需要执行任何操作,若是运算符号,则取出栈顶的两个数组进行运算并将结果重新加入到栈中,直到tokens数组遍历完,栈中最后只存...