Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 很基本的一道换顺序的题目,先建一个pre 指针,设为Non…

Given a non-empty, singly linked list with head node head, return a middle node of linked list. If there are two middle nodes, return the second middle node. Example 1: Input: [1,2,3,4,5] Output: Node 3 from this list (Serialization: [3,4,5]) The ret…

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Given linked list -- head = [4,5,1,9], which looks like following: 4 -> 5 -> 1 -> 9 Example 1: Input: head = [4,5,1,9], node = 5 Output:…

206. Reverse Linked List 之前在牛客上的写法: 错误代码: class Solution { public: ListNode* ReverseList(ListNode* pHead) { if(pHead == NULL) return NULL; ListNode* p1 = pHead; ListNode* p2 = pHead->next; ListNode* p3 = pHead->next->next; pHead->next = NULL;…

面试题16:反转链表 提交网址: http://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168 或 https://leetcode.com/problems/reverse-linked-list/ Total Accepted: 101523  Total Submissions: 258623  Difficulty: Easy Reverse a singly linked lis…

Reverse a singly linked list. 题目标签:Linked List 题目给了我们一个链表,要求我们倒转链表. 利用递归,新设一个newHead = null,每一轮 把下一个 node 存入 next 记住,把目前的点 cursor 连到 newHead: 然后把 next 当作新的 cursor:把 cursor 当作新的 newHead 递归下去. 换句话说,每到一个点,把下一个点(右边的)记住,传到下一轮,然后把目前的点反向链接(向左链接). Java Solut…

Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 反向链表,分别用递归和迭代方式实现. 递归Iteration: 新建一个node(value=任意值, next = None), 用一个变量 next 记录head.next,head.next指向新node.next,新 node.next…

Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 递归的办法: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { v…

Reverse Linked List,一道有趣的题目.给你一个链表,输出反向链表.因为我用的是JavaScript提交,所以链表的每个节点都是一个对象.例如1->2->3,就要得到3->2->1. 1.数组构造 一个很容易想到的方法是用数组保存新构造每个节点,然后反向构造链表,输出: var reverseList = function(head) { var ans = []; while (head) { var node = new ListNode(head.val);…

Reverse a singly linked list. 解题思路: 用Stack实现,JAVA实现如下: public ListNode reverseList(ListNode head) { if(head==null) return null; Stack<ListNode> stack =new Stack<ListNode>(); ListNode temp=head; while(temp!=null){ stack.push(temp); temp=temp.ne…