Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod…

问题描述:1->2->3->4,假设k=2进行反转,得到2->1->4->3:k=3进行反转,得到3->2->1->4 算法思想:基本操作就是链表反转,将k个元素当作滑动窗口,依次进行反转. public class ReverseNodesInKGroup { public ListNode reverseKGroup(ListNode head, int k) { if (k == 1 || head == null || head.next ==…

[LeetCode]863. All Nodes Distance K in Binary Tree 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/description/ 题目描述: We are given a binary tree (with root no…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in…

题目描述 Leetcode 24 题主要考察的链表的反转,而 25 题是 24 的拓展版,加上对递归的考察. 对题目做一下概述: 提供一个链表,给定一个正整数 k, 每 k 个节点一组进行翻转,最后返回翻转后的新链表. k 的值小于或等于链表的长度,如果节点总数不是 k 的整数倍,将最后一组剩余的节点保持原有顺序. 注意: 算法只能使用常数的空间 不能单纯的改变节点内部的值,需要进行节点交换. 举例: Example: Given 1->2->3->4->5. For k = 2,…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 25: Reverse Nodes in k-Grouphttps://oj.leetcode.com/problems/reverse-nodes-in-k-group/ Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.If th…

Reverse Nodes in k-Group Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the valu…

1. Reverse Nodes in k-Group Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple…

［抄题］: 给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下.链表元素个数不是k的倍数,最后剩余的不用翻转. ［思维问题］: ［一句话思路］: // reverse head->n1->..->nk->next.. // to head->nk->..->n1->next.. // return n1 每k个转一次,再递归 ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: h…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod…