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;…

表不支持随机查找,通常是使用next指针进行操作. 206. 反转链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ //时间:O(n) 只遍历了一遍链表 //空间:O(1) 开了三个指针的空间 class Solution { public: ListNode*…

206. Reverse Linked List[easy] Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 解法一: class Solution { public: ListNode* reverseList(ListNode* head) { ListNode * pre = NULL;…

Question 206. Reverse Linked List Solution 题目大意:对一个链表进行反转 思路: Java实现: public ListNode reverseList(ListNode head) { ListNode newHead = null; while (head != null) { ListNode tmp = head.next; head.next = newHead; newHead = head; head = tmp; } return new…

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 Linked List 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? 解法一:非递归 /** * Definition for singly-linked list. * struct ListNode { * int va…

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 Linked List 描述 反转一个单链表. 示例: 输入: 1->2->3->4->5->NULL    输出: 5->4->3->2->1->NULL 进阶: 你可以迭代或递归地反转链表.你能否用两种方法解决这道题? 解析 设置三个节点pre.cur.next (1)每次查看cur节点是否为NULL,如果是,则结束循环,获得结果 (2)如果cur节点不是为NULL,则先设置临时变量next为cur的下一个节点 (3)让cur…

要求 反转一个链表 不得改变节点的值 示例 head->1->2->3->4->5->NULL NULL<-1<-2<-3<-4<-5<-head 思路 设置三个辅助指针 实现 50实现反转,51-52实现后移 1 #include <iostream> 2 using namespace std; 3 4 struct ListNode { 5 int val; 6 ListNode *next; 7 ListNode(…

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…

题目描述: Reverse a singly linked list. 解题思路: 使用递归或者迭代的方法. 代码如下: 方法一:递归 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode reverseLi…

题目: Reverse a singly linked list. 提示: 此题不难,可以用迭代或者递归两种方法求解.记得要把原来的链表头的next置为NULL: 代码: 迭代: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { pub…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Reverse a singly linked list. (二)解题 题目大意:反转一个单向链表 解题思路:题目要求用迭代和递归分别实现 迭代版本: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNod…

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? 解法一:(C++)利用迭代的方法依次将链表元素放在新链表的…

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. 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? -----------------------------…

题目要求 Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL 题目分析及思路 给定一个单链表,要求得到它的逆序.可以使用列表对链表结点进行保存,之后新建一个列表对链表的逆序进行保存.最后返回新建列表的第一个元素即可. python代码 # Definition for singly-linked list. #…

Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Sol…

Reverse a singly linked list. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(ListNode* head) { if(!head) retur…

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…

Reverse a singly linked list. click to show more hints. Subscribe to see which companies asked this question. 利用循环. public class Solution { public ListNode reverseList(ListNode head) { if(head == null ||head.next == null) return head; ListNode pre =…

Reverse a singly linked list. 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? 题意: 如题 思路: 无 代码: /** * Definition for…

题目: 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? 分析: 分别用迭代和递归来实现. 迭代就是新建一个…

翻译 反转一个单链表. 原文 Reverse a singly linked list. 分析 我在草纸上以1,2,3,4为例.将这个链表的转换过程先用描绘了出来(当然了,自己画的肯定不如博客上面精致): 有了这个图(每次把博客发出去后就会发现图怎么变得这么小了哎!仅仅能麻烦大家放大看或者另存为了.这图命名是1400X600的).那么代码也就自然而然的出来了: ListNode* reverseList(ListNode* head) { ListNode* newHead = NULL; wh…

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. 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? Approach #1 (Iterative) [Acce…

题目描述 Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL 参考答案 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), nex…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 迭代 递归 日期 [LeetCode] 题目地址:https://leetcode.com/problems/reverse-linked-list/ Total Accepted: 105474 Total Submissions: 267077 Difficulty: Easy 题目描述 Reverse a singly linked list.…

▶ 关于单链表翻转的两个问题. ▶ 206. 翻转整个单链表. ● 自己的代码,9 ms,使用了递归. class Solution { public: ListNode* reverseList(ListNode* head) { if (head == nullptr) return nullptr; ListNode *p; for (p = head; p->next != nullptr; p = p->next);// 找到末元(翻转后的首元) reverseNode(head);…

The task is reversing a list in range m to n(92) or a whole list(206). All in one : U need three pointers to achieve this goal. 1) Pointer to last value 2) Pointer to cur p value 3) Pointer to next value Here, showing my code wishes can help u. Of co…