Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) space complexi…

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.You should try to do it in place. The program should run in O(1) space complexit…

每天一算:Odd Even Linked List 描述 给定一个单链表,把所有的奇数节点和偶数节点分别排在一起.请注意,这里的奇数节点和偶数节点指的是节点编号的奇偶性,而不是节点的值的奇偶性. 请尝试使用原地算法完成.你的算法的空间复杂度应为 O(1),时间复杂度应为 O(nodes),nodes 为节点总数. 示例 1: 输入: 1->2->3->4->5->NULL  输出: 1->3->5->2->4->NULL 示例 2: 输入: 2-…

这道题是LeetCode里的第707到题.这是在学习链表时碰见的. 题目要求: 设计链表的实现.您可以选择使用单链表或双链表.单链表中的节点应该具有两个属性:val 和 next.val 是当前节点的值,next 是指向下一个节点的指针/引用.如果要使用双向链表,则还需要一个属性 prev 以指示链表中的上一个节点.假设链表中的所有节点都是 0-index 的. 在链表类中实现这些功能: get(index):获取链表中第 index 个节点的值.如果索引无效,则返回-1. addAtHead(…

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…

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. The program should run in O(1) space complexity and O(nodes) time complexity. E…

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…

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…

题目标签:Linked List 题目让我们自己设计一个 linked list,可以是单向和双向的.这里选的是单向,题目并不是很难,但要考虑到所有的情况,具体看code. Java Solution: Runtime:  56 ms, faster than 21.21% Memory Usage: 45.2 MB, less than 88.89% 完成日期:07/08/2019 关键点:edge cases public class ListNode { int val; ListNode…

去掉链表中相应的元素值 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* removeElements(ListNode* head, int val) { if(!head) return NULL…