之前的文章「递归反转链表的一部分」讲了如何递归地反转一部分链表,有读者就问如何迭代地反转链表,这篇文章解决的问题也需要反转链表的函数,我们不妨就用迭代方式来解决. 本文要解决「K 个一组反转链表」,不难理解: 这个问题经常在面经中看到,而且 LeetCode 上难度是 Hard,它真的有那么难吗? 对于基本数据结构的算法问题其实都不难,只要结合特点一点点拆解分析,一般都没啥难点.下面我们就来拆解一下这个问题. 一.分析问题 首先,前文学习数据结构的框架思维提到过,链表是一种兼具递归和迭代性质的数…

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…

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…

题目: 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…

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…

题意:给一个单链表,每k个节点就将这k个节点反置,若节点数不是k的倍数,则后面不够k个的这一小段链表不必反置. 思路:递归法.每次递归就将k个节点反置,将k个之后的链表头递归下去解决.利用原来的函数接口即可,不用重新定义. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; *…

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…

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…

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def reverseKGroup(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode "…

题目: 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…

2018-09-11 22:58:29 一.Reverse Linked List 问题描述: 问题求解: 解法一:Iteratively,不断执行插入操作. public ListNode reverseList(ListNode head) { if (head == null) return null; ListNode dummy = new ListNode(-1); dummy.next = head; ListNode cur = head; ListNode then = nul…

Leetcode 25. Reverse Nodes in k-Group 以每组k个结点进行链表反转(链表) 题目描述 已知一个链表,每次对k个节点进行反转,最后返回反转后的链表 测试样例 Input: k = 2, 1->2->3->4->5 Output: 2->1->4->3->5 Input: k = 3, 1->2->3->4->5 Output: 3->2->1->4->5 详细分析 按照题目要求…

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…

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…

题目链接: https://leetcode.com/problems/reverse-nodes-in-k-group/?tab=Description   Problem :将一个有序list划分为k个组,并且每个组的元素逆置   链表操作 :递归算法  每次寻找到该组的尾部,然后进行逆置操作,返回头部,这样每次递归操作之后能够进行下一次的逆置操作. 链表操作画图比较形象!!!! 对于递归算法:找到共同点,找到程序退出点,注意特殊情况 本题中,共同点为每组为k个节点,并且每组进行的操作均为逆…

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…

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…

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…

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…

We are given a binary tree (with root node root), a target node, and an integer value K. Return a list of the values of all nodes that have a distance K from the target node.  The answer can be returned in any order. Example 1: Input: root = [3,5,1,6…

题目 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…

题目描述: 中文: 给你一个链表,每 k 个节点一组进行翻转,请你返回翻转后的链表. k 是一个正整数,它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,那么请将最后剩余的节点保持原有顺序. 示例 : 给定这个链表:1->2->3->4->5 当 k = 2 时,应当返回: 2->1->4->3->5 当 k = 3 时,应当返回: 3->2->1->4->5 说明 : 你的算法只能使用常数的额外空间. 你不能只是单纯的改…

We are given a binary tree (with root node root), a target node, and an integer value K. Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned in any order. Example 1: Input: root = [3,5,1,6,…

给出一个链表,一次翻转 k 个指针节点,并返回修改后的链表.k 是一个正整数,并且小于等于链表的长度.如果指针节点的数量不是 k 的整数倍,那么最后剩余的节点应当保持原来的样子.你不应该改变节点的值,只有节点位置本身可能会改变.题目应当保证,仅使用恒定的内存.例如,给定这个链表:1->2->3->4->5当 k = 2时,应当返回: 2->1->4->3->5当 k = 3时,应当返回: 3->2->1->4->5详见:https:/…

我用String代替了链表显示,本题的大意是每k个进行逆序处理,剩下的不够k个的就按照原顺序保留下来. public class ReverseNodes { public static void main(String[] args) { String str = "1->2->3->4->5->6->7->8->9->10->11->12->13->14->15"; String[] strArra…

题目: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…

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…

一天一道LeetCode系列 (一)题目 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 i…

前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html   1.题目描述 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 no…